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I am wondering if I have made a wrong jump in logic with my answering to this question.

The original statement goes "There exists an integer $n$ such that $n^2=n$, $n$ is even and $n$ is greater than $0$".

I have taken the approach of first defining that if $n$ is even, than $n=2k$ where $k$ is of the set of integers.

I have then substituted this into $n^2=n$ by saying that $(2k)^2=(2k)$, and as such $2k=1$. Given this I have concluded it to be false as there is no integer $k$ such that $2k=1$.

I am worried about this logic though because if $n$ is equal to $0$ (which is an even number that fits the definition of $2k$) then $n^2=n$ is true which seems to go against my previous logic (which is where the final comparison to $n$ having to be greater than $0$ would come in).

Am I right in being worried about the direction I have taken this logic? If so where have I gone wrong in my line of thinking? Thank you for all help.

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    $\begingroup$ To get $2k=1$ you divided both sides by $k$. But if $n=0$ then $k=0$. $\endgroup$ – halrankard Aug 25 at 12:43
  • $\begingroup$ You can't divide by $2k$ if $2k=0$, because we don't divide things by $0$. $\endgroup$ – Eod J. Aug 25 at 12:47
  • $\begingroup$ Thank you for clarifying on that, I hadn't considered that important detail at all in my assumptions. $\endgroup$ – James Aug 25 at 12:47
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You did (implicitly) make use of the assumption that $n > 0$. Since $n >0$, your $k$ cannot be $0$. Thus it is okay to divide by $k$, which you did in your proof when you moved from $4k^2=2k$ down to $2k=1$.

Good for you for thinking about this!

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  • $\begingroup$ Thank you for the answer. I see now the implicit assumption that I made. $\endgroup$ – James Aug 25 at 12:49
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We have $n^2=n$ implies $n^2-n=0$, i.e., $$n(n-1)=0.$$ Thus either $n=0$ or $n=1$, but, by hypothesis, both these options are impossible. (They are the only possible solutions by inspection and the fundamental theorem of algebra; moreover, by the quadratic formula,

$$\begin{align} n&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(0)}}{2(1)}\\ &=\frac{1\pm 1}{2}\\ &=0\text{ or }1.) \end{align}$$

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  • $\begingroup$ I am not quite sure I fully understand this implication? I see that if n^2=n then n=1 then n-1=0 where n is not equal to 0. How were you able to get n(n-1)=0 from this may I ask? $\endgroup$ – James Aug 25 at 13:03
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    $\begingroup$ I've edited my answer, @James. Does that help? $\endgroup$ – Shaun Aug 25 at 13:10
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    $\begingroup$ That does help, thank you for the clarification. $\endgroup$ – James Aug 25 at 13:15
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    $\begingroup$ You're welcome, @James. $\endgroup$ – Shaun Aug 25 at 14:40
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    $\begingroup$ $$n^2=0$$ $$n^2 - n = 0$$ Factorizing, $$n(n - 1) = 0$$ and you can do the rest. $\endgroup$ – SAGNIK UPADHYAY Aug 25 at 15:44
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The flaw lies in the implication

$$(2k)^2=2k\implies 2k=1.$$

This claim is indeed false when $k=0$.

$$0=0\not\Longrightarrow 0=1.$$

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  • $\begingroup$ Notice that no "division by zero" argument is necessary. $\endgroup$ – Yves Daoust Aug 25 at 14:44

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