I have to rewrite the following in symbolic language:

(a) Every integer of the form 4y + 2, y 2 Z, is even. (b) There is a positive integer that is prime, divides 6 and divides 9. (c) There is a positive integer that is both prime and irrational.

Here are my answers and just want to check that that are correct.

(a) ($\forall\mathrm{x}\in\mathbb{Z}$)($\exists\mathrm{y}\in\mathbb{Z}$)(x=4y+2 $\in\mathbb{E}$)

(b) ($\exists\mathrm{x}\in\mathbb{P}$) (x|6 $\wedge$ x|9)

(c)($\exists\mathrm{x}\in\mathbb{Z}$ > 0) (x $\in\mathbb{P}\wedge$ x $\in\mathbb{R}$ - $\mathbb{Q}$)

Then I had to negate (a) and (c).

(a) $\neg$(($\forall\mathrm{x}\in\mathbb{Z}$)($\exists\mathrm{y}\in\mathbb{Z}$)(x=4y+2 $\in\mathbb{E}$)) $\iff$ ($\exists\mathrm{x}\in\mathbb{Z}$)$\neg$($\exists\mathrm{y}\in\mathbb{Z}$)(x=4y+2 $\in\mathbb{E}$) $\iff$ ($\exists\mathrm{x}\in\mathbb{Z}$)($\forall\mathrm{y}\in\mathbb{Z}$)(x=4y+2 $\notin\mathbb{E}$)

(c) $\neg$(($\exists\mathrm{x}\in\mathbb{Z}$ > 0) (x $\in\mathbb{P}\wedge$ x $\in\mathbb{R}$ - $\mathbb{Q}$)) $\iff$ ($\forall\mathrm{x}\in\mathbb{Z}$>0) ($\mathrm{x}\notin\mathbb{P}\vee\mathrm{x}\notin\mathbb{R}$ - $\mathbb{Q}$)

  • $\begingroup$ Please refer to this link for information on how to format math on this site. $\endgroup$ – Regret Mar 10 '15 at 22:27
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    $\begingroup$ Edit your post to include your own answers, and we'll be happy to check if they are correct, and if not, steer you in the right direction. What's the point of asking others to do work you may have already done correctly? $\endgroup$ – Namaste Mar 10 '15 at 22:28
  • $\begingroup$ It seems to me (trivially) that "a positive integer that is both prime and irrational” is impossible. Is it not? $\endgroup$ – Moritz Mar 10 '15 at 22:41
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    $\begingroup$ Well, false sentences can be written down in symbols just like true ones. $\endgroup$ – Crostul Mar 10 '15 at 22:42
  • $\begingroup$ Very true, worth it for me to learn how to post properly here too. $\endgroup$ – ajs512 Mar 11 '15 at 2:25

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