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 Feb 1 answered Could I do this to an infinite series? Feb 1 comment Could I do this to an infinite series? Typically, we define $\sum_{k=1}^{\infty} x_k$ as $\lim_{n \rightarrow \infty} \sum_{k=1}^n x_k$ if and only if the limit exists, so this doesn't quite work. It would be correct to say that the partial sums are equal to the resulting expression, i.e. $\sum_{k=1}^n k + \sum_{k=1}^n k = \sum_{k=1}^n 2k$, but neither of these sums converge so we can't really talk about their limits. Jan 17 comment Finding at least 2 elements in a set that satisfys an equation For each $n$, temporarily consider the additive group modulo $n$, that is $\mathbb{Z}_n$. By "$-1$", we merely mean the inverse of the element $1$. In $\mathbb{Z}_5$, we have $-1 = 4$, because $1 + 4 = 0$ (and in general, $-1 = n-1$). Now, try out squaring this additive $-1$ in the multiplicative group $U(n)$ (first verify that it's always in there), and show that it works. Jan 17 comment Finding at least 2 elements in a set that satisfys an equation Namely, $1$ and $-1$ are the two elements in $\mathbb{Z}$ under usual multiplication that square to give the identity. So, what are the analogues of $1$ and $-1$ in the finite group $U(n)$? Jan 17 awarded Commentator Jan 17 comment Finding at least 2 elements in a set that satisfys an equation Okay, so the group operation is multiplication mod $n$. Then actually $U(n) = (\mathbb{Z}_n)^\times$ as a group. Got it. Jan 17 comment Finding at least 2 elements in a set that satisfys an equation What group structure are you giving $U(n)$ when we determine if there are two elements $x,y \in U(n)$ that satisfy $x^2 = y^2 = 1$? If we're working in the group $\mathbb{Z}$, the only element that satisfies $x^2 = 1$ is 1. Oct 19 comment Using Pumping Lemma to show a language is not regular This is correct, you could probably be a little more concise on the ending. Just say that the fact that $xyyz$ is not in the language means the language does not satisfy the pumping lemma. So our assumption that the language is regular must be incorrect. It also couldn't hurt to write out $xyyz$ to be more explicit, that is, $xyyz = 0^{p+k}1^p 1^p$ for some $k \geq 1$, noting that $2(p+k) > 2p$. Apr 30 answered When to use the $\equiv$ symbol (such as in $5^{6}$ $\equiv$ 1 mod 7) vs = Apr 25 awarded Popular Question Dec 15 answered To which group is $G/\ker \phi$ isomorphic to? Nov 30 awarded Custodian Nov 30 reviewed Approve Meaning of the word “conjugate” across mathematics? Nov 30 asked Meaning of the word “conjugate” across mathematics? Nov 24 accepted Solutions of $a^{2} - 2b^{2} \equiv 0$ mod $p$ Nov 24 comment Solutions of $a^{2} - 2b^{2} \equiv 0$ mod $p$ Thank you very much! The terminology really helps when researching this! Nov 24 asked Solutions of $a^{2} - 2b^{2} \equiv 0$ mod $p$ Jul 2 awarded Curious Jun 29 awarded Enthusiast Jun 19 awarded Organizer