# lec1109 - Basics of Number Theory I have used the...

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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 non-zero 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. Since 13 | B, we can express B=13B’, where B’ is an integer.
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## This document was uploaded on 06/09/2011.

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lec1109 - Basics of Number Theory I have used the...

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