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8h
comment How many $3$ integer subsets have no consecutive integers, where integers are less than $20$?
The answer to your title is: "infinitely many."
16h
revised Chosen maximal subject is a subgroup
added 43 characters in body
17h
revised Chosen maximal subject is a subgroup
added 22 characters in body
17h
answered Chosen maximal subject is a subgroup
17h
comment Working with an ellipse
A good place to start would be here.
17h
comment Chosen maximal subject is a subgroup
@Derek: Thank you for clarifying that for me.
1d
answered Is there a way to simplify…
1d
revised Inequtions problem - how to calculate total of sales for a determined ROI?
added 481 characters in body
1d
revised Inequtions problem - how to calculate total of sales for a determined ROI?
added 413 characters in body
1d
answered Inequtions problem - how to calculate total of sales for a determined ROI?
1d
comment Is $S_5$ isomorphic with the direct product $A_5 \times Z_2$?
@dREaM: It doesn't, it has one element: $e$. Alex is suggesting that you could use a similar "argument" to claim that $$\{e,e,e,...,e\}$$ (with $5389$ $e$'s) had $5389$ elements, but again, this is false. It has one element: $e$.
1d
revised Finding the adherent points of $A=\left\{\left(1/n,1/m\right)|n,m\in\mathbb{N}\right\}$
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1d
comment Notation: $f(A)$ when $f$ is a function $f:A\to B$.
If you wish to say that $\forall x\in A\bigl(f(x)=c\bigr),$ then you would be better to say $$f(A)=\{c\}.$$
1d
revised Show $R \setminus S$ is a union of prime ideals
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1d
answered Show $R \setminus S$ is a union of prime ideals
1d
comment Prove the limit is $\sqrt{e}$.
+1: Very nice approach!
1d
awarded  Nice Answer
1d
reviewed Reject Prove the limit is $\sqrt{e}$.
1d
answered Prove the limit is $\sqrt{e}$.
1d
answered interpreting $(1-t)f(a)+tf(b)$ from $f((1-t)a+tb)\leq (1-t)f(a)+tf(b)$