Positive and Negative Integers

This blog now begins a series reviewing the basic concepts of algebra.

First we start with the integers, which are a set of whole numbers.

Negative integers are those less than 0
Positive integers are those greater than 0

i.e.
...-3,-2,-1,0,1,2,3...

The sign < means less than
The sign > means greater than

Thus
-3 < 1
3 > 2
3 > -2

and so on.

Prove if x+y is irrational then x or y are irrational.

For real numbers x and y prove if x+y is irrational then x or y are irrational.

This is a proof by contraposition, so we assume x and y are rational.

then x=m/k and y=n/t where m,k,n,t are integers.

so x+y = m/k + n/t = (mt + nk)/(kt) which is rational since m t n k are integers.

So x+y is rational. Therefore, by contraposition (~q=>~p therefore p=>q), if x+y is irrational then x or y are irrational.

Prove that if x^2 is odd then x is odd. (Direct proof)

Assume x is an integer. If x^2 (x squared) is odd then x is odd.

First lets look at some examples

3^2 is 9
5^2 is 25
7^2 is 49
9^2 is 81

So it does appear that for any odd integer, its square is also an integer, how can we prove this for all cases?

First we assume x is odd, then x=2y+1 for some integer y.

So now x^2 = (2y+1)(2y+1) = 4y^2 + 4y + 1= 2(2y^2 + 2) + 1 which is odd.

Prove if x*y = 1 then x = y = 1

Assume that x and y are positive integers, prove that if x * y = 1 then x = y = 1

If x * y = 1 then x = (1/y) and y = (1/x)

Yesterday we proved that if an integer x divides an integer y then x<=y.

So since y divides 1 or (1/y) then y<=1

and since x divides 1 or (1/x) then x<=1

Since both x and y are integers then y=1 and x=1, so x = y = 1

Prove that if x is even and y is odd, then x+y is odd

Let x and y be integers. Prove that if x is even and y is odd, then x+y is odd.

x is even, therefore x=2k for some integer k.
y is odd, therefore y=2j+1 for some integer j.

by substitution into x+y=odd we get

2k + 2j + 1 = odd

2k + 2j + 1 = 2(k+j) + 1, which is odd.