Learning Mathematics
This is a guide on learning mathematics.
Table of Contents
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 positve, 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 commutivity 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 \((3/5)^{4} = \frac{3^{4}}{5^{4}} = \frac{81}{525}\)