Linked Questions

19
votes
8answers
3k views

Is a number meeting these conditions divisible by forty-nine?

I am not a mathematician, I'm a linguistics PhD student. As part of my research I need to put various convoluted sentences through various syntactic transformations and see then check whether people ...
32
votes
5answers
1k views

Looking to Acquire Intution Regarding the Fundamental Theorem of Arithmetic

I'm sorry ahead if time if this is overly trivial for this site. Currently in school, much of what I enjoy is number theory - based. Currently, I lean pretty heavily on the FTA for a good deal of my ...
8
votes
2answers
34k views

If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$

Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$. This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was ...
0
votes
3answers
344 views

Are there any integer solutions to $a^3=b^2$?

I was wondering if there were any two integers $a$ and $b$ where $a^3=b^2$.
9
votes
1answer
2k views

If $\gcd(a,b)=d$, then $\gcd(ac,bc)=cd$?

$A$ an integral domain, $a,b,c\in A$. If $d$ is a greatest common divisor of $a$ and $b$, is it true that $cd$ is a greatest common divisor of $ca$ and $cb$? I know it is true if $A$ is a UFD, but can'...
11
votes
2answers
324 views

Is my proof that the square root of all imperfect squares are irrational correct?

I was answering a Quora question about whether $\sqrt{13}$ is irrational or not (link if needed), and I tried to prove that, in fact, the square root of all imperfect squares are irrational. This is ...
3
votes
3answers
229 views

Rings where divisors of $mn$ are product of divisors of $m$ and $n$; relation to UFDs

Using the fundamental theorem of arithmetic, it's easy to prove this proposition: Proposition. Every divisor of $mn$ can be written as the product of a divisor of $m$ to a divisor of $n$. My ...
4
votes
2answers
362 views

About a type of integral domains

I was reading the notes about factorization written by Pete L. Clark and I found that he uses the name "EL-domain" to refer to those integral domains in which every irreducible element is prime (page ...
18
votes
1answer
406 views

Prove or disprove that $2^{\frac{\pi}{2}}$ is an irrational number.

Problem Prove or disprove that $2^{\frac{\pi}{2}}$ is an irrational number. My Try According to our mathematical intuition, we may want to apply Gelfond–Schneider theorem, which states that if $\...
5
votes
1answer
213 views

gcd's in non-UFD rings

In a UFD ring we have that for coprime $a,b \in R$, i.e. $(a,b)=1$: $$ a|cb \Rightarrow a|c $$ Does this property hold for non-UFD rings? I think not but do not recall a standard counter-example. ...
2
votes
3answers
195 views

Proof that a prime can't divide a multiplication of two reminders of it

Let $p$ be a prime number. Let $r_1$, $r_2$ be integers such that $1\leq r_1 < p$ and $1 \leq r_2 < p$. How to prove that $p \nmid r_1r_2$? I know one way to prove this. It can be proved by ...
1
vote
0answers
274 views

Interesting or creative applications of Bezout’s (Bachet’s) Identity

Bezout’s Identity (but discovered earlier, in a restricted form, by Bachet) is the following result. Theorem: Let $a$ and $b$ be nonzero integers and let $d$ be their greatest common divisor. Then ...
2
votes
1answer
232 views

Commutative Cancellative Semigroup: When is an irreducible element prime?

Suppose $(S,\cdot)$ is a semigroup with neutral element $e$ (i.e. $xe=ex=x$ for all $x\in S$) and the following properties: S is commutative: $xy=yx$. S is cancellative: $xy = xz$ implies $y = z$. ...
3
votes
2answers
122 views

The largest algebraic structure in which every irreducible is prime

We know that in a UFD every irreducible is prime. But this document rise up in my mind a question that; is there a known algebraic structure (larger than a UFD) in which every irreducible is prime? ...
1
vote
1answer
103 views

How do you prove $\sqrt{n}$ is an integer or it is irrational? [duplicate]

I have tried this problem five times but I keep getting stuck. I keep following the proof for $\sqrt{2}$. I know that I have to prove that the set is nonempty. Which I do by induction. $2^1 > 1$ ...

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