Basics of Number Theory
I have used the divisibility definition several times. Here I will
present it again, as we delve more deeply into its uses. We will
say that an integer a divides an integer b evenly without a
remainder, like this: a  b. This implies that there exists an
integer c such that b = ac. We will only define division by non
zero integers. Hence, it is not permissible to write 0  a.
Here are some rules that division of integers follow. (Note, a, b
and c are always nonzero integers.)
1) 1  a
2) a  0
3) if a  b, and b  c, then a  c.
4) if a  b and b  a, then a = +b or a = b
5) if x = y + z, and we have a  y and a  z, then a  x as well.
6) if a  b and a  c, then we have a  bx + cy for all ints x and y.
An example of how we can use these rules is as follows.
Are there any integer solutions to the equation
5x + 10y = 132?
The answer is no. We know that 5 must divide 5x and it must
also divide 10y, thus 5  (5x+10y). We can see this clearly by
factoring the expression as 5(x+2y). But we know that 5  132 is
false, thus, there is no solution. One more way we can see this is
by the following:
5(x+2y) = 132
x+2y = 132/5, since x and y are ints, x+2y is, but 132/5 is not, so
there are no integer solutions that satisfy the equation.
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Here is another example of how to use these rules:
Prove that if 13 divides 3x+4y, that 13 also divides 7x+5y. We
can rewrite 7x+5y as 13x + 13y – 2(3x+4y). So we have:
7x + 5y = 13(x+y) – 2(3x + 4y)
Let A = x+y, B=3x+4y
We have 7x+5y = 13A – 2B.
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 Fall '09
 Number Theory, Division, Remainder, 7x+5y

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