# If x + y is odd for integers x and y, then x * y is even?

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• Sep 5th 2013, 05:10 AM
lamentofking
If x + y is odd for integers x and y, then x * y is even?
Prove the following statement:

If x + y is odd for integers x and y, then x * y is even

My response:

If the sum of two numbers is odd, then their product is even.
• Sep 5th 2013, 05:15 AM
FelixFelicis28
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by lamentofking
Prove the following statement:

If x + y is odd for integers x and y, then x * y is even

My response:

If the sum of two numbers is odd, then their product is even.

That doesn't constitute as "proof" of the statement.

Firstly, note that an odd number will always have the form $\displaystyle 2n - 1, \ n \in \mathbb{Z^{+}}$ and an even number of the form $\displaystyle 2m, \ m \in \mathbb{Z^{+}}$.

You have to first prove that $\displaystyle x+y$ will be odd iff $\displaystyle x$ is odd and $\displaystyle y$ is even or $\displaystyle y$ is odd and $\displaystyle x$ is even. Then prove that their product must be even i.e. $\displaystyle xy \equiv 0 \pmod{2}$.
• Sep 5th 2013, 05:36 AM
lamentofking
Re: If x + y is odd for integers x and y, then x * y is even?
So could I say, "if the sum is odd, then either m is odd and n is even or m is even and n is odd. The product of an odd and even number is an even product."
• Sep 5th 2013, 05:38 AM
FelixFelicis28
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by lamentofking
So could I say, "if the sum is odd, then either m is odd and n is even or m is even and n is odd. The product of an odd and even number is an even product."

You can say it, and that's what the statement means in terms of 'words', but it still doesn't constitute as proof. Look at what I said in my last post.
• Sep 5th 2013, 05:48 AM
lamentofking
Re: If x + y is odd for integers x and y, then x * y is even?
The 3 line equal sign means "defined as"? Well then I can say that x + y is odd because either x is odd and y is even or x is even and y is odd. Their product is even because the product (mod) 2 will yield a remainder of 0.
• Sep 5th 2013, 05:54 AM
Hartlw
Re: If x + y is odd for integers x and y, then x * y is even?
x+y=2n+1 -> one is odd and one is even.
evenxodd=even
• Sep 5th 2013, 05:56 AM
FelixFelicis28
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by lamentofking
The 3 line equal sign means "defined as"?

It means 'equivalent to' or in this case, 'congruent to'.

Quote:

Well then I can say that x + y is odd because either x is odd and y is even or x is even and y is odd.
Tbh, I'm not sure how rigorously this question wants you to prove your statement as to whether you can assume odd + even = odd for all integers. I personally would check each case:

a) x, y = even
b) x, y = odd
c) x = odd, y = even OR y = odd, x = even.

Do this by writing them in the aforementioned form i.e. say for the first case a) if x, y = even, then we can write $\displaystyle x = 2n, \ y = 2m$ for some $\displaystyle n, m \ \in \mathbb{Z^{+}}$

Now obviously $\displaystyle x + y = 2n + 2m = 2(n + m)$ and as $\displaystyle n, \ m \ \in \mathbb{Z^{+}}$ then $\displaystyle x + y$ is in the form $\displaystyle 2k, \ k \in \mathbb{Z^{+}}$ hence is even, so that eliminates this case.

Quote:

Their product is even because the product (mod) 2 will yield a remainder of 0.
You will need to prove this the way I have above - once you've checked each case and confirmed that only case c) holds for the statement, write $\displaystyle x = 2n, \ y = 2m - 1$ or vice versa for some $\displaystyle n, m \ \in \mathbb{Z^{+}}$ and multiply them together and the results falls out fairly quickly.
• Sep 5th 2013, 06:04 AM
Hartlw
Re: If x + y is odd for integers x and y, then x * y is even?
Reply to post #7. Why so complicated and obtuse?
What's wrong with post #6?
• Sep 5th 2013, 06:08 AM
FelixFelicis28
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by Hartlw
Reply to post #7. Why so complicated and obtuse?
What's wrong with post #6?

Tbh, I have no idea what level of rigour OP needs in his proof of the statement so better safe than sorry, I suppose. There's nothing wrong with post #6 but it assumes odd + even = odd and odd x even = even and I have no idea whether OP can assume that or not, hence the long post. But I dunno.

Edit: I wasn't trying to undermine your post or anything, if that's what you mean. I was in the process of typing my post out when you posted your's.
• Sep 5th 2013, 06:22 AM
Hartlw
Re: If x + y is odd for integers x and y, then x * y is even?
Nothing is assumed.
x+y = odd is given. -> x+y=2n+1 -> x and y can't both be even or odd.
oddxeven=(2n+1)(2m)=even. Sorry, thought that was obvious.
• Sep 5th 2013, 06:38 AM
FelixFelicis28
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by Hartlw
Nothing is assumed.
x+y = odd is given. -> x+y=2n+1 -> x and y can't both be even or odd.

Tbh, I agree, this part may have been a bit OTT but as I've said, I have no idea what OP can assume or not so wrote it out on my post.

Quote:

oddxeven=(2n+1)(2m)=even. Sorry, thought that was obvious.
Regardless of how obvious it is, it doesn't constitute as proof, especially if the question asks you to prove even x odd = even (which is what I interpreted what the question was getting at, anyway). Apologies if that isn't the case.
• Sep 5th 2013, 07:13 AM
Hartlw
Re: If x + y is odd for integers x and y, then x * y is even?
"oddxeven=even is obvious" is not a proof. oddxeven=(2n+1)x(2m)=even is.

But anyhow, my apologies. I get cranky when I see congruences because I don't understand them (I do understand the formal definition).

If we are dealing with a set of objects for which congruence is defined,
1) What is the definition that determines members of the set.
2) What are the members of the set?
3) If a and b belong to the set, what is the definition of axb and a+b.

Or they are a relation. Is multiplication and division defined for a relation (subset definition). If so, what is the definition of multiplication and division for congruences as a relation?

If my questions are too obtuse, you can just say so. Can't be more precise because I don't understand congruences.
• Sep 5th 2013, 07:24 AM
HallsofIvy
Re: If x + y is odd for integers x and y, then x * y is even?
Yes, if m+ n is odd then one of m or n is even, the other odd. Whether then saying that "therefore mn is even" is a good proof or not depends upon whether or not you have previously proved, and can use, the statements "if m+ n is odd then one is even, the other odd" and "if an integer is even, then its product with any other integer is even".

If not, a complete proof of "if m+ n is odd then mn is even" would have to include proofs of those.
• Sep 5th 2013, 08:03 AM
Hartlw
Re: If x + y is odd for integers x and y, then x * y is even?
Quote:

Originally Posted by HallsofIvy
Yes, if m+ n is odd then one of m or n is even, the other odd. Whether then saying that "therefore mn is even" is a good proof or not depends upon whether or not you have previously proved, and can use, the statements "if m+ n is odd then one is even, the other odd" and "if an integer is even, then its product with any other integer is even".

If not, a complete proof of "if m+ n is odd then mn is even" would have to include proofs of those.

That's exactly what I did:
x+y = odd is given. -> x+y=2n+1 -> x and y can't both be even or odd.
oddxeven=(2n+1)(2m)=even. Sorry, thought that was obvious.

Do you have a problem with "x+y=2n+1 -> x and y can't both be even or odd."

OK, I'll spell it out: if x and y are odd: x=2n+1, y=2m+1 -> x+y=2(m+n) + 2 = even
If x and y are even, x=2n, y=2m, x+y=2(m+n) = even. But x+y is given as odd. So x and y can't both be even or odd.

I'm really confused by all these responses to a very simple question and answer.
• Sep 5th 2013, 04:08 PM
Deveno
Re: If x + y is odd for integers x and y, then x * y is even?
An instructor would probably want a proof such as this:

Suppose x + y = 2m + 1 (given).

Case 1: x = 2k (x is even)

Then y = 2m + 1 - 2k = 2(m-k) + 1, that is, if x is even, y is odd.

Case 2: x = 2k + 1 (x is odd)

Then y = 2m + 1 - (2k + 1) = 2(m - k ) + 1 - 1 = 2(m - k), that is, y is even.

Thus in either case, we have one odd integer, and one even integer.

So without loss of generality we may take x = 2k, y = 2m + 1 (or else consider y + x).

Then: xy = (2k)(2m + 1) = 4km + 2k = 2(2km + k), that is: xy is even.

The "completeness" of this proof lies in the fact that an integer (x in this case) must be either even or odd (and not both), something which might be considered obvious (but perhaps not).

An alternative proof (based on divisibility):

If x is even, then 2|x, so 2|xy thus xy is even (since if x = 2k, then xy = (2k)y = 2(ky)) (no matter whether y is even OR odd). It follows from 2 does not divide x+y, and 2 does not divide y that 2 does not divide y:

for if 2|x, 2|y, then x+y = 2x + 2y = 2(x+y), so that 2|x+y, contradicting the given that x+y is odd.

If x is odd, then given that 2 does not divide x + y, we must have x + y = 2n + 1 (0 and 1 are the only possible remainders upon division by 2, and thus x = 2k + 1), and it follows that 2|y:

for if 2 does not divide y, then x = 2k + 1, y = 2m + 1 and x + y = 2k + 2m + 2 = 2(k + m + 1), which clearly is divisible by 2, a contradiction.

So if x is odd, 2|y, whereupon 2|xy since xy = x(2t) = 2(xt).

Now, while both of these are apparently long-winded explications of an apparently intuitively obvious fact, their approach generalizes very well to the more general setting of a unique factorization domain (ring) (such a polynomials) where we no longer have such a clear intuitve idea of what the individual elements look like.

Proofs differ in their level of rigor, for some, it's a stylistic affair. None of the proofs given (including mine) qualify as a "formal" proof, but rather are "semi-formal" designed to CONVINCE. For example, when Hartlw says:

"If x+y is odd, x and y can't both be even or odd"

A person seeing such a statement may ask: "well, WHY?"