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Let $x\in\mathbb{R}$. I want to find for which $x$ the statement $$ \neg(x\le 21\iff x>15) $$ holds.

I believe it is true when $x\in(-\infty,15)\cup[21,\infty)$, but I don't know how to write this down, as I feel that there is nothing to write down.

Can someone give me some tips? Is my answer correct and how should I write down the answer?

Daniel W. Farlow
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jacob
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    I believe you're right except $(-\infty,15)$ should be $(-\infty,15]$ and $[21,\infty)$ should be $(21,\infty)$. – Daniel W. Farlow Jan 30 '15 at 15:20
  • @induktio: yes, please – jacob Jan 30 '15 at 15:22
  • As Tim, MJD, and myself just discussed in Tim's answer, it will be more beneficial for you to work through most of it yourself. The answer just posted by drhab substantiates my first comment on your question. Good luck! – Daniel W. Farlow Jan 30 '15 at 15:32

2 Answers2

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$A\iff B$ is true when either both $A$ and $B$ are true, or when $A$ and $B$ are false. So $\neg(A\iff B)$ is true if $A$ is true and $B$ is false, or when $A$ is false and $B$ is true.

Tim Raczkowski
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  • OP asked for "tips", which is what this is. In any event if it is a homework problem, as seems likely, for OP to receive a complete solution might put them in violation of their school's ethics code. – MJD Jan 30 '15 at 15:24
  • @induktio accepted :) – Tim Raczkowski Jan 30 '15 at 15:28
  • It may be worth noting that $\neg(p\Leftrightarrow q)\equiv p\oplus q$ (which is just putting into symbols what you stated).. – Conor O'Brien Feb 01 '15 at 07:58
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Note that the following statements are equivalent:

  • $\neg\left(p\iff q\right)$

  • $\neg\left(p\Rightarrow q\wedge q\Rightarrow p\right)$

  • $\neg\left(\left(\neg p\vee q\right)\wedge\left(p\vee\neg q\right)\right)$

  • $\neg\left(\neg p\vee q\right)\vee\neg\left(p\vee\neg q\right)$

  • $\left(p\wedge\neg q\right)\vee\left(\neg p\wedge q\right)$

The last line tells us explicitly that these statements are true if and only if exactly one of the statements $p$ and $q$ is true.

So you need to find for wich $x$ you have $x\leq21$ and not $x>15$ (wich is the case if $x\leq15$) or not $x\leq21$ and $x>15$ (which is the case if $x>21$).

drhab
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