Prove if c divides a and c divides b, then c divides any combination an+bm, where n and m are integers

Prove if c divides a and c divides b, then c divides any combination an+bm, where n and m are integers.

Since c divides both a and b then there must exist integers j and t such that a=cj and b=ct.

So by substitution into an+bm we get

an+bm = (cj)n+(ct)m = c (jn + tm). So c divides (an + bm) ~~~~

Thus, if c divides a and b, then c also divides any multiple of a and b.

Prove that there do not exist integers k and j such that 2j + 6k = 9

Prove that there do not exist integers k and j such that 2j + 6k = 9

Assume that j and k are integers. Then 2j + 6k = 2(j+3k) which is even, but 9 is odd, so it is evident that no integers j and k can be used to equal an odd number.

Prove if x is even, then x+1 is odd.

Let x be an integer and prove if x is even, then x+1 is odd.

We will use contraposition for this proof, so we start by assuming that x + 1 is not odd.

Then we can say there exists an integer k such that x + 1 = 2k

so x = 2k - 1 = 2(k-1) + 1

Thus x = 2(k-1) + 1 which is odd. So we can no say if x + 1 is not odd, then x is not even. Thus, by contraposition, if x is even, x+1 is odd.

If x^2 (x squared) is odd then x is odd. (Contraposition)

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


This is the same proof as we did yesterday, except this time we will prove the statement using another form of logic called contraposition.


To do this, we must first state the contrapositive of what we set out to prove.

so

If x^2 (x squared) is odd then x is odd

now becomes

If x^2 (x squared) is even then x is even

which is the contrapositive, the idea is that if we can prove that x^2 is even for all x that is even, it stands to reason that x^2 can only be odd for all x that is odd.

The way to write this symbolically is ~Q => ~P so therefore P => Q.

So now we prove if x^2 (x squared) is even then x is even

x is even so x=2y for some integer y.

then x^2 = 2y(2y) = (4y^2) = 2(2y^2) which is even.

Therefore if x^2 is even x is even, so by contraposition, if x^2 is odd, x is odd.