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8h
comment Comma placement inline math
@BrianO The first fragment can be read that way as well, imagine there's an implicit "where" or "with" after $f(x_j)$.
8h
comment Comma placement inline math
@BrianO I believe the OP is saying that you can read the sentence as "Given $f(x_j)$, where $j=1, ... N$, - our goal is to...". In this case the comma is clearly recommended.
1d
comment Can an infinite sum of irrational numbers be rational?
And the linear independence of the terms follows from the transcendence of $\pi$, which is the cleverest part of the answer. In fact any power series with rational coefficients, evaluated at a transcendental value but having a rational sum, would have done. It makes me wonder if there are "super-transcendental" numbers, which are not roots of any power series with rational coefficients.
2d
comment How limiting/ heavy is the “triangle inequality” assumption?
If you drop the triangle inequality, basically all you have left is a symmetric function from $X^2$ into the positive reals. The triangle inequality is responsible for every important property of a metric.
Feb
8
comment Is it accurate to say that multiplication of two integers yields an integer?
@user7530 The important issue that needs to be dealt with here is that the OP thinks they can trust a computer's floating point circuits to tell them about math. This will just be over their head.
Feb
8
comment Is it accurate to say that multiplication of two integers yields an integer?
I just put $\sqrt 5^2$ into my computer and got $5.000000001$. Computers only do arithmetic approximately.
Feb
6
comment Squaring both sides when units are different?
@TonyK But what does $\sqrt{9 \text{inches}}$ mean? You can't take the square root of an inch any more than you can take the square root of a tomato. The point I was making in my second comment is that your kinematics example has the units outside of the square root, because having them inside is meaningless.
Feb
6
comment Squaring both sides when units are different?
@TonyK But anyway, with that interpretation the statement in the OP isn't correct - it's not true that $\sqrt 9$ inches equals $\sqrt{0.25}$ yards.
Feb
6
comment Squaring both sides when units are different?
@TonyK Right, but $x$ here refers to the amount of inches, not the inches themselves, so it's still $\sqrt{2x}\ \text{inches}$, not $\sqrt{2(x\text { inches})}$.
Feb
6
comment Squaring both sides when units are different?
Are you sure it's not $9^{1/2}\text{inches}=0.25^{1/2}\text{yards}$? Because I don't know what it would mean to take the square root of of an actual physical inch.
Feb
5
comment Is there a solution for this problem ??
I think you have these wrong. First of all there's no number (other than 0) which increases by one when you put an extra digit in front of it. Second of all, conditions 1 and 3 together imply that the number must be 9876543210 (well, maybe 0123456789 if you allow leading zeros).
Feb
5
comment How did the rule of addition come to be and why does it give the correct answer when compared empirically?
It seems to me that the way you've phrased the question shows a little confusion. You ask "how" this "came to be", as if it's an arbitrary rule that someone had to invent. 25 + 19 is 44 because if you have 25 coins in one pocket and 19 in the other, then if you tally up the total you find you have 44. What you should be asking is, why is it that "carrying the one" always gives you the correct answer.
Feb
2
comment Law of Excluded Middle Controversy
At its heart, constructivism is about what it means for a mathematical object to exist. A materialist might already object that mathematical objects don't "exist" - they're just thoughts. At a stretch, you might say they exist because you can imagine them, that is, construct them. And now you come along with an object that not only is purely imaginary, but you can't even describe it or construct it in any way. What sort of "existence" is that? If I understand correctly, excluding the LEM makes non-constructive existence proofs impossible. Someone else will have to explain why.
Feb
2
comment Classification of homomorphisms $\mathbb Q \to \mathbb C^\times$
Does the situation change drastically if the injective condition is dropped?
Feb
1
comment Is this iteration involving primes known?
You have the iteration $$x_{n+1} = x_n + p_n + 1$$ Where $(p_n)$ is some sequence of primes (this is what you call $y$ in your question). Therefore $$x_n = x_0 + n + \sum_{k=0}^n p_n$$ You're asking if for any/for all prime $x_0$, we can find a sequence of primes $(p_n)$ such that every $x_n$ is prime.
Jan
26
comment Find depth of three node tree
Does every node of the tree have three children?
Jan
25
comment How to find irrational numbers between rationals. (And is my method correct?)
Yes, your method is correct. Basically, given $a<b$, we know that if $a^2<r<b^2$, then $a<\sqrt r < b$, so the problem is to show that there is always at least one $r$ which is not a rational square, which you've done in probably the simplest way possible. I prefer your construction to the ones in the answers, which are only "simpler" if you don't think about the amount of machinery that has to go into defining what it means to multiply irrationals by rationals, what it means for one real number to be greater than another, etc. Your method is simpler and more direct.
Jan
22
comment Distributive Law and how it works
Every question of the form "Why X?" should be answered by "Why do you expect not-X?". Why should it be the case that the relationship between addition and multiplication is symmetric in this way? Are you saying that just because A bears relationship R to B, that therefore it's reasonable to expect that B should bear relationship R to A? Do you find it surprising that cats eat mice, and yet mice don't eat cats?
Jan
22
comment ODE Maximal solution
It indeed somewhat trivial, which is why the proofs of this theorem are never all that complicated, conceptually. Formally they may require a bit of set-theoretic machinery, but that's to be expected when you're working with infinite sets of functions like this. These proofs by Zorn's Lemma or what have you are just giving names to your intuitions about why this is "obviously" true. Similarly, it seems "trivial" that from an infinite collection of sets you can construct a set which chooses one element from each set, but the formal details can lead you down a very deep rabbit hole.
Jan
15
comment convergence of continued nested function
@user55622 That isn't a proof of convergence, though.