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Prove $\forall a,b\in \mathrm{Z^{even}},$ if $a\not\equiv b (\mod 3)$ then $\frac{b-a}{2}\equiv 1 (\mod3)$

I think my above proof is wrong, I proceed to try some values, like

$2 \mod 3 = 2$,

$4 \mod 3 = 1$

$-2 \mod 3 = 1$,

$-4 \mod 3 = 2$

So if I let $a = -2$ and $b = 2$, $-2\not\equiv 2 (\mod 3)$ but $\frac{2-(-2)}{2} = 2\not\equiv 1 (\mod3)$. Is this a counter example to this question? So hence I just disprove this statement?

Probability_Sarah
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  • What does this mean $Z^{even}$. Is it the same as this $Z^2$ – Moustapha_M_I Mar 29 '23 at 11:36
  • Yes, of course , That can be refuted . – Moustapha_M_I Mar 29 '23 at 11:39
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    Please do not rely on pictures of text. – Shaun Mar 29 '23 at 12:32
  • Use $a\equiv b\pmod{n}$ for $a\equiv b\pmod{n}$. – Shaun Mar 29 '23 at 12:32
  • The line "since $a$, $b$ are even they have remainder $0$ or $2$ after division by three" is incorrect. For example, $4$ is even but $4 \equiv 1\pmod{3}$. – JD1874 Mar 29 '23 at 12:44
  • $\bmod 3!:\ 2\equiv-1\Rightarrow n/2\equiv -n,,$ so your claim is $,a\not\equiv b\Rightarrow a-b\equiv 1,,$ clearly false (e.g. $,a=4,b=2),$ Checking your proof line-by-line with this counterexmaple shows your first claim (2nd line) is already false, i.e. $,4\not\equiv 0,2\pmod{3}.,$ This is a general method of debugging proofs - see the linked dupe. – Bill Dubuque Mar 29 '23 at 19:33
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Mar 29 '23 at 19:37
  • @JackDavidson Yes I understand, it can also be 4 mod 3 is 1, the written proof is wrong. What about my counter example? Is that a valid counter example to disprove the statement? Thanks. – Probability_Sarah Mar 30 '23 at 07:07
  • @BillDubuque Yes I apologise I knew that line 2 was incorrect, as 1 could also be the result of (mod 3), which is why I wrote down a counter example, If my counter example valid? – Probability_Sarah Mar 30 '23 at 07:09
  • @SarahV.P Yep! Finding a counterexample is always a valid way to disprove a statement. – JD1874 Mar 30 '23 at 08:57
  • @JackDavidson Is mine valid? Like picking $a = -2$ and $b = 2? $ Did I disproved it? – Probability_Sarah Mar 31 '23 at 07:19
  • @SarahV.P Yes it is, since $\frac{2-(-2)}{2}=2 \equiv 2\pmod{3}$. – JD1874 Mar 31 '23 at 08:25
  • Let $,a,b = 2j,2k.$ $!\bmod 3!:\ \overbrace{2j\not\equiv 2k}^{\large a\ \not\equiv\ b}!\iff! j\not \equiv k$ which doesn't imply $!!\overbrace{k!-!j}^{\large (b-a)/2 }!!!\equiv 1$ (e.g. $,k!=!2,j!=!0)$ – Bill Dubuque Apr 02 '23 at 05:01
  • @BillDubuque What do you mean? – Probability_Sarah Apr 02 '23 at 05:43
  • My prior comment shows how to deduce that the claim in the title is false. – Bill Dubuque Apr 02 '23 at 05:57

1 Answers1

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Here is my solution. Let $E$ be the set of even numbers. Then, if $(a, b) \in E$, we have $a = 2k, b = 2m, (k, m) \in \mathbf{Z}$ and we have $\frac{b-a}{2} = m - k$. Now, $\neg \ a \equiv b \ (\mod3) \iff \neg \ m \equiv k \ (\mod 3)$. Now, if for example $m \equiv 0 (\mod 3), k \equiv 1 (\mod 3)$, we have $m - k = -1 \equiv 2 (\mod 3)$, which shows that your proof is flawed. A big mistake I can see in your proof is assuming that $k$ and $m$ are either $0$ or $2$, they can be $1$ as well, as in $m = 3l + 1 (l \in \mathbf{Z})$...

Martin Westin
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