1,429 reputation
726
bio website biditacharya.wordpress.com
location Berkeley, CA
age 19
visits member for 3 years, 2 months
seen Aug 18 at 8:01

Freshman at UC Berkeley. Originally from Nepal

The scientist does not study nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. - Henri Poincaré.


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May
2
comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $ \{1\})$
@user129120 Wow, My bad! For some reason, I thought you were asking for a function $f:\mathbb R \setminus \{1\}\to \mathbb R$. However, Since the function I proposed is a bijection, its inverse is a function from $\mathbb R \to \mathbb R \setminus \{1\}$. Kind of a cheat but still!
May
1
comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $ \{1\})$
Is there a particular reason why you would need a piece-wise function? Then you could use $ \ f:x\to \frac 1{1-x}$.
Dec
12
comment How to figure out the Argument of complex number?
@mt_ I am sorry, I don't know how I forgot to mention that. I have edited my answer. I hope it is better than the previous version.
Aug
26
comment Proving $n^4 + 4 n^2 + 11$ is $16k$
sos440's comment is better justified by noticing that if $n$ is even, then so is $n^4+4n^2$. Hence $n^4+4n^2+11$ has to be odd (since $11$ is an odd number).
Aug
23
comment Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.
So if $2^{3n}$, when divided by $7$ leaves a remainder of $1$, what remainder must $2^{3n}-1$ (which is, as you may have noticed, $1$ less than $2^{3n}$) leave when divided by 7?
Aug
20
comment Why isn't math on the sine of angles the same as math on the angles in degrees?
@tomasz, "The thing is, before applying a „rule”, you should verify if it is actually true." that is exactly my point. I don't know why you disagree :)
Aug
20
comment Why isn't math on the sine of angles the same as math on the angles in degrees?
(+1) for the 'In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.'
Aug
14
comment Proving that $2^{2^n} + 5$ is always composite by working modulo $3$
First of all, please do note that I am not comparing my answers with others. Now that I read my answer, I realize that I badly phrased my little disclaimer :) What I mean to say is I have included parts like "...this means that any even power of 2 is 1 greater than some multiple of 3..." in my answer which are obvious. So I guess I need to edit/ delete the disclaimer, eh?
Aug
14
comment Proving that $\mu$ is $\sup S$
(+1) Neat answer!
Aug
14
comment Proving that $\mu$ is $\sup S$
@PeterTamaroff Done!
Aug
14
comment Proving that $\mu$ is $\sup S$
My bad! I wrote the definition wrong. I did not realize.
Aug
14
comment Proving that $\mu$ is $\sup S$
but $1.5$ is not an upper bound
Aug
14
comment Proving that $\mu$ is $\sup S$
I am sorry, I don't follow
Aug
14
comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition?
oh, okay thanks
Aug
14
comment Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition?
I'm sorry, but what does $B(x)$ mean in general?
Aug
14
comment Proving that $\mu$ is $\sup S$
Spot on! :) The question says that $\mu$ is the supremum iff there is an element of $S$ in the interval. But what we just did was started out by assuming that there is no element of $S$ in the interval and proved that if this is the case, then $\mu \ne \sup S$. So, for $\mu = \sup S$, there has to be an element of $S$ in the interval
Aug
14
comment Proving that $\mu$ is $\sup S$
I could elaborate more if you'd like to
Aug
14
comment Are all infinities equal?
I do not mean to do any self-marketing but if you want to learn about this from the beginning, try out this blog post that was recently wrote by me wp.me/p2aEXv-2N . I am writing a follow up article to this and it will be out very soon
Aug
14
comment Proving that $\mu$ is $\sup S$
$\lambda \not \in S$ means that $\lambda$ is a upper bound for $S$ which is less than $\mu$. If you remember, one of the properties of the least upper bound is that if there exists a quantity that is less than the least upper bound (say $l$), that quantity has to be in the set under consideration. Or else, $l$ can't be the least upper bound
Aug
14
comment Proving that $\mu$ is $\sup S$
Assume that for some $\epsilon >0$, $\not{\exists} x \in [\mu -\epsilon , \mu]$, such that $x\in S$. This implies that $\exists \lambda < \mu$ where $\lambda \not \in S \implies \mu \ne \sup S$. So if $\mu = \sup S$ then has to be an element of S in the interval $[μ−ϵ,μ]$