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May
28
comment Prove by combinatorial method that $ \frac{(2m)! \cdot (2n)!}{(m)! \cdot (n)! \cdot (m+n)!} $ is an integer
What you describe at the end is not really making full use of the situation, because the separate factors $(2m)!/m!^2$, $(2n)!/n!^2$, and $(2m)!/((m-n)!(m+n)!)$ if $m \geq n$ (or analogous expression if $n \geq m$) are all integers.
May
28
comment A finite group which has a unique subgroup of order $d$ for each $d\mid n$.
How much group theory do you know already? And is this a homework problem?
May
28
revised A finite group which has a unique subgroup of order $d$ for each $d\mid n$.
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May
27
comment The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$
Certainly. Otherwise it is false that $|u/v - 1|_p = |u-v|_p$; they differ by the factor $|v|_p$. That equation is symmetric in $u$ and $v$ precisely when they both have absolute value $1$. Without that equality, you can't say the ratio $u/v$ is as close to $1$ as the difference $u-v$ is to $0$, which is the reason that we can think of the multiplicative group $(1+p\mathbf Z_p)/(1 + p^n\mathbf Z_p)$ as a subset of the additive group $\mathbf Z_p/p^n\mathbf Z_p$ for the purpose of counting its size.
May
27
revised What are statements about the natural numbers where induction is impossible or unnecessary to prove?
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May
27
comment What branch of Mathematics does the study of Algebraic/Transcendental Numbers lie in?
@Trogdor, there is indeed a subject within number theory called "transcendental number theory." It is concerned with transcendence of special values of functions like exponential, elliptic, or modular functions.
May
27
comment “Rationalizing the denominator” of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$?
Yes, but to make it explicit is a mess. Do you understand why ${\mathbf Q}(\sqrt[3]{2}) \cong {\mathbf Q}[x]/(x^3-2)$? If so, you're asking to invert $a + bx + cx^2$ in ${\mathbf Q}[x]/(x^3-2)$, and that is a linear algebra problem (inverting a $3 \times 3$ matrix) by turning $(a + bx + cx^2)(A+Bx+Cx^2) \equiv 1 \bmod x^3-2$ into $3$ linear equations in $3$ unknowns $A$, $B$, and $C$.
May
27
revised Basic question about finite fields and characteristic
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May
26
revised What are statements about the natural numbers where induction is impossible or unnecessary to prove?
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May
26
answered What are statements about the natural numbers where induction is impossible or unnecessary to prove?
May
26
comment What are statements about the natural numbers where induction is impossible or unnecessary to prove?
I think it's more important to make students realize the truly wide scope of induction! It can be used not just to prove theorems about positive (or nonnegative) integers, but also theorems about polynomials by induction on the degree or about vector spaces by induction on the dimension or about square matrices by induction on the size of the matrix (related to the second example), and so on. Also, some theorems really require a stronger form of induction: that every integer greater than $1$ has a prime factorization doesn't get proved by ordinary induction but can by strong induction.
May
26
comment What are statements about the natural numbers where induction is impossible or unnecessary to prove?
artificial and uninspiring when ripped from the context where they naturally arise, so it's hard to see that anyone would really find those examples compelling if they're a beginner. (I am thinking of violations of the Hasse principle, for instance.)
May
26
comment What are statements about the natural numbers where induction is impossible or unnecessary to prove?
Why do you consider this important? What I mean is, if you actually found students in your class trying to use induction in a setting where they shouldn't then you would have a real example from your teaching to provide in future courses. And if you don't really find students attempting induction in a setting where it doesn't seem viable then the whole thing is a non-issue. Induction is a very useful idea, so why try to discourage it so soon after introducing it? I can think of lots of examples in number theory that don't proceed by induction on $n$, but the statements will sound [contd.]
May
26
comment What are statements about the natural numbers where induction is impossible or unnecessary to prove?
Yes, but I don't see how this will convince students that there are some results about positive integers for which induction is not a viable option (as far as we can tell, of course) since it is quite accessible to a beginner's level of understanding.
May
25
comment What are statements about the natural numbers where induction is impossible or unnecessary to prove?
Why would you consider induction not easy for that summation identity? It is the example used to illustrate induction in any textbook I have read that discusses induction.
May
25
comment If I know the order of every element in a group, do I know the group?
@DerekHolt, look at arxiv.org/pdf/math/9210219.pdf.
May
24
comment Let $X$ be a set of primes $p$ so that $5^{p^2}+1 \equiv 0 \pmod {p^2}$ Which of these sets is $X$ equal to?
Fermat's little theorem tells us for every prime $p$ and integer $a$ that $a^p \equiv a \bmod p$. For all primes $p$, if $x \equiv y \bmod p$ then $x^p \equiv y^p \bmod p^2$. So from $5^p \equiv 5 \bmod p$ we get $5^{p^2} \equiv 5^p \bmod p^2$. Therefore the hypothesis of the question implies $-1 \equiv 5^p \bmod p^2$. Reducing both sides mod $p$, $-1 \equiv 5 \bmod p$, so $p$ divides 6. Hence $p$ is $2$ or $3$. But $p = 2$ is impossible since for any odd number $x$ we have $x^2 \equiv 1 \bmod 4$. You can check directly that $p = 3$ works, so the only choice is $\{3\}$.
May
24
comment Structure of the group $\{1+p\mathbb Z_p \}$
What Serre is essentially saying is that $1 + p$ (for $p \not= 2$) or $5 = 1 + 4$ (for $p = 2$) are topological generators of $U_1$ (for $p \not= 2$) or $U_2$ (for $p = 2$). That is, the homomorphism $f \colon \mathbf Z \rightarrow U_1$ where $f(n) = (1+p)^n$ for $p \not= 2$ is $p$-adically continuous and it extends by (uniform) continuity to an isomorphism $\mathbf Z_p \rightarrow U_1$ (similar result for $f(n) = 5^n$ when $p = 2$). If you understand inverse limits then the commutative diagram says the projective system on the right can be identified with the projective system on the left.
May
24
comment Structure of the group $\{1+p\mathbb Z_p \}$
And I too took a course with that book. I stand by what I wrote before that trying to make sense of the isomorphism using the $p$-adic exponential and logarithm is more intuitive. I didn't understand that proof when I first saw it and I was able to handle the rest of the book.
May
24
comment Smallest field containing $F$ and $a \in K$
God no, just try an actual example. Take $F = \mathbf Q$, $K = \mathbf R$, and $a = \sqrt{2}$.