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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
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Jun
29
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Jun
19
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