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I'm reading D'Angelo and West, first edition for recreation. In example 20, it states "If integers x and y are odd, then x+y is even." (I took this to mean, P(x,y both odd) -> Q (x+y is even.) Easy proof - no problem. The example continues with "The converse of this conditional is false." (I took this to mean Q(x,y both odd) does not imply P(x+y is even). The reason is that x+y could be even but x and y are not both odd. (For example, both x and y are even).

In the example that follows, it is stated that an integer is even if and only if it is the sum or two odd integers, that is, P->Q and Q->P. The sum of two odd integers is always even - got that. But why is the other statement, an integer is even if it is the of two odd integers. That is true for every integer (2k = 2k-1 +1 shows that) but it's not always true in every case is it? For example the sum of two even integers is even 2k = 2p + 2q = 2(p+q).

What am I missing?

Mario Krenn
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Gary Hagan
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  • I think the confusion is this: "is the sum of two odd integers" means the same as "can be written as the sum of two odd integers". As you observed yourself, every even number has this property. However, the fact that even numbers can also be written as a sum of even numbers is neither here nor there. –  Nov 15 '13 at 14:22

3 Answers3

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The statement can be more clearly written as:

$x$ is an even integer if and only if there exists odd integers, $y$ and $z$, such that $y+z=x$.

Alraxite
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The if and only if statement does not state anything about the sum of two even integers.

Therefore you cannot assume that it is true or false.

Regards to the first paragraph you wrote, the converse of the statement is actually (x+y is even) implies (x, y both odd) which is FALSE. (I'm sure you just made a typing mistake).

Jack Jiang
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It's saying: $ P(x$ is even$) \iff \exists(y,z\in \mathbb{N}_{odd}): x=y+z$ It's not saying that it must be the sum of two odd numbers, just that it can be expressed that way. You can't do that for odd numbers.