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There are topics with the same name but my question is not as abstract as in those.

My question is as follows: taken a generic definition like $x\;\mathbf{ is\; something}$ if $y$ it could be written like $x\;\mathbf{ is\; something}\Leftarrow y$ so everytime you have $y$ then you can deduce $x\;\mathbf{ is\; something}$. But if you start with $x\;\mathbf{ is\; something}$ as an axiom you could not deduce $y$ which should be deducible. Like if you have $x\;\mathbf{ is\; even}\Leftarrow x \equiv 0 \pmod{2}$ you SHOULD have $x \equiv 0 \pmod{2}$ starting by $x\;\mathbf{is\;even}$. That should be resolved if you say by saying $x\;\mathbf{is\;even}$ iff $x\equiv0\pmod{2}$ instead of if (so you would have $\Leftrightarrow$ instead of $\Leftarrow$) but I definitely feel that I'm missing something...

Can you help me? :D

Sorry for my bad English and if my question was stupid.

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  • $\begingroup$ say, $y$ is "$x$ is divisible by 4" then you can claim one way, but not another. If $x $ is even was taken as an axiom, you still could not deduce $y$ from here. $\endgroup$ – tortue Aug 4 '18 at 20:15
  • $\begingroup$ For example [here] (en.wikipedia.org/wiki/Parity_(mathematics)) it says An integer is even if it is evenly divisible by two and in many other places definitions are given like this. $\endgroup$ – Mega-X Aug 4 '18 at 20:17
  • $\begingroup$ See in particular Carl Mummert's answer to the linked question. $\endgroup$ – Hans Lundmark Aug 4 '18 at 20:17
  • $\begingroup$ Yes @HansLundmark that question is definitely what I was looking for. Sorry for opening a duplicate topic. $\endgroup$ – Mega-X Aug 4 '18 at 20:19
  • $\begingroup$ @Mega-X: Don't worry, it's not a big deal! $\endgroup$ – Hans Lundmark Aug 4 '18 at 20:20

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