# Tagged Questions

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### Proof that doesn't exists a rational $s$ such that $s^2 = 6$

Well, I solved it, and I would like to know if there is anything that can be corrected or improved here. I think that the proof ended up too long, and with too many letters. Surely there is a better ...
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### Alternative Proof of Infinitely Many Primes? [duplicate]

I've seen Euclid's proof of infinitely many primes, what are other approaches to proving there are infinitely many primes?
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### Prove $\log_a(b)$ is irrational given that $a, b$ are positive distinct primes.

I know this is a classical proof by contradiction exercise, and there are full solutions else where, doing a quick search I didn't find any, but I would approach this question like this: Suppose ...
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### If $p$ is irreducible and $p \not \mid a$, then $\text{gcd}(p,a)=\pm 1$.

I will be taking a Rings and Fields course in the Fall, so I figured I would read ahead in the textbook (A First Course in Abstract Algebra, by Anderson and Feil) to prepare. Recall the following ...
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### Why are only fractions with denominator 2 and 5 non-repeating?

Given a rational number $\frac{n}{d}$, I understand that in the base $10$ number system, the number can be represented as a non-repeating decimal number if and only if $d$ has only prime factors of ...
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### Combinatorial proof of the fact $p$ doesn't divide $n \choose p^k$

Let $p^k | n$ and $p^{k+1} \nmid n$. Is there any combinatorial proof of the fact that $p \nmid {n \choose p^{k}}$ ?
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### $x^2\mid27 \implies x\mid9$ : Prove

$x$ is given as a natural number. I was trying this by direct proof: assume $27\mid x^2$, then $x^2 = 27m \Longrightarrow x=3\cdot \sqrt{3m} \Longrightarrow \sqrt{3m}$ must be integer ...
### Proof that $1^3+2^3+\cdots +n^3 = (1+2+\cdots+n)^2$ without using induction [duplicate]
Possible Duplicate: Intuitive explanation for the identity $\sum\limits_{k=1}^n {k^3} = \left(\sum\limits_{k=1}^n k\right)^2$ How to prove this without using mathematical induction? ...
### If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$.
In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that: If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$? I'm not familiar with these kind of calculations, so I'd like ...