What would be the contrapositive of this statement?

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Homework Statement


The original statement is Prove that if xy and x+y are even then both x and y are even.


Homework Equations





The Attempt at a Solution


I think it goes like "If x or y is odd then xy and x+y are odd"? I'm not too sure though because the first "and" in the hypothesis is confusing. I'm not sure if its supposed to go like"If x or y is odd then xy or x+y is odd?
 
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bonfire09 said:

Homework Statement


The original statement is Prove that if xy and x+y are even then both x and y are even.

The Attempt at a Solution


I think it goes like "If x or y is odd then xy and x+y are odd"? I'm not too sure though because the first "and" in the hypothesis is confusing. I'm not sure if its supposed to go like"If x or y is odd then xy or x+y is odd?
You want to negate the statement "xy and x+y are even". If this statement is not true, it means that either xy is not even or x+y is not even, or both. And "not even" is the same as "odd". So indeed "If x or y is odd then xy or x+y is odd" is correct.
 
The statement is "If ((x is even) and (y is even)) then ((xy is even) and (x+y is even))".

As you suggest, the contrapositive of "if A then B" is "if not B then not A". So the contrapositive of this statement is
"if NOT ((xy is even) and (x+y is even)) then NOT ((x is even) and (y is even))".

The thing you are missing is that "NOT A and B" is "(Not A) or (not B)".
So the contra positive is
"if xy is odd or x+y is odd, then either x is odd or y is odd".
 
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I would think the sentence would have to be " If x is odd or y is odd then xy is odd or x+y is odd." I think you have it flipped around.
 
bonfire09 said:
I would think the sentence would have to be " If x is odd or y is odd then xy is odd or x+y is odd." I think you have it flipped around.
To whom are you responding?

(We have a "QUOTE" feature to take the guess-work out of this.)

By the Way: " If x is odd or y is odd then xy is odd or x+y is odd." is a valid contrapositive to the original statement.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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