Alex
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 Sep 24 awarded Autobiographer Dec 2 accepted True or False question about continuous functions Dec 2 comment True or False question about continuous functions Well, my first thought would be $\infty$ (if I could set a function value to $\infty$...), but after looking closer at the function, as $x$ gets closer to 0.5 from the negative side, it approaches $-\infty$, and $\infty$ from the positive side. Wouldn't any value that I set create a discontinuity between the positive and negative sides? Dec 2 comment True or False question about continuous functions Nevermind, @DonAntonio answered the question I was asking! Thank you, so simply setting a random value for $f(0.5)$ is sufficient to define it there, but still not be continuous. Dec 2 comment True or False question about continuous functions @ChrisEagle Ok, since f is not defined at $f(0.5)$, what else could I use to find something that has an image which is not an interval, but still defined everywhere in $[0,1]$? Maybe it would have something to do with continuity, since the problem does not say "continuous function"? Dec 2 asked True or False question about continuous functions Nov 15 awarded Supporter Oct 24 awarded Teacher Oct 24 answered Find a number $x<100$ that satisfies three congruences. Oct 23 accepted How to prove that an integer exists matching the criteria Oct 23 comment How to prove that an integer exists matching the criteria Thank you, that helps a lot. Because $a-\epsilon < a_n < a+\epsilon$ and $a-\epsilon>0$, we then know that, for all $n>N_\epsilon$ that $0N$ that $a_n>0$! It took me a bit of re-reading your answer to fully get that ha! Thank you. Oct 23 asked How to prove that an integer exists matching the criteria Feb 12 awarded Scholar Feb 12 accepted Order of $ab$ if $a$ and $b$ commute and $\langle a\rangle\cap\langle b\rangle=\{1\}$ Feb 12 awarded Student Feb 12 comment Order of $ab$ if $a$ and $b$ commute and $\langle a\rangle\cap\langle b\rangle=\{1\}$ Thank you, I was looking at the whole problem incorrectly. Thanks to everyones' comments, I see what I actually need to do. $\langle ab\rangle$ would contain $(ab)^1, (ab)^2,$ ... . So, by using the fact that it's commutative, you can get $ab, a^2 b^2,$... . So, if you have $(ab)^k$, if $k$ is a multiple of m, then $a = e$, and if $k$ is a multiple of $n$, then $b = e$, so to get $a = b = e$, $k$ would need to be a multiple of both $a$ and $b$. Thus the order is lcm($m$, $n$). Now all I have left to do is fine a counter-example. Feb 12 comment Order of $ab$ if $a$ and $b$ commute and $\langle a\rangle\cap\langle b\rangle=\{1\}$ Dylan Moreland I see what you mean. $\langle ab\rangle$; would contain $ab$, $(ab)^2$, ... . $(ab)^2 = (ab)(ab) = a^2 b^2$ (because of commutativity). Ok, I think I have a new approach to solving this problem than before, thank you! Feb 12 asked Order of $ab$ if $a$ and $b$ commute and $\langle a\rangle\cap\langle b\rangle=\{1\}$