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Oct
18
comment Prove that $\mathcal B(\mathbb R)\times \mathcal B(\mathbb R)\subseteq \mathcal B (\mathbb R^2)$
@saz Yes.${{}}$
Oct
18
asked Prove that $\mathcal B(\mathbb R)\times \mathcal B(\mathbb R)\subseteq \mathcal B (\mathbb R^2)$
Oct
18
revised Is there an introductory book on Genetic Sequencing Theory for mathematicians?
edited tags
Oct
18
answered Why isn't area under curve from 0 to infinity of $\frac{1}{x^2}$ equal to 3?
Oct
17
answered “Introducing something extra”
Oct
13
comment What is a “formal definition” of a set?
Could you provide more context? What is the exact formulation of the question?
Oct
12
answered Adjusting weight of a body of water by substituting part of it with a lighter liquid
Oct
10
accepted Is there a geometrical definition of a tangent line?
Oct
8
comment Is there a geometrical definition of a tangent line?
Could you clarify what you mean by "one direction" and "other direction"?
Oct
5
asked Is there a geometrical definition of a tangent line?
Oct
4
reviewed Approve Prove all multiples of $U$ contain all the digits $0$ to $9$
Oct
4
revised Prove all multiples of $U$ contain all the digits $0$ to $9$
edited title
Oct
3
comment Is $x^{1-\frac{1}{n}}+ (1-x)^{1-\frac{1}{n}}$ always irrational?
@Slade You're right - this started out as a proof that there are no counterexamples for $n>2$ of that specific type, and then I forgot what I was trying to do. There may be counterexamples of a more general form.
Oct
3
answered Is $x^{1-\frac{1}{n}}+ (1-x)^{1-\frac{1}{n}}$ always irrational?
Oct
3
comment Is $x^{1-\frac{1}{n}}+ (1-x)^{1-\frac{1}{n}}$ always irrational?
@RossMillikan $(a+b)^n\neq a^n+a^b$
Sep
30
awarded  Explainer
Sep
30
comment Prove (or disprove) this $\sum_{n=1}^{\infty}\dfrac{a_{n}}{n}$ is convergence?
Can't you just let $a=1$?
Sep
25
comment What is the meaning of “true”?
The answers here may be helpful.
Sep
25
comment Really advanced techniques of integration (definite or indefinite)
@NikosM. Monte Carlo does numerical integration, not symbolic.
Sep
23
comment Hard-to-put-together but easy-to-prove results
Maybe something like ${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}$? Once you think of the argument, it's simple and obvious, but thinking of the argument isn't entirely obvious if you're new to combinatorics. I still think this question is too vague, though.