Mathematical Proofs Problem?

Question

This is what the problem is:
(x+1)y^2 is even if and only if x is odd or y is even. How do i go about proving this?

Answer 1

To prove it, you show it's true for both of your cases -- for x odd and for y even -- that proves the "if" part of the statement. To prove the "only if" part, you also have to show that it's not true for your other cases -- i.e., for when x is NOT odd and for when y is NOT even.

CASE 1 - x is odd

If x is odd, then you can write x as 2n + 1 for some integer n. Substitute this into your formula, and solve:

(x+1) * y^2 =

(2n + 1 + 1) * y^2 =

(2n + 2) * y^2 =

2 * (n+1) * y^2

Now, regardless of the values of n and y here, this expression will always be even because of the 2. Therefore in this case #1, x is odd, (x+1)y^2 will always be even.

You would do similar work for all the other cases. I won't do all your work, as I'd like you to try this for yourself but I will get you started on each of the other cases:

CASE 2 - y is even - let y = 2n then substitute and solve

CASE 3 - x is even - let x = 2n then substitute and solve

CASE 4 - y is odd - let y = 2n+1 then substitute and solve

Good luck!

If you try this and get stuck, email me and post your work here and I will come back and try to help you some more.

Answer 2

Part 1 Suppose x is odd or y is even, let us prove that (x+1)y^2 is even.

a) Statement 1:If x is odd then x+1 is even. Proof. If x is odd, it can be written down as x = 2n+1 for some natural number n. Then x+1 equals (2n+1)+1 = 2n+2= 2(n+1), for the same n. Then x+1 is divisable by 2, so x+1 is even, as requested.
Statement 2: If x+1 is even, x+1 multiplied by any number z, is even.
Explanation: x+1 is divisable by 2, so x+1 multiplied by z is still diviasable by 2. Then (x+1)z is even for any z, in particular, for z=y^2.
Thus, if x is odd, we proved that x+1 is even, so (x+1)y^2 is even, for any y.
b)If y is even, it is divisable by 2, so y^2 is divisable by 4, in particular, y^2 is divisable by 2, so y^2 is even. But an even number multipling any number is even, so y^2(x+1) is even for any x.
From a) and b) combined it follows that if x is odd or y is even, (either or, let alone both) then (x+1)y^2 is even.

Part 2 Suppose that (x+1)y^2 is even. We need to prove that then either x is odd or y is even.
Since (x+1)y^2 is even, it has to be divisable by 2. [Any even number is always divisable by 2, by its very definition]
But then, at least one of the expressions, x+1 or y^2 has to be divisable by 2. Otherwise, if neither is divisable by 2, both of the expressions would be odd(according to the definition of odd numbers) and the product of any two odd numbers is always odd. So, if neither expression were even, the result would not be even. From our assumptions it stems that the product is even, so it is impossible for both expression be odd. Then, at least one of them is even, and possibly both, is even.
c)If x+1 is even, then x+1-1 has to be odd[for any even number k, k-1 would be odd, check for yourself]. (x+1)-1=x. Then x is odd. That was under the assumption of x+1 even.
d)Now, suppose it is y^2 which is even. Then y has to be even. Why? Because if y were odd, its product with the self would be odd, which would contradic our knowledge of y^2 being even. So y is even under the assumption of y^2 being even. And we already proved that at least one of them has to be even, x+1 or y^2. That, as we have seen, results in x being odd or y being even depending on the circumstannces. Now we have covered both cases in part 2, so we proved part 2.
Part 1 and part 2 together present you with the entire proof of the question. Hope it helps.

Answer 3

I ♥ AUG is correct just thought i'd confirm it and get the two points

Answer 4

sure, I can help you with that...
Bleh, nevermoing. other people did it first.