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For a simple question like

Let $x, y \in Z$. If $3 | x$ or $3 | y$ then $3 | x y$.

Is it alright to assume all $x$ and all $y$ exist in $Z$?

I am trying to negate the statement but since it does not say 'each' explicitly, I am not sure.

gebruiker
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Navy Seal
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  • Yes, it should be "for all" in the statement. – Timbuc Feb 23 '15 at 12:18
  • Okay thanks. Quick follow up please, in that case, when I negate I have (3 is not divisible by xy) AND ( 3 | y OR 3 | x y)

    So isnt this always true because of the first part (before AND) since I can choose any x and y say 3 and 1 that are divisible by 3 ? @Timbuc

     – Navy Seal Feb 23 '15 at 12:25
  • Read my hinted answer below. – Timbuc Feb 23 '15 at 12:26

1 Answers1

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$$P = \forall\,x,y\in\Bbb Z\;,\;\;\left[\;\left(3\mid x\;\;\vee\;\;3\mid y\right)\longrightarrow 3\mid xy\;\right]$$

and from here

$$\neg P=\exists\,x,y\in\Bbb Z\;,\;\;\neg\left[\;\left(3\mid x\;\;\vee\;\;3\mid y\right)\longrightarrow 3\mid xy\;\right]$$

You now may want to use

$$A\longrightarrow B\equiv \neg A\vee B$$

and also

$$\neg (A\vee B)\equiv \neg A\wedge \neg B$$

Timbuc
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  • Okay, my order was (not B and not A) but my question is- if I can prove any one of the parts false, then the original statement is correct right? – Navy Seal Feb 23 '15 at 12:33
  • If you meant to prove that in $;\neg A\wedge\neg B;$ any of these two parts is false then the answer is yes, as conjunction is true only if both its parts is true. – Timbuc Feb 23 '15 at 12:36
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    Thanks a lot, especially for being patient with a noob. Cool bio too! @timbuc – Navy Seal Feb 23 '15 at 12:38