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Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
Indeed, there are only counterexamples to the claim: There are no real numbers $a$, $b$ such that $\log_ab=-\log_ba$.
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@Arthur: But $1/x\ne -x$
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@corciacandy: No, $b^{-x} = (a^x)^{-x} = a^{-x^2} \ne a$ unless $x=0$ and $a=1$.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Never. See joois answer for a more general proof.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
BTW, I think generalization to non-positive functions could be done by simply splitting the domain into the parts where $f>0$ and $f<0$ (parts where $f=0$ obviously don't contribute, since they are subset of the $x$ axis which is of measure zero). Obviously $f$ and $-f$ have the same measure.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Indeed, it is. +1
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
@Ali: Actually, the bounded domain cannot be absolutely general; if the domain is a non-measurable set, I'm sure the graph on that domain will also be unmeasurable; however if it is unmeasurable, it certainly will not have a measure greater than zero either.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
@Ali: Well, the proof uses that the sine is bounded, so you can only use it on bounded functions. For unbounded functions (like $1/x$) you'll need other arguments. I think the best way to prove it for the general bounded domain would be to decompose the open interval into countably many closed intervals and use the $\sigma$-additivity of the Lebesgue measure.
Apr
19
answered Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Apr
19
comment Error in my proof?
@PVanchinathan: Yes, that's a simple proof that the sequence converges. But the answer is missing such a proof.
Apr
18
comment Error in my proof?
That argumentation only works if you already know that the sequence converges. Otherwise you'll find that $1+2+4+8+\ldots=-1$ because it obviously fulfils the equation $2x+1=x$, and $-1$ is the only solution to that equation.
Apr
18
answered Number Theory - Multiple of $36$ problem
Apr
18
answered Mathematical induction
Apr
17
answered How do pocket calculators calculate exponents?
Apr
17
comment Largest eigenvalues of AA' and A'A
Are you assuming that $A$ is a real matrix?
Apr
17
answered Need help with proof of existence of $\sqrt{2}$
Apr
17
comment Is $2^{\aleph_0} = \aleph_1$?
math.stackexchange.com/q/178069/34930
Apr
17
answered Is $2^{\aleph_0} = \aleph_1$?
Apr
17
comment Metric spaces are completely normal
@bot: Of course I meant $M=Y, N=Z$. However I indeed genuinely went wrong in constructing your $M$ and $N$.
Apr
17
comment Total Ordering on the Rationals
You have to also prove that $\le$ is a partial order to begin with. That is, you have to prove transitivity and antisymmetry. Reflexivity is already implied by what you've proven.