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I've been working at this for about an hour and cannot figure it out.

Problem: Prove that if $x+5$ is odd, then $x^2$ is even.

I've tried induction, contrapositive and contradiction methods. I cannot assume things like $1$+even=odd, only that $2k+1$ is odd and $2k$ is even where $k$ = integer.

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    $\begingroup$ First try to prove that x is even. $\endgroup$ – PM 2Ring Jan 13 '15 at 7:11
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    $\begingroup$ When $x+5$ is odd, then $x$ is even and therefore $x^2$ is even. $\endgroup$ – Redundant Aunt Jan 13 '15 at 7:13
  • $\begingroup$ No one has answered my question. Is anyone familiar with logic and proofs? I don't need intuitive answers. $\endgroup$ – user2665637 Jan 13 '15 at 7:23
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    $\begingroup$ I added an answer, but others were faster. However, you could easily take those "intuitive answers" and make it rigorous. Take it as an extra exercise to prove that $1 + \text{even} = \text{odd}$ and $1 + \text{odd} = \text{even}$. The same can be done for $\text{odd} + \text{odd} = \text{even}$, etc. $\endgroup$ – Eff Jan 13 '15 at 7:40
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If $x+5$ is odd, then by defintion, $\exists k\in \mathbb{Z}: x+5=2k+1.$

$$x+5 = 2k+1 \implies x=2k+1-5=2k-4$$ $$x=2k-4 = 2(k-2)$$ $$\implies x^2=4(k-2)^2=2(2(k-2)^2)$$ $$p=2(k-2)^2 \in \mathbb{Z} \implies x^2=2p$$ Thus, $x^2$ is even.

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  • $\begingroup$ @user2665637 This is the accepted answer, but I'd say you better also use beedge89's answer's second line "Even numbers squared are even". If you are allowed to use the unique prime factorization of natural numbers, you can then skip on the explicit squaring of $x$ above. $\endgroup$ – polynomial_donut Sep 29 '15 at 19:13
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$x+5$ odd implies $x$ is even (odd minus odd) is even.

Even numbers squared are even.

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since " $x+5$ is odd" so "$x$ is even". therefore "$x^2$ is even":
for the last part note that if "$x$ is even" then we can write $x$ as: $x=2k$. so $x^2=4k^2=2\times 2k^2$, which is even.

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If $x+5$ is odd, then $x+4$ is even, so $x$ is even.

And if $x$ is even, then $x \cdot x = x^2$ must be even, too.

This follows from these statements:

  • If $n$ is odd, then $n+1$ is even
  • If $n$ is even, then $n+c$ is even, where $c$ is also even.
  • An even number multiplied by another even number gives an even number.
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To prove $x^2$ is even, you must first prove that $x$ is even.

You must always remember the following rules: $$odd + odd = even$$ $$odd + even = odd$$ $$even + even = even$$ Given $x+5$ is odd. Since odd + odd can never be odd, $x$ should be even .

EDIT

Continuing from where you left.

Case 1: Let $x$ be even.$$x=2k$$ $$x+5 = 2k+5 = even + odd = odd$$

Case 2: Let $x$ be odd.$$x=2k+1$$ $$x+5 = 2k+1+5 = 2k+6 = 2(k+3) = even$$

Hence $x$ should be even which implies $x^2$ should be even.

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  • $\begingroup$ Can you recreate that with only, even = 2k and odd = 2k+1 where k are integers. $\endgroup$ – user2665637 Jan 13 '15 at 7:20
  • $\begingroup$ @user2665637 Is this what you wanted ? $\endgroup$ – lsp Jan 13 '15 at 7:25
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If you would prefer:

$$x+5=2k+1$$ for some $k\in \mathbb{Z}$. Then $$x=2k-4=2(k-2)$$ where we find $$x^2=(2(k-2))^2=4(k-2)^2=2(2(k-2)^2)=2c$$ where $c\in \mathbb{Z}$. Hence $x^2$ is even.

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Let $x+5$ be odd. Then there exists an integer $k$ such that $2k+1 = x+5$, hence $2k = x + 4$, so $x+4$ is even. It follows that $x = 2k-4 = 2(k-2)$. Since $k-2$ is an integer it follows that $x$ is even. Now let $x = 2c$ for an integer $c$, then $x^2 = (2c)^2 = 4c^2 = 2(2c^2)$ is even since $2c^2$ is an integer.

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Sorry but this is not right because $x$ is not a number ... replace $x$ with $2$ and therefore $2 + 5 = 7$ which is odd so you're right

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If $x^2$ is even,then $x$ is even and $x+4$ is even

You'll know that $(x+4)+1$ is odd

Q.E.D.

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  • $\begingroup$ Whilst it's true that if x^2 is even, then x is even we can't use that fact without proving it first. Actually, (as others have noted) we really need to show that x is even implies x^2 is even. $\endgroup$ – PM 2Ring Jan 13 '15 at 8:02
  • $\begingroup$ Let prove by contrapositive method $(p\rightarrow q \equiv ~q \rightarrow ~p)% $\endgroup$ – Yuko Jan 13 '15 at 13:31
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    $\begingroup$ Contrapositive would be "if $x^2$ is odd, then...." $\endgroup$ – Squirtle Feb 21 '15 at 7:18
  • $\begingroup$ You assumed what you wanted to prove and proved that it implies what you should've begun with. The contrapositive proof would be: "if $x^2$ is odd, then $x+5$ is even". $\endgroup$ – user26486 Feb 21 '15 at 7:51

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