Prove for every odd integer n, 4n^2 + 3n + 2 is odd

Prove for every odd integer n, 4n^2 + 3n + 2 is odd

Let n be an odd integer, so n = 2h + 1 for some integer h.

So 4n^2 + 3n + 2 = 4(2h+1)^2 + 3(2h+1) + 2 =

16h^2 + 16h + 4 + 6h + 3 + 2 =

16h^2 + 22h + 9 =

2(8h^2 + 11h + 4) + 1, which is odd, and therefore

4n^2 + 3n + 2 is odd for every odd integer n.

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 ab is odd, then both a and b are odd. (Contradiction)

Assume a and b are positive integers, prove if a*b is odd then a and b are odd.

Proof by contradiction: where we want to prove Q => P we take ~P=> Q Λ ~ Q which is a contradiction, so therefore we conclude P.

This is a proof by contradiction which means that we have to assume the opposite consequent, which is the say that a and b are not both odd. So either a is even or b is even.

If a is even, then a=2k for some integer k, and b*2k will always be even.

If b is even, then b=2m for some integer m, and a*2m will always be even.

So either case leads to a contradiction that a*b is odd, and therefore if a*b is odd, then both a and b are odd.

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.

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.