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This question already has an answer here:

I am trying to prove or disprove that if $x$ is even, then $x + 5$ is odd.

This is what I have thus far, but I am stuck:

  • Assume that the chose variable (x) are in the domain: (x) is an integer

  • Assume the IF part of the statement: (x) is even

  • Prove the THEN part: Let a be a dummy variable that we assume is an integer. $x=2a+1$ (Definition of an Odd Integer)

Now find $x+5$.

$x+5=2a+1+5$ (Because we added the same value to both sides of the equation)

$x+5=2a+6$

$x+5=2(a+3)$ (Because of the Distributive Property of Multiplication over Addition)

Let $z=a+3$;

$x + 5 = 2z,$ but this is the definition of an even integer; where did I mess up?

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marked as duplicate by MathOverview, wythagoras, Najib Idrissi, Siminore, user223391 Oct 3 '15 at 17:08

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  • $\begingroup$ You're assuming that x is even, and therefore x=2a+1 for some integer a? In other words, you've assumed that x is even, and also that x is odd... $\endgroup$ – immibis Oct 3 '15 at 0:27
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In general when faced with "even" or "odd" assumptions in proof for some variable $x \in \mathbb{Z}$, it's helpful to express $x$ in terms of some other integer $k \in \mathbb{Z}$.

In your case, if $x$ is even, then there must exist some integer- we'll call it $k$- such that $x=2k$. We don't know exactly what value $k$ is, and frankly, we don't need to know.

Since $x=2k$, and we want to show that $x+5$ is odd, simply add 5 to the first value:

$$x+5= 2k+5$$ $$=2k+4+1$$ $$=2(k+2)+1$$

In general, we know that an odd number $m$ is one where for some integer $n$, $m=2n+1.$

Note that regardless of what $k$ is, $k+2$ will maintain its evenness or oddness (depending on $k$'s value); however, $2(k+2)$ will always be even.

And thus, $2(k+2)+1$ is odd.

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Your first line of proof does not agree with your assumption that x is even:

Assume the IF part of the statement: (x) is even
x=2a+1 (Definition of an Odd Integer)
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given $x$ is even so $x=2k$ where you took $x=2k+1$

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  • $\begingroup$ codex = 2k code2k + 5 = 2k + 6 code2k + 5 = 2(k+3) code2k + 5 = 2r $\endgroup$ – Joshua Mathews Oct 2 '15 at 20:32
  • $\begingroup$ $2k + 5 = 2k + 4 + 1 = 2(k +2) + 1$ which is odd. $\endgroup$ – user70962 Oct 2 '15 at 20:33
  • $\begingroup$ where did the extra 1 come from? $\endgroup$ – Joshua Mathews Oct 2 '15 at 20:34
  • $\begingroup$ I just broke it up: $5 = 4 + 1.$ $\endgroup$ – user70962 Oct 2 '15 at 20:35
  • $\begingroup$ @JoshuaMathews it isn't extra; 5=4+1 $\endgroup$ – daOnlyBG Oct 2 '15 at 20:37
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Notice, since $x$ is even then it can be rewritten as $x=2m$

Where, $m$ is some integer

Now, we have $$x+5=2k+5=2k+4+1$$$$=2(k+2)+1=\color{blue}{2n+1}$$ Where, $n=k+2$ is an integer

Thus, we get $\color{red}{x+5}=\color{blue}{2n+1}$ which is an odd integer i.e. $\color{red}{x+5}$ is not even.

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    $\begingroup$ Thank you! I messed up from the very beginning. $\endgroup$ – Joshua Mathews Oct 2 '15 at 20:39

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