dashdart
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 Oct28 comment finding numbers to make both sides equal @bof Sorry! I meant the largest. Silly error! Oct28 comment finding numbers to make both sides equal Either by solving $1000=x(x+1)(x+2)$ or by brute-force, you'll find out that $45$ is the smallest $x$ that works. So, anything below $45$ is fair game. Oct17 comment Solving an exercise in Pinter's Abstract Algebra The set you describe is called the normalizer of $H$ in $G$. Check out proofwiki.org/wiki/Normalizer_is_Subgroup May2 comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $\{1\})$ @user129120 Wow, My bad! For some reason, I thought you were asking for a function $f:\mathbb R \setminus \{1\}\to \mathbb R$. However, Since the function I proposed is a bijection, its inverse is a function from $\mathbb R \to \mathbb R \setminus \{1\}$. Kind of a cheat but still! May1 comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $\{1\})$ Is there a particular reason why you would need a piece-wise function? Then you could use $\ f:x\to \frac 1{1-x}$. Dec12 comment How to figure out the Argument of complex number? @mt_ I am sorry, I don't know how I forgot to mention that. I have edited my answer. I hope it is better than the previous version. Aug26 comment Proving $n^4 + 4 n^2 + 11$ is $16k$ sos440's comment is better justified by noticing that if $n$ is even, then so is $n^4+4n^2$. Hence $n^4+4n^2+11$ has to be odd (since $11$ is an odd number). Aug23 comment Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. So if $2^{3n}$, when divided by $7$ leaves a remainder of $1$, what remainder must $2^{3n}-1$ (which is, as you may have noticed, $1$ less than $2^{3n}$) leave when divided by 7? Aug20 comment Why isn't math on the sine of angles the same as math on the angles in degrees? @tomasz, "The thing is, before applying a „rule”, you should verify if it is actually true." that is exactly my point. I don't know why you disagree :) Aug20 comment Why isn't math on the sine of angles the same as math on the angles in degrees? (+1) for the 'In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.' Aug14 comment Proving that $2^{2^n} + 5$ is always composite by working modulo $3$ First of all, please do note that I am not comparing my answers with others. Now that I read my answer, I realize that I badly phrased my little disclaimer :) What I mean to say is I have included parts like "...this means that any even power of 2 is 1 greater than some multiple of 3..." in my answer which are obvious. So I guess I need to edit/ delete the disclaimer, eh? Aug14 comment Proving that $\mu$ is $\sup S$ (+1) Neat answer! Aug14 comment Proving that $\mu$ is $\sup S$ @PeterTamaroff Done! Aug14 comment Proving that $\mu$ is $\sup S$ My bad! I wrote the definition wrong. I did not realize. Aug14 comment Proving that $\mu$ is $\sup S$ but $1.5$ is not an upper bound Aug14 comment Proving that $\mu$ is $\sup S$ I am sorry, I don't follow Aug14 comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition? oh, okay thanks Aug14 comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition? I'm sorry, but what does $B(x)$ mean in general? Aug14 comment Proving that $\mu$ is $\sup S$ Spot on! :) The question says that $\mu$ is the supremum iff there is an element of $S$ in the interval. But what we just did was started out by assuming that there is no element of $S$ in the interval and proved that if this is the case, then $\mu \ne \sup S$. So, for $\mu = \sup S$, there has to be an element of $S$ in the interval Aug14 comment Proving that $\mu$ is $\sup S$ I could elaborate more if you'd like to