Learning Mathematics

This is a guide on learning mathematics.

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Integers

The most common numbers are those used for counting, namely the numbers
1,2,3,4,5... 
which are called the positive integers. Even for counting, we need
at least one other number, zero.

For instance, we may wish to count the number of right answers you may get on a
test, out of a possible 100. If you get 100, then all of your answers were
correct. If you get 0, then no answer was correct.

The positive integers and zero can be represented geometrically on a line, in a
manner similar to a ruler or a measuring stick.

For this we first have to select a unit of distance, say the inch, and then on
the line we mark off the inches to the right.

For convenience, it is useful to have a name for the positive integers together
with zero, and we shall call these the natural numbers. Thus 0 is a natural
number, so is 2, and so is 124,521. The natural numbers can be used to measure
distances, as with the ruler.

By definition, the point represented by 0 is called the origin. The natural
numbers can also be used to measure other things. For example, a thermometer is
like a ruler which measures temperature. 

However, the thermometer shows us that
we encounter other types of numbers besides the natural numbers, because there
may be temperatures which may go below 0. Thus we encounter naturally what we
shall call negative integers which we call minus 1, minus 2, and so on and which
we write as:

-1, -2,-3, -4...

We represent the negative integers on a line as being on the other side of 0
from the positive numbers.

The positive integers, negative integers, and zero all together are called the
integers. Thus -9,0,10, -5 are all integers.

If we view the line as a thermometer, on which a unit of temperature has been
selected, say the degree Fahrenheit, then each integer represents a certain
temperature. The negative integers represent temperatures below zero.

Our discussion is already typical of many discussions which will occur,
concerning mathematical objects and their applicability to physical situations.
In the present instance, we have the integers as mathematical objects, which are
essentially abstract qualities.

We also have different applications for them, for instance measuring distance or
temperatures. These are of course not the only applications. Namely, we can use
the integers to measure time. We take the origin 0 to represent the year of the
birth of Christ. 

Then the positive integers represent years after the birth of Christ, while the
negative integers can be used to represent BC years. With this convention, we
can say that the year -500 is the year 500 BC.

Adding a positive number, say 7, to another number, means that we must move 7
units to the right of the other number. For instance:

5 + 7 = 12.

Seven units to the right of 5 yields 12. On the thermometer, we would of course
be moving upward instead of right. For instance, if the temperature at a given
time is 5 and it goes up by 7, then the new temperature is 12.

Observe the very simple rule for addition with 0, namely:
\(0 + a = a + 0 = a\)

What about adding negative numbers? Look at the thermometer again. Suppose the
temperature at a given time is 10 degrees, and the temperature drops by 15
degrees. The new temperature is then -5 degrees, and we can write:
\(10 -15 = -5\)

Thus -5 is the result of subtracting 15 from 10, or of adding -15 to 10. In
terms of points on a line, adding a negative number, say -3, to another number
means that we must move 3 units to the left of this other number. For example:
\(5 + (-3) = 2\)
because starting with 5 and moving 3 units to the left yields 2. Similarly:
\(7 + (-3) = 4\) and \(3 + (-5) = -2\)
Note that we have:
\(3 + (-3) = 0\) or \(5 + (-5) = 0\).
We can also write these equations in the form:
\((-3) + 3 = 0\) or \((-5) + 5 = 0\).

For instance, if we start 3 units to the left of 0 and move 3 units to the
right, we get 0. Thus, in general, we have the formulas:
\(a + (-a) = 0\) and \(-a + a = 0\)

In the representation of integers on the line, this means that a and -a lie on
opposite sides of 0 on that line. We can now write:
\(3 = (-3)\) or \(5 = (-5)\)

We use the name 'minus a' for '-a' rather than the words 'negative a'. I find
the words 'negative a' confusing because they suggest that -a is a negative
number. This is not true unless a itself is positive. For instance:
\(3 = -(-3)\)
is a positive number, but 3 is equal to -a, where -a = -3, and a is a negative
number. Because of this property:
\(a + (-a) = 0\)
one also calls '-a' the additive inverse of 'a'.

The sum and product of integers are also integers.

Rules for Addition

integers follow very simple rules for addition. These are commutativity and
associativity.

Commutativity: If a,b are integers, then \(a + b = b + a\)
For instance, \(3 + 5 = 5 + 3\) or \(-2 + 5 = 3 = 5 + (-2)\)

Associativity: If a,b,c are integers then \((a + b) + c = a + (b + c)\)
In view of this, it is unnecessary to use parentheses in such a simple context,
and we simply write \(a + b + c\)
For instance: \((3 + 5) + 9 = 8 + 9 = 17\) and \(3 + (5 + 9) = 3 + 14 = 17\)
We write simply \(3 + 5 + 9 = 17\)

Associativity also holds with negative numbers. For example:
\((-2 + 5) + (-3) = -3 + (-3) = -6\) and \(2 + (-5 + (-3)) = 2 + (-8) = -6\)

The rules of addition mentioned above will not be proved, but we shall prove
other rules from them. Note that:
If a + b = 0 then b = -a and a = -b.
To prove this, add -a to both sides of the equation \(a + b = 0\).
We get \(-a + a + b = -a + 0 = -a\)
Since \(-a + a + b = 0 + b = b\) we find \(b = -a\).
Similarly, we find \(a = -b\). We could also conclude that \(-b = -(-a) = a\)
As a matter of convention, we shall write \(a - b\) instead of \(a + (-b)\).

Thus a sum involving three terms may be written in many ways, as follows:
\((a-b) + c = (a + (-b)) + c\) by associativity
\(= a + (-b + c)\) by commutativity
\( (a + c) - b\) by associativity
and we can also write this sum as: \(a - b + c = a + c - b\)
omitting the parentheses. Generally, in taking the sum of integers, we can take
the sum in any order by applying associativity and commutativity repeatedly.

As a special case, for any integer we have: \(a = -(-a))\).
This is true because \(a + (-a) = 0\)
and we can apply it with \(b = -a\). Remark that this formula is true whether
'a' is positive, negative, or 0. If 'a' is positive, then '-a' is negative. If
'a' is negative then '-a' is positive. In the geometric representation of
numbers on the line, 'a' and '-a' occur symmetrically on the line opposite sides
of 0. Of course, we can pile up minus signs and get other relationships, like:
\(-3 = -(-(-3))\) or \(3 = -(-3) = -(-(-(-3)))\)

Thus when we pile up the minus signs in front of 'a', we obtain 'a' or '-a'
alternatively. For the general formula with the appropriate notation.

For any integers 'a,b' we have \(-(a + b) = -a + (-b)\)
or, in other words \(-(a + b) = -a - b\)

Proof: Remember that if 'x,y' are integers, then \(x = -y\) and \(y = -x\) mean
that \(x + y = 0\). Thus to prove our assertion, we must show that:
\((a + b) + (-a - b) = 0\)
This comes out immediately:
\((a + b) + (-a - b) = a + b - a - b\) by associativity
\(                   = a - a + b - b\) by commutativity
\(                   = 0 + 0        \)
\(                   = 0            \)
This proves our formula.

Examples:
\(-(3 + 5) = -3 - 5 = -8\)
\(-(-4 + 5) = -(-4) - 5 = 4 - 5 = -1\)
\(-(3 - 7) = -3 - (-7) = -3 + 7 = 4\)

You should be very careful when you take the negative of a sum which involves
itself in negative numbers, taking into account that:
\(-(-a) = a\)

The following rule concerning positive numbers is so natural that you probably
would not even think it worth while to take special notice of it. We will state
it explicitly:

"If a,b are positive integer, then a + b is also a positive integer."

For instance, 17 and 45 are positive numbers, and their sum, 62, is also a
positive integer.

We assume this rule concerning positivity. We shall see later that it also
applies to positive real numbers. From it we can prove:
"If a,b are negative integers, then a + b is negative."

Proof: We can write \(a = -n\) and \(b = -m\), where m and n are positive.
Therefore: \(a + b = -n - m = -(n + m)\)
which shows that 'a + b' is negative, because 'n + m' is positive.

Example:
If we have the relationship between three numbers,
\(a + b = c\)
then we can derive other relationships between them. For instance, add '-b' to
both sides of this equation and we get:
\(a + b - b = c - b\)
whence \(a + 0 = c - b\) or
\(a = c - b\)
Similarly, we can conclude \(b = c - a\)

For instance, if \(x + 3 = 5\)
then \(x = 5 - 3 = 2\)
if \(4 - a = 3\)
then adding 'a' to both sides yields \(4 = 3 + a\)
and subtracting 3 from both sides yields \(1 = a\)
If \(-2 - y = 5\)
then \(-7 = y\) or \( y = -7\)

Rules for Multiplication

We can multiply integers, and the product of two integers is again an integer.
We shall list the rules which apply to multiplication and to its relation with
addition.

We again have the rules of commutativity and associativity:
\(ab = ba\) and \((ab)c = a(bc)\)

We emphasize that these apply whether 'a,b,c' are negative, positive, or zero.
Multiplication is also denoted by a dot. For instance:
\(3 \cdot 7 = 21\)
\((3 \cdot 7) \cdot 4 = 21 \cdot 4 = 84\)
\(3 \cdot (7 \cdot 4) = 3 \cdot 28 = 84\)

For any integer 'a', the rules of multiplication by 1 and 0 are:
\(1a = a\) and \(0a = 0\)

Example:
\((2a)(3b) = 2(a(3b))\)
\(         = 2(3a)b\)
\(         = (2 \cdot)ab\)
\(         = 6ab\)

In this example we have done something which is frequently useful, namely we
have moved to one side all the explicit numbers like '2,3' and put on the other
side those numbers denoted by a letter like 'a or b'. Using commutativity and
associativity, we can prove similarly:
\((5x)(7y) = 35xy\) or \((2a)(3b)(5x) = 30abx\)

We suggest that you carry out the proof of this equality completely, using
associativity and commutativity for multiplication.

Finally, we have the rule of distributivity, namely:
\(a(b + c) = ab + bc\) and \((b + c)a = ba + ca\)

These rules will not be proved, but will be used constantly. We shall, however,
make some comments on them, and prove other rules from them.

First, observe that if we just assume distributivity on one side, and assuming
distributivity on the left, we have:
\((b + c)a = a(b + c) = ab + ac = ba + ca\)
which is the proof of distributivity on the right.

Observe also that our rule \(0a = 0\) can be proved from the other rules
concerning multiplication and the properties of addition. We carry out the proof
as an example. We have:
\(0a + a = 0a + 1a = (0 + 1)a = 1a = a\)
So: \(0a + a = a\)
Adding '-a' to both sides, we obtain:
\(0a + a - a = a - a = 0\)

The left hand side is simply \(0a + a - a = 0a + 0 = 0a\)
so that we obtain \(0a = 0\) as desired.

We can also prove \((-1)a = -a\)
Proof: We have :
\((-1)a + a = (-1)a + 1a = (-1 + 1)a = 0a = 0\)
By definition, \((-1)a + a = 0\) means that \(-1)a = -a\), as was able to be
shown, \(-(ab) = (-a)b\)

Proof:

We must show that '(-a)b' is the negative of 'ab'. This amounts to
showing that \(ab + (-a)b = 0\)
But we have by distributivity \(ab + (-a)b = (a + (-a)b = 0b = 0\)
thus proving what we wanted. 

Examples:
\(-(3a) = (-3)a = 3(-a)\)
\(4(a - 5b) = 4a - 20b\)
\(-3(5a - 7b) = -15a + 21b\)

In each of the above cases, you should indicate specifically each one of the
rules we have used to derive the desired equality. Again, we emphasize that you
should be especially careful when working with negative numbers and repeated
minus signs. This is one of the most frequent sources of error when we work with
multiplication and addition.

Examples:
\((-2a)(3b)(4c) = (-2) * 3 * 4abc = -24abc\)
\((-4x)(5y)(-3c) = (-4)5(-3)xyc = 60xyc\)
Note that the product of two minus signs gives a plus sign.

Examples
\((-1)(-1) = 1\)
To see this, all we have to do is apply our rule
\(-(ab) = (-a)b = a(-b)\)
Then we find:
\((-1)(-1) = -(1(-1)) = -(-1) = 1\)

Example:
More generally, for any integers 'a,b' we have:
\((-a)(-b) = ab\)

From this we can see that a product of two negative numbers is positive,
because, if 'a,b' are positive and '-a,-b' are therefore negative, then (-a)(-b)
is the positive number 'ab'. For instance, '-3' and '-5' are negative but
\((-3)(-5) = -(3(-5)) = -(-(3*5)) = 15\)

Example
A product of a negative number and a positive number is negative. For instance,,
'-4' is negative, '7' is positive, and 
\((-4) * 7 = -(4*7) = -28\)
so that '(-4) * 7' is negative.

When we multiply a number with itself several times, it is convenient to use a
notation to abbreviate this operation. So we write:
\(aa = a^2\), \(aaa = a^3\), \(aaaa = a^4\)
and in general, if 'n' is a positive integer,
\(a^n = aa \codt \cdot \cdot a\) The product is taken 'n' times.
We say that \(a^n\) is the n-th power of 'a'. So \(a^2\) is the second power of
'a', and \(a^5\) is the fifth power of 'a'. If 'm,n' are positive integers,
then: \(a^{m+n} = a^ma^n\)

This simply states that if we take the product of 'a' with itself 'm+n' times,
then this amounts to taking the product of 'a' with itself 'm' times and
multiplying this with the product of 'a' with itself 'n' times.`

Example
\(a^2a^3 = (aa)(aaa) = a^{2+3} = aaaaa = a^5\)

Example
\((4x)^2 = 4x * 4x = 4 * 4xx = 16x^2\)

Example
\((7x)(2x)(5x) = 7 * 2 * 5xxx = 70x^3\)

We have this rule for powers,
\((a^m)^n = a^{mn}\)

This means that if we take the product of 'a' with itself 'm' times, and then
take the product of \(a^m\) with itself 'n' times, then we obtain the product of
'a' with itself 'mn' times.

Example
\((a^3)^4 = a^12\)

Example
\((ab)^n = a^nb^n\)

Example
\(2a^3)^5 = 2^5(a^3)^5 = 32a^15\)

Example
The population of a city is 300 thousand in 1930, and doubles every 20 years.
What will be the population after 60 years?

This is a case of applying powers. After 20 years, the population is 2*300,000.
After 40 years the population is 4 * 300,000. After 60 years, the population is
8 * 300,000, which is a correct answer. Of course, we can also say that the
population will be 2,400,000.

The following three formulas are used constantly. They are so important that
they should be thoroughly memorized by reading them out loud and repeating them
like a poem, to get an aural memory of them.

1. \((a + b)^2 = a^2 + 2ab + b^2\)
2. \((a-b)^2 = a^2 - 2ab + b^2\)
3. \((a + b)(a - b) = a^2 - b^2\)

Proofs:
The proofs are carried out by applying repeatedly the rules for multiplication.
\((a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) \)
\(= aa + ab + ba + bb = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2\)
which proves the first formula.

\((a - b)^2 = (a - b)(a - b) = a(a - b) - b(a - b)\)
\(= aa - ab - ba + bb = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2\)
This proves the second formula.

\((a + b)(a - b) = a(a - b) + b(a - b) = aa - ab + ba - bb\)
\(= a^2 - ab + ab - b^2 = a^2 - b^2\)
This proves the third formula.

Example
\((2 + 3x)^2 = 2^2 + 2*2*3x + (3x)^2 = 4 + 12x + 9x^2\)

Example
\(3 - 4x)^2 = 3^2 - 2*3*4x + (4x)^2 = 9 - 24x + 16x^2\)

Example
\((-2a + 5b)^2 = 4a^2 + 2(-2a)(5b) + 25b^2 = 4a^2 - 20ab + 25b^2\)

Example
\(4a - 6)(4a + 6) = (4a)^2 - 36 = 16a^2 - 36\)

We have discussed so far examples of products of two factors. Of course, we can
take products of more factors using associativity.

Example - Expand the expression
\((2x + 1)(x - 2)(x + 5)\)
as a sum of powers of x multiplied by integers.
We first multiply the first two factors and obtain:
\((2x + 1)(x - 2) = 2x(x - 2) + 1(x - 2) = 2x^2 - 4x + x - 2 = 2x^2 - 3x - 2\)
We now multiply this last expression with 'x+5' and obtain:
\((2x + 1)(x - 2)(x + 5) = (2x^2 - 3x - 2)(x + 5)\)
\(= (2x^2 - 3x - 2)x + 2x^2 - 3x - 2)5 = 2x^3 - 3x^2 - 2x + 10x^2 - 15x - 10\)
\(2x^3 + 7x^2 - 17x - 10\)

Even and Odd Integers

We consider the positive integers 1,2,3,4,5,..., and we shall distinguish
between two kinds of integers. 

We call 1,3,5,7,9... the odd integers.

We call 2,4,6,8,10, the even integers.

So, the odd integers go up by 2 and the even integers go up by 2. The odd integers
start with 1, and the even integers start with 2. Another way of describing an
even integer is to say that it is a positive integer which can be written in the
form 2n for some positive integer n. For instance, we can write:
\(2 = 2 * 1\)
\(4 = 2 * 2\)
\(6 = 2 * 3\)
\(8 = 2 * 4\)
and so on. Similarly, an odd integer is an integer which differs from an even
integer by 1, and so can be written in the form '2m-1' for some positive integer
m. 

For instance:
\(1 = 2 * 1 - 1\)
\(3 = 2 * 2 - 1\)
\(5 = 2 * 3 - 1\)
\(7 = 2 * 4 - 1\)
\(8 = 2 * 5 - 1\)

Note that we can also write an odd integer in the form \(2n + 1\)
if we allow 'n' to be a natural number, allowing 'n = 0'. For instance, we can
have:
\(1 = 2 * 0 + 1\)
\(3 = 2 * 1 + 1\)
\(5 = 2 * 2 + 1\)
\(7 = 2 * 3 + 1\)
\(9 = 2 * 4 + 1\)

Theorem 1
Let 'a,b' be positive integers
If 'a' is even and 'b' is even, then 'a + b' is even
If 'a' is even and 'b' is odd, then 'a + b is odd 
If 'a' is odd and 'b' is even, then 'a + b' is odd
If 'a' is odd and 'b' is odd, then 'a + b' is even

Proof
We shall prove the second statement. Assume that 'a' is even and that 'b' is
odd. Then we can write:
\(a = 2n\) and \(b = 2k + 1\) for some positive integer 'n' and some natural
number 'k'. Then:
\(a + b = 2n + 2k + 1 = 2(n + k) + 1 = 2m + 1\) (m = n + k)

This proves that \(a + b\) is odd.

Theorem 2
Let 'a' be a positive integer. If 'a' is even then \(a^2\) is even. If 'a' is
odd, then \(a^2\) is odd.

Proof
Assume that 'a' is even. This means that \(a = 2n\) for some positive integer
'n'. Then \(a^2 = 2n * 2n = 2(2n^2) = 2m\) where \(m = 2n^2\) is a positive
integer. So, \(a^2\) is even.

Next, assume that 'a' is odd, and write \(a = 2n + 1\) for some natural number
n. Then \(a^2 = (2n + 1)^2 = (2n)^2 + 2(2n)1 + 1^2\)
\(= 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1 = 2k + 1\) where \(k = 2n^2 + 2n\)
Hence, \(a^2\) is odd, so proving our theorem. 

Corollary
Let 'a' be a positive integer. If \(a^2\) is even, then 'a' is even. If \(a^2\)
is odd, then 'a' is odd.

Proof
This is really only a reformulation of the theorem, taking into account ordinary
logic. If \(a^2\) is even, then 'a' cannot be odd because the square of an odd
number is odd. If \(a^2\) is odd, then 'a' cannot be even because the square of
an even number is even.

We can generalize the property used to define an even integer. Let 'd' be a
positive integer and let 'n' be an integer. We shall say that 'd' divides 'n',
or that 'n' is divisible by 'd' if we can write \(n = dk\) for some integer 'k'.
So an even integer is a positive integer which is divisible by 2. According to
our definition, the number 9 is divisible by 3 because \(9 = 3 * 3\).

Also, 15 is divisible by 3 because \(15 = 3 * 5\).
Also, -30 is divisible by 5 because \(-30 = 5(-6)\).
Note that every integer is divisible by 1, because we can always write:
\(n = 1 * n\)
Furthermore, every positive integer is divisible by itself.

Rational Numbers

BY a rational number we shall mean simply an ordinary fraction, that is a
quotient m/n, where m,n are integers and n is not equal to zero, and thus we
must always be sure that n does not equal zero. 

For instance, \(\frac{1}{4},\frac{2}{3},-\frac{3}{4},-\frac{5}{7}\)
are rational numbers. Finite decimals also give us examples of rational numbers. 

For instance, \(1.4 = \frac{14}{10}\) and \(1.41 = \frac{141}{100}\)

Just as we did with the integers, we can represent the rational numbers on the
line. For instance, 1/2 lies one-half of the way between 0 and 1, while 2/3 lies
two-thirds of the way between 0 and 1.

The negative rational number -3/4 lies on the opposite side of 0 at a distance
3/4 from 0.

There is no unique representation of a rational number as a quotient of two
integers. For instance, we have \(\frac{1}{2} = \frac{2}{4}\)

We can interpret this geometrically on the line. If we cut up the segment
between 0 and 1 into four pieces, and we take two-fourths of them, this this is
the same as taking one-half of the segment.

We need a general rule to determine when two expressions of quotients of
integers give the same rational numbers. We assume this rule without proof.

Rule for cross-multiplying 
Let m, n, r, s be integers and assume that n does not
equal 0 and s does not equal 0. Then:
\(\frac{m}{n} = \frac{r}{s}\) if and only if \(ms = rn\).

Example
We have \(\frac{1}{2} = \frac{2}{4}\) because \(1*4 = 2*2\)
Also we have \(\frac{3}{7} = \frac{9}{21}\) because \(3*21 = 9*7\)

We shall make no distinction between an integer m and the rational number m/1.
So we write \(m = m/1 = \frac{m}{1}\)

With this convention, we see that every integer is also a rational number. For
instance, \(3 = 3/1\) and \(-4 = -4/1\) 

Observe the special case of cross-multiplying when one side is an integer.
For instance: \(\frac{2n}{5 = \frac{6}{1}, \frac{2n}{5} = 6, 2n = 30, n = 15\)
are all equivalent formulations of a relation involving n.

Of course, cross-multiplying also works with negative numbers.
For instance: \(\frac{-4}{5} = \frac{8}{-10}\) because \((-4)(-10) = 8 * 5\)

For the moment, we are dealing with quotients of integers and describing how
they behave. In the next section we shall deal with multiplicative inverses.
There, you can see how the rule for cross-multiplication can in fact be proved
from properties of such an inverse.

Some people view this proof as the reason why cross-multiplication works.
However, in some contexts, one wants to define the multiplicative inverse by
using the rule for cross-multiplication. This is the reason for emphasizing it
here independently.

Cancellation rule for fractions:
Let a be a non-zero integer. Let m,n be integers, \(n \not= 0\). 
\(\frac{am}{an} = \frac{m}{n}\)
Proof = To test equality, we apply the rule for cross-multiplying. We must
verify that: \((am)n = m(an)\)
which we see is true by associativity and commutativity.

The examples which we gave are special cases of this cancellation rule.
For instance: \(\frac{-4}{5} = \frac{(-2)(-4)}{(-2)(5)} = \frac{8}{-10}\)

In dealing with quotients of integers which may be negative, it is useful to
observe that:
\(\frac{-m}{n} = \frac{m}{-n}\)
This is proved by cross-multiplying, namely we must verify that:
\((-m)(-n) = mn\)
which we already know is true.

The cancellation rule leads us to use the notion of divisibility already
mentioned. Indeed, suppose that d is a positive integer and m,n are divisible by
d. Then we can write:
\(m = dr\) and \(n = ds\)
for some integers r and s, so that:
\(\frac{m}{n} = \frac{dr}{ds} = \frac{r}{s}\)
We see that our cancellation rule is applicable.

Example
We have \(\frac{10}{15} = \frac{2*5}{3*5} = \frac{2}{3}\)
because 10 and 15 are both divisible by 5.

We say that a rational number is positive if it can be written in the form m/n
where m,n are positive integers. Let a be a positive rational number. We shall
say that a is expressed in lowest form as a fraction
\(a = \frac{r}{s}\)
where r,s are positive integers if the only common divisor of r and s is 1. 

Theorem 3
Any positive rational number has an expression as a fraction in lowest form.

Proof:
First write a given positive rational number as a quotient of positive integers
m/n. We know that 1 is a common divisor of m and n. furthermore, any common
divisor is at most equal to m or n. So, among all common divisors there is a
greatest one, which we denote by d. So we can write:
\(m = dr\) and \(n = ds\)
with positive integers r and s. Our rational number is equal to:
\(\frac{m}{n} = \frac{dr}{ds} = \frac{r}{s}\)

All we have to do now is to show that the only common divisor of r and s is 1.
Suppose that e is a common divisor which is greater than 1. Then we can write
\(r = ex\) and \(s = ey\)
with positive integers x and y. So,
\(m = dr = dex\) and \(n = ds = dey\)

Therefore 'de' is a common divisor for m and n, and is greater than d since e is
greater than 1. This is impossible because we assumed that d was the greatest
common divisor of m and n. Therefore 1 is the only common divisor of r and s,
and our theorem is proved.

Example
Any positive rational number can be expressed as a quotient m/n, where m,n are
positive integers which are not both even, because if m/n is the expression of
this rational number in lowest form, then 2 cannot divide both m and n, and
therefore at least one of them must be odd.

Let m/n and r/s be rational numbers, expressed as quotients of integers. We can
put these rational numbers over a common denominator ns by writing:
\(\frac{m}{n} = \frac{ms}{ns}\) and \(\frac{r}{s} = \frac{nr}{ns}\)

For instance, to put 3/5 and 5/7 over the common denominator \(5 * 7 = 35\),
\(\frac{3}{5} = \frac{3*7}{5*7} = \frac{21}{35}\) and \(\frac{5}{7} =
\frac{5*5}{7*5} = \frac{25}{35}\)

This leads us to the formula for the addition of rational numbers. Consider
first a special case, when the rational numbers have a common denominator, for
instance 
\(\frac{3}{5} + \frac{8}{5} = \frac{11}{5}\)

This is reasonable just from the interpretation of rational numbers: If we have
3/5 of something, and add 8/5 of that same thing, then we get 11/5 of that
thing. In general, we canw rite the rule for addition when the rational numbers
have a common denominator as:
\(\frac{a}{d} + \frac{b}{d} = \frac{a+b}{d}\)

Example
\(\frac{-5}{8} + \frac{2}{8} = \frac{-3}{8}\)
When the rational numbers do not have a common denominator, we get the formula
for their addition by putting them over a common denominator. Namely, let m/n
and r/s be rational numbers, expressed as quotients of integers m,n and r,s with
neither equal to zero. Then we have seen that:
\(\(\frac{m}{n} = \frac{sm}{sn}\) and \(\frac{r}{s} = \frac{nr}{ns}\)

So our rational numbers now have the common denominator sn, and so the formula
for addition in this general case is:
\(\frac{m}{n} + \frac{r}{s} = \frac{ms + rn}{ns}\)

Example
\(\frac{3}{5} + \frac{4}{7} = \frac{3*7 + 4*5}{35} = \frac{21 + 20}{35} =
\frac{41}{35}\)

Example
\(\frac{-5}{2} + \frac{3}{7} = \frac{(-5)*7+2*3}{14} = \frac{-29}{14}\)

Example
\(\frac{3}{-4} + \frac{5}{7} = \frac{21-20}{-28} = \frac{1}{-28}\)

Using our rule for adding rational numbers, we conclude at once:
The sum of positive rational numbers is also positive.
Observe that our number 0 has the property that
\(\frac{0}{n} = \frac{0}{1} = 0\)
for any integer \(n \not= 0\). Indeed, applying our test for the equality of two
fractions, we must verify that:
\(0*1 = 0*n\)
and this is true because both sides are equal to 0.

For any rational number a, we have:
\(0 + a = a + 0 = a\)
This is easily seen using the analogous property for integers. Namely, write
\(a = m/n\), where m,n are integers and \(n \not= 0\). Then:
\(0 + a + \frac{0}{n} + \frac{m}{n} = \frac{0 + m}{n} = \frac{m}{n} = a\)
and similarly on the other side,
Let a = m/n be a rational number, where m,n are integers and \(n \not= 0\).
Then we have:
\(\frac{-m}{n} + \frac{m}{n} = \frac{-m + m}{n} = 0\)

For this reason, we shall write:
\(\frac{-m}{n} = -\frac{m}{n}\)

By a previous remark,
\(\frac{-m}{n} = -\frac{m}{n}\)

This shows how a minus sign can be moved around the various terms of a fraction
without changing the value of the fraction.
A rational number which can be written as a fraction:
\(-\frac{m}{n} = \frac{-m}{n} = \frac{m}{-n}\)
where m,n are positive integers will be called negative. For example the number:
\(\frac{3}{-5} = \frac{-3}{5} = -\frac{3}{5}
is negative. Using the definition of addition of rational numbers, you can
easily verify for yourselves that a sum of negative rational numbers is
negative.

Addition of rational numbers satisfies the properties of commutativity and
associativity.

Just as we did for integers, the above statement will be accepted without proof.
It is in fact a general property of much more general numbers, which will be
restated again for these numbers.

This remark will again be made later whenever we meet a similar feature.
For instance, we see as before that:
\(a + b = 0\) then \(b = -a\)

We just add -a to both sides of the equation \(a + b = 0\). In words, we can
say: To test whether a given rational number is equal to minus another, all we
need to verify is that the sum of the numbers is equal to 0. We shall now give
the formula for multiplication of rational numbers:
\(\frac{m}{n} * \frac{r}{s} = \frac{mr}{ns}\)

Thus to take the product of two rational numbers, we multiply their numerators
and multiply their denominators, and the denominator of the product is the
product of the denominators.

Example
We have: \(\frac{3}{5} * \frac{7}{8} = \frac{21}{40}\)

Example
\(\frac{2}{7} * \frac{11}16} = \frac{22}{112}\)

As you can see, sometimes it is best not to carry out a multiplication before
looking at the possibility of cancellations.

Example
We have \(\frac{-4}{5} * \frac{7}{-3} = \frac{(-4)7}{5(-3)} = \frac{-28}{-15} =
\frac{28}{15}\)

Example
Let \(a = m/n\) be a rational number expressed as a quotient of integers. Then:
\(a^{2} = (m/n)^{2} = \frac{m}{n}*\frac{m}{n} = \frac{m^{2}}{n^{2}}\)

Similarly, \(a^{3} = \frac{m}{n} \frac{m}{n} \frac{m}{n} = \frac{m^{3}}{n^{3}}\)

In general, for any positive integer k, we have:
\(a^{k} = (m/n)^{k} = \frac{m^{k}}{n^{k}}\)

Example
We have \((1/2)^{3} = \frac{1}{2^{3}} = \frac{1}{8}\)

Example
We have \((3/5)^{4} = \frac{3^{4}}{5^{4}} = \frac{81}{525}\)

Multiplicative Inverses

Rational numbers satisfy one property which is not satisfied by integers.
If a is a rational number that is not equal to zero, then there exists a
rational number denoted by \(a^{-1}\).

If a = m/n where m and n are integers that are not equal to zero, then the
inverse of a is n/m.

We call \(a^{-1}\) the multiplicative inverse of a.

Example
The multiplicative inverse of 1/2 is 2/1 or 2.
\(2 * \frac{1}{2} = 1\)

The multiplicative inverse of 2/3 is 3/2.
The multiplicative inverse of -5/7 is -7/5.

Proof=We multiply both sides of the relation ab = 1 by inverse of a and get:
\(a^{-1}ab = a^{-1} * 1 = a^{-1}\)

Using associativity on the left,
\(a^{-1}ab = 1b = b\)
so that we do find b = \(a^{-1}\)

From the existence of an inverse for non-zero rational numbers:
\(ab = 0 | a = 0, b = 0\)

Proof: Suppose a is not equal to zero. Multiply both sides of the equation \(ab
= 0\) by \(a^{-1}\) and we get \(a^{-1}ab = 0a^{-1} = 0\)

On the other hand, \(a^{-1}ab = 1b\), so that we find b = 0.

We use the same notation as for quotients of integers in taking quotients of
rational numbers.
\(a/b\) instead of \(b^{-1}a\)

Example
Let a = 3/4 and b = 5/7
\(\frac{3/4}{5/7} = \frac{3}{4} * (\frac{5}{7})^{-1} = \frac{3}{4}\frac{7}{5} =
\frac{21}{20}\)

Our rule for cross-multiplication which applied to quotients of integers applies
as well when we want to cross-multiply rational numbers. We state it, and prove
it using only the basic properties of addition, multiplication, and inverses.

If \(\frac{a}{b} = \frac{c}{d}\) then \(ad = bc\)
If \(ad = bc\) then \(\frac{a}{b} = \frac{c}{d}\)

Proof: We can rewrite this relation in the form \(b^{-1}a = d^{-1}c\)
Multiply both sides by \(db\)
We get: \(dbb^{-1}a = bdd^{-1}c\)
so that: \(da = bc\)
because \(bb^{-1}a = 1a = a\) and \(dd^{-1}c = 1c = c\)

Conversely, assume that \(ad = bc\).
Multiply both sides by \(b^{-1}d^{-1}\), which is equal to \(d^{-1}b^{-1}\)
and \(ab^{-1} = d^{-1}c\)
This means that a/b = c/d

Example
By crpss-multiplying, we have:
\(\frac{3}{x-1} = 2\)
if \(3 = 2(x-1) = 2x - 2\)
which is the same as \(3 + 2 = 2x\)

We can solve for x and get \(x = \frac{5}{2}\)

Example
By cross-multiplying we have:
\(\frac{4+x}{1/2 x} = 5\)
This is the same as:
\(4 + x = 5 * 1/2 x = \frac{5x}{2}\)
This is the same as
\(2(4 + x) = 5x\) or
\(8 + 2x = 5x\)
Subtracting 2x from both sides, we solve for x, and get:
\(x = \frac{8}{3}\)

Cancellation law for multiplication:
If \(ab = ac\) then b = c
Proof: Multiply both sides of the equation \(ab = ac\) by \(a^{-1}\)
We get: \(a^{-1}ab = a^{-1}ac\)

We also have the cancellation law similar to that for quotients of integers.
\(\frac{ab}{ac} = \frac{b}{c}\)
This can be verified by cross-multiplication
\(abc = bac\)

We can operate with fractions formed with rational numbers much as we could
operate with fractions formed with integers.

Example
If a/b and c/d are two quotients of rational numbers, then we can put them over
a common denominator:
\(\frac{a}{b} = \frac{ad}{bd}\) and \(\frac{c}{d} = \frac{bc}{bd}\)

Example
We can add quotients in a similar manner to the addition for quotients of
integers
\(\frac{x}{b}+\frac{y}{b} = b^{-1}x+b^{-1}y\)
\(=b^{-1}(x+y) = \frac{x+y}{b}\)
Combining this with the common denominator procedure of the preceding example,
we find:
\(\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\)

This formula is entirely analogous to the formula expressing the sum of two
rational numbers.

Example
Show that: \(\frac{1}{x-y}+\frac{1}{x+y} = \frac{2x}{x^{2}-y^{2}\)
To do this, we add the two quotients on the left by our general formula which we
just derived and get:
\(\frac{1(x+y)+1(x-y)}{(x-y)(x+y)}=\frac{x+y+x-y}{x^{2}-y^{2}}\)

In the preceding example, the quotients 1/(x-y) and 1/(x+y) make no sense if
x-y=0 or x+y=0. In such instances, we assume that x and y are such that x-y do
not equal zero and x+y do not equal zero. In the sequel we shall sometimes omit
the explicit mention of such conditions if there is no danger of confusion.

Example
Solve for x
\(\frac{3x+1}{2x-5}=4\)
We cross multiply
\(3x + 1 = 4(2x - 5) = 8x - 20\)
So: \(8x - 3x = 1 - (-20) = 1 + 20 = 21\)
This gives us:
\(5x = 21\)
\(x = \frac{21}{5}\)

Example
We give an example from the physical world. Suppose that an object is moving
along a straight line at constant speed. Let s denote the speed, let d denote
the distance traveled by the object, and let t denote the time taken to travel
the distance d. 

\(d = st\)

Of course we must select units of time and distance before we can associate
numbers with these. For instance, suppose that the distance traveled is 5 mi,
and the time taken is 1/2 hr. Then the speed is:
\(s = d/t = \frac{5 mi}{1/2 hr} = 2*5mi/hr = 10 mi/hr\)

Example
A person takes a trip and drives 8 hr, a distance of 400 mi. His average speed
is 60 mph on the freeway and 30 mph when he drives through a town. How long did
the person drive through towns during his trip?

To solve this, let x be the length of time the person drives through towns. Then
the length of time the person is on the freeway is 8-x. The distance driven
through towns is equal to 30x, and the distance driven on freeways is 60(8-x).
Since the total distance driven is 400 mi, we have:
\(30x + 60(8-x) = 400\)
This is the same as:
\(30x + 480 - 60x = 400\)
\(80 = 30x\)
\(x = \frac{80}{30} = \frac{8}{3}\)
The person spent 8/3 hours driving through towns.

Example
The radiator of a car contains 8 qt of liquid, consisting of water and 40%
antifreeze. How much should be drained and replaced by antifreeze if the
resultant mixture should have 90% antifreeze?

Let x be the number of quarts which must be drained. After draining this amount,
we are left with (8-x) qt of liquid, of which 40% is antifreeze. We are left
with:
\(\frac{40}{100}(8-x)qt\)
Since we now add x qt of antifreeze, we see that x satisfies
\(x + \frac{40}{100}(8-x) = \frac{90}{100}*8\)
From this we can solve for x, transforming this equation into equivalent
equations:
\(x + \frac{40}{100} * 8 - \frac{40}{100}x = \frac{90}{100} * 8\)
\(\frac{60}{100}x = \frac{50}{100}* 8\)
\(x = \frac{400}{60} = \frac{20}{3}\)

Solving Systems of Two Linear Equations in Two Unknowns

When faced with two linear equations involving the same two variables, the goal is to determine the values of those variables that satisfy both equations simultaneously. Such a pair of equations is called a system of simultaneous linear equations. Consider the following system as an example:

$$2x+y=1$$$$3x-2y=4$$

The task is to find values for $x$ and $y$ that make both equations true at the same time. One of the most straightforward and reliable techniques for accomplishing this is known as the elimination method. The central idea of elimination is to combine the two equations in such a way that one of the variables is removed, leaving a single equation in one variable that can be solved directly. Once that variable is known, it can be substituted back into either of the original equations to find the other variable.

Applying the Elimination Method

In the system above, the variable $x$ appears with a coefficient of 2 in the first equation and a coefficient of 3 in the second equation. To eliminate $x$, the coefficients of $x$ in the two equations must be made equal in magnitude. This is accomplished by multiplying each equation by a suitable number. Specifically, the first equation can be multiplied by 3, and the second equation can be multiplied by 2. This yields:

$$6x+3y=3$$$$6x-4y=8$$

Now that the coefficients of $x$ are identical in both equations, subtracting the second equation from the first will cause the $x$-terms to cancel. It is important to subtract each side of the second equation from the corresponding side of the first equation. Carrying out this subtraction gives:

$$3y-(-4y)=3-8$$

Simplifying the left-hand side:

$$3y+4y=7y$$

And simplifying the right-hand side:

$$3-8=-5$$

Thus:

$$7y=-5$$

Dividing both sides by 7 produces:

y=−57y=−75​

With the value of $y$ now known, the next step is to determine $x$. This can be done by substituting y=−57y=−75​ into either of the original equations. Using the first equation, $2x+y=1$, the substitution gives:

2x−57=12x−75​=1

Adding 5775​ to both sides:

2x=1+57=77+57=1272x=1+75​=77​+75​=712​

Dividing both sides by 2:

x=127÷2=1214x=712​÷2=1412​

Thus the solution is:

x=1214,y=−57x=1412​,y=−75​

The fraction 12141412​ can be reduced to lowest terms by dividing the numerator and denominator by their greatest common divisor, which is 2. This gives x=67x=76​. Both 12141412​ and 6776​ are correct representations of the same value; reducing to lowest terms is simply a matter of convention and convenience.

An Alternative Approach: Eliminating $y$ First

The order in which the variables are eliminated is a matter of choice. As a variation, one could eliminate $y$ first instead of $x$. To do this, multiply the first equation by 2 and leave the second equation unchanged:

$$4x+2y=2$$$$3x-2y=4$$

Adding the two equations together causes the $y$-terms to cancel:

$$4x+3x=6$$

This simplifies to:

$$7x=6$$

Dividing both sides by 7 gives:

x=67x=76​

This is the same value for $x$ that was found previously. To find $y$, substitute x=67x=76​ into the first original equation, $2x+y=1$:

y=1−2x=1−127=77−127=−57y=1−2x=1−712​=77​−712​=−75​

Once again, the solution is x=67x=76​ and y=−57y=−75​, confirming that either elimination strategy leads to the same result.

Systems with No Solution

Not every system of two linear equations has a solution. There are cases in which no pair of values for $x$ and $y$ can satisfy both equations simultaneously. For example, consider the system:

$$2x-y=6$$$$2x-y=7$$

The left-hand sides of these two equations are identical, yet the right-hand sides differ. It is impossible for the same expression $2x-y$ to equal both 6 and 7 at the same time. Therefore, this system has no solution.

A similar situation arises with the system:

$$2x-y=5$$$$6x-3y=7$$

Observe that the second equation can be divided by 3 to obtain:

2x−y=732x−y=37​

This means that any solution of $6x-3y=7$ must also satisfy 2x−y=732x−y=37​. However, the first equation requires $2x-y=5$. Since 7337​ and 5 are not equal, no pair of values can satisfy both equations at once. Consequently, this system also has no solution.

These examples illustrate the two typical ways in which a system of two linear equations can fail to have a solution: either the equations contradict each other directly, or they represent parallel relationships that never intersect. The purpose here is not to develop a comprehensive theory of when solutions exist or do not exist, but rather to become comfortable with the basic techniques for solving systems that do have solutions. In practice, most simple systems encountered will have a unique solution unless they are constructed in a way that is essentially equivalent to one of the contradictory examples above.

Verifying the Solution

When a solution is found by the elimination method, it is always possible to verify that the values obtained actually satisfy both original equations. This verification can be performed explicitly each time. Substituting x=67x=76​ and y=−57y=−75​ into the first equation:

2(67)+(−57)=127−57=77=12(76​)+(−75​)=712​−75​=77​=1

And into the second equation:

3(67)−2(−57)=187+107=287=43(76​)−2(−75​)=718​+710​=728​=4

Both equations are satisfied, confirming that the solution is correct. While it is possible to prove in general that the elimination method always produces a valid solution when one exists, such a proof requires setting up convenient notation and using general letters for the coefficients. That level of abstraction is not necessary here. The primary objective is to develop a simple, efficient, and reliable approach to solving systems of two equations in two unknowns.

Summary of Key Steps

• Multiply one or both equations by suitable constants so that the coefficients of one variable become equal in magnitude.

• Add or subtract the equations to eliminate that variable, leaving a single equation in one unknown.

• Solve the resulting equation for the remaining variable.

• Substitute the value found back into either original equation to determine the other variable.

• Verify the solution by substituting both values into both original equations.

Conclusion

Simultaneous linear equations in two variables arise naturally in a wide variety of problems, including those that might otherwise be posed in terms of a single variable. The elimination method provides a systematic and efficient way to find the unique solution when it exists. By mastering this technique, one gains a practical tool for solving problems involving two unknown quantities. The method is straightforward, requires only basic arithmetic, and can be applied consistently to any system of two linear equations that has a solution.

Solving Equations with Three Unknowns

When working with a system of three linear equations involving three variables, the objective is to determine the values of those variables that satisfy all equations simultaneously. The method presented here follows a systematic process of successive elimination, in which one variable is removed at a time until a single equation in one unknown remains. Once that value is found, it can be substituted back into the earlier equations to determine the remaining unknowns. This approach is both reliable and efficient, provided that the order of elimination is chosen thoughtfully to simplify the arithmetic.

Consider the following system of equations:

$$3x+2y+4z=1$$$$-x+y+2z=2$$$$x-3y+z=-1$$

The goal is to find the values of $x$, $y$, and $z$ that satisfy all three equations at once. To begin, it is helpful to examine the equations and decide which variable can be eliminated most conveniently. In this case, adding the second and third equations eliminates $x$ immediately, because the coefficients of $x$ in those two equations are opposites of one another. Carrying out this addition yields:

$$y-3y+3z=2-1$$

which simplifies to:

$$-2y+3z=1$$

This newly obtained equation involves only $y$ and $z$, reducing the complexity of the problem.

Next, the goal is to eliminate $x$ from another pair of equations so that a second equation in only $y$ and $z$ can be formed. To accomplish this, the second equation is multiplied by 3 and then added to the first equation. This operation produces:

$$2y+3y+4z+6z=1+6$$

which simplifies to:

$$5y+10z=7$$

At this stage, two equations involving only $y$ and $z$ have been obtained. These two equations form a smaller system that can be solved using the same elimination strategy applied previously. The two equations are:

$$-2y+3z=1$$$$5y+10z=7$$

To eliminate $y$, the first of these two equations is multiplied by 5 and the second is multiplied by 2, and then the results are added together. This gives:

$$15z+20z=5+14$$

which simplifies to:

$$35z=19$$

Solving for $z$ produces:

z=1935z=3519​

With the value of $z$ now known, it becomes possible to determine $y$ by substituting this value back into one of the equations involving only $y$ and $z$. Using the equation $-2y+3z=1$, the following steps are carried out:

2y=3z−1=3⋅1935−12y=3z−1=3⋅3519​−1

This becomes:

57−3535=22353557−35​=3522​

Dividing by 2 then yields the value of $y$:

y=1135y=3511​

Finally, with both $y$ and $z$ determined, the value of $x$ can be found by substituting these values into any one of the original three equations. Using the third original equation, $x-3y+z=-1$, the following calculation is performed:

$$x=-1+3y-z$$

Substituting the known values gives:

x=−3535+3335−1935x=−3535​+3533​−3519​

which simplifies to:

x=−9135x=−3591​

However, this fraction can be reduced by dividing both the numerator and denominator by the common factor of 7, resulting in:

x=−135x=−513​

It is worth noting that the final solution may be expressed in equivalent fractional forms depending on the stage at which simplification is applied. In the present case, the solution to the system is:

• x=−2135x=−3521​, which reduces to −35−53​

• y=1135y=3511​

• z=1935z=3519​

To summarize, the process of solving a system of three linear equations in three unknowns involves selecting an efficient order of elimination, systematically removing one variable at a time, solving the resulting single-variable equation, and then back-substituting to find the remaining variables. This method is general and can be applied to a wide variety of linear systems, though the arithmetic may vary depending on the coefficients involved. With practice, the selection of the most convenient elimination order becomes increasingly intuitive, allowing for both accuracy and efficiency in obtaining the final solution.

Real Number System

The integers and rational numbers are part of a larger system of numbers. As we
know, the integers and rational numbers correspond to some points on the line.
The real numbers are those numbers which correspond to all points on the line.
Another way of describing them is to say that they consist of all numbers which
have a decimal expansion, possibly infinite.

It is a rather long and tedious process to develop the theory of the real
numbers systematically. We shall first summarize some of the algebraic
properties which they satisfy. Unless otherwise specified, number means real
number.

The sum and product of two numbers are numbers, and they satisfy the following
properties, similar to those of rational numbers.

Addition is commutative and associative, meaning that for all real numbers,
a,b,c, we have:

\(a + b = b + a\)
and
\(a + (b + c) = (a + b) + c\)

We also have:

\(0 + a = a\)

To each real number a there is associated a number denoted by -a such that :

\(a + (-a) = a\)

As with integers, or rational numbers, we represent a and -a on opposite sides
of 0 on the line.

The number -a is called the additive inverse of a. We also read it as minus a.
If b is a number such that \(a + b = 0\), then we must have \(b = -a\), as we
see by adding -a to both sides of the equations \(a + b = 0\)

Multiplication is commutative and associative, meaning that for all real numbers
a,b, and c, we have:

\(ab = ba\) 
and
\(a(bc) = (ab)c\)

We also have:
\(1a = a\)
and 
\(0a = 0\)

Multiplication is distributive with respect to addition, meaning that:

\(a(b + c) = ab + ac\)
and
\((b + c)a = ba + ca\)

So far, these properties are the same as those satisfied by the integers and
rational numbers. In particular, further properties which were proved using only
these basic ones are now valid for the real numbers. For instance we recall the
important formulas:

\((a + b)^2 = a^2 + 2ab + b^2\)

\((a - b)^2 = a^2 - 2ab + b^2\)

\((a + b)(a - b) = a^2 - b^2\)

These are true if a,b, are real numbers because their proofs used only
commutativity and associativity.

If a is a real number that does not equal 0, then there exists a real number
denoted by \(a^{-1}\) such that:

\(a^{-1}a = aa^{-1} = 1\)

As with rational numbers, this number \(a^{-1}\) is called the multiplicative
inverse of a. Instead of writing \(a^{-1}\), we write \(\frac{1}{a}\) and we
write: a/b instead of \(b^{-1}a\) or \(ab^{-1}\).

The proofs of properties concerning inverses before depended only on the basic
we have mentioned so far, and are applicable to the real numbers. We have
cross-multiplication, cancellation rules and so on. We have the uniqueness of
the multiplicative inverse. Namely, if multiplying both sides by \(a^{-1}\)
shows that \(b = a^{-1}\).

Positive Real Numbers

The positive real numbers can be understood geometrically as those numbers that are not equal to zero and that lie to the right of zero on the real number line. This geometric picture provides an intuitive foundation for the algebraic properties that follow. When a number $a$ is positive, this fact is denoted by the inequality $a>0$. The concept of positivity is fundamental to the structure of the real numbers, and its basic properties serve as the foundation from which many other useful results can be derived.

Fundamental Properties of Positivity

The theory of positive numbers rests upon two foundational properties. The first property concerns closure under addition and multiplication: if $a$ and $b$ are both positive numbers, then their product $ab$ and their sum $a+b$ are also positive. This means that the set of positive real numbers is closed under these two operations, a fact that has profound consequences throughout mathematics.

The second foundational property establishes a trichotomy for every real number. For any real number $a$, exactly one of the following three statements must hold: $a$ is positive, $a$ equals zero, or $-a$ is positive. These three possibilities are mutually exclusive, meaning that no two of them can be true simultaneously. This property ensures that every real number has a well-defined sign classification.

The Definition of Negative Numbers

Building upon these foundational properties, it is possible to define negative numbers. A number is said to be negative if it is neither positive nor equal to zero. In other words, if a number is not positive and is not zero, then it is negative. The relationship between positive and negative numbers is elegantly symmetric: if $a$ is a negative number, then its additive inverse $-a$ is positive. This definition allows for a complete classification of all real numbers as either positive, negative, or zero.

Deriving Further Properties of Positivity

From the two basic properties of positivity, several other important properties can be readily proved. These derived properties expand the understanding of how positive and negative numbers interact through multiplication and division.

• If $a$ is positive and $b$ is negative, then the product $ab$ is negative.

• If $a$ is negative and $b$ is negative, then the product $ab$ is positive.

• If $a$ is positive, then its multiplicative inverse $1/a$ is positive.

• If $a$ is negative, then its multiplicative inverse $1/a$ is negative.

These properties are essential tools for working with inequalities and for understanding the behavior of algebraic expressions involving positive and negative quantities.

The Existence of Square Roots

One of the fundamental properties of the real numbers, assumed without proof, is that every positive real number has a square root. This means that if $a>0$, then there exists a real number $b$ such that b2=ab2=a. This property guarantees that equations of the form x2=ax2=a always have solutions when $a$ is positive, a fact that is crucial for many areas of mathematics.

A notable consequence of this property is that a number whose square is 2 exists as a real number, even though it is irrational. This demonstrates that the real number system is rich enough to contain solutions to equations that cannot be solved within the rational numbers alone.

Determining All Numbers with a Given Square

Given that every positive real number has at least one square root, a natural question arises: how many real numbers have a square equal to a given positive number? For instance, what are all the real numbers $x$ such that x2=2x2=2?

This question can be answered precisely. There are exactly two such numbers: one is positive and the other is negative. To prove this, suppose that b2=2b2=2, and let $x$ be any real number such that x2=2x2=2 as well. Then:

x2−b2=0x2−b2=0

The left side can be factored as a difference of squares:

$$(x+b)(x-b)=0$$

By the zero product property, this implies that either $x+b=0$ or $x-b=0$. Therefore, either $x=-b$ or $x=b$. This shows that any number whose square is 2 must be either $b$ or $-b$.

Furthermore, the square of $-b$ is also equal to 2, since:

(−b)2=(−b)(−b)=b2=2(−b)2=(−b)(−b)=b2=2

Thus, there are precisely two numbers whose square is 2. Of these two numbers, exactly one is positive. By convention, the positive number whose square is 2 is denoted by 22​. The other number, which is negative, is denoted by −2−2​. Therefore, the two numbers whose square is 2 are 22​ and −2−2​, and it follows that 2>02​>0.

Generalization to Arbitrary Positive Numbers

The same reasoning applies to any positive number $a$. Given any positive number $a$, there are precisely two numbers whose square is $a$. If $b$ is one of them, then $-b$ is the other. This can be shown by replacing 2 with $a$ in the preceding argument. By convention, the symbol aa​ denotes the unique positive number whose square is $a$. The other number whose square is $a$ is therefore −a−a​.

This relationship can be expressed by saying that the solutions of the equation x2=ax2=a are x=±ax=±a​. This notation is read as "$x$ equals plus or minus the square root of $a$." It is important to recognize that the symbol aa​ always refers to the positive square root, while the negative square root is denoted by −a−a.

Implications for Equations Involving Squares

A useful consequence of the preceding discussion is that if $x$ and $y$ are numbers such that x2=y2x2=y2, then either $x=y$ or $x=-y$. It is not possible to conclude that $x=y$ in all cases, because the negative possibility must also be considered. This fact is frequently used in solving equations and simplifying expressions.

Additionally, for any real number $x$, the number x2x2​ is always greater than or equal to zero. This is because the square root function, by convention, returns the nonnegative square root. Therefore, x2x2​ represents the absolute value of $x$, often written as $|x|$. This observation reinforces the idea that the square root symbol always denotes a nonnegative quantity, even when the number being squared is negative.

Summary

The theory of positive real numbers and square roots provides a foundation for understanding the real number system. The basic properties of positivity establish that positive numbers are closed under addition and multiplication, and that every real number is either positive, negative, or zero. From these properties, further results about the signs of products and quotients can be derived. The existence of square roots for positive numbers leads to the conclusion that every positive number has exactly two square roots: one positive and one negative. By convention, the symbol aa​ denotes the positive square root, while the negative square root is written as −a−a​. These conventions and results are essential tools for solving equations and working with algebraic expressions involving squares and square roots.