Let that   $\frac{1}{3}=0.333...$, then multiply by $3$:  $\frac{1}{3}\times 3=1\tag1$  $0.333...\times3=0.999...\tag2$  $1-0.999...=0.000...0001$  Here it seems that $1 \ne 0.999...$  So what is the correct view concerning $1$ and $0.999...$  Are they equal or not ?
1055
I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?
69977
$$\frac{1}{3}=.33\bar{3}$$    is a rational number, but the $3$ keeps on repeating indefinitely (infinitely?). How is this a ratio if it shows this continuous pattern instead of being a finite ratio?    I understand that $\pi$ is irrational because it extends infinitely without repetition, but I am confused about what makes $1/3=.3333\bar{3}$ rational. It is clearly repeating, but when you apply it to a number, the answers are different: $.33$ and $.3333$ are part of the same concept, $1/3$, yet:  $.33$ and $.3333$ are different numbers:    $.33/2=.165$ and $.3333/2=.16665$, yet they are both part of $1/3$.   How is $1/3=.33\bar{3}$ rational?
84743
$$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$ The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
121303
Solving $\frac{1}{x} &lt; 4$ gives me $x &gt; \frac{1}{4}$. The book however states the answer is: $x &lt; 0$ or $x &gt; \frac{1}{4}$.  My questions are:  Why does this inequality has two answers (preferably the intuition behind it)?  When using Wolfram Alpha it gives me two answers, but when using $1 &lt; 4x$ it only gives me one answer. Aren't the two forms equivalent?
5370
I was surprised when I heard zero is plural, and even 1.00001 is plural.  Then, what about following numbers ?  (1) 1.00  which means the value measured  between 0.995-1.004, having possibility that it was exactly 1.  (2) 0.99999.....  which is mathematically equal to 1. Also, are others that are mathemattically equal to 1 than this, treated completely same as 1?  (3) -1  Is this singular? Or negative numbers are generally plural?  (4) i = square root of -1  I know it's weird to count things with imaginary numbers, nonetheless, if you were to count, which form, plural or singular, would you be to use?  That's it.
112232
If percentage points can equal decimal points, then $1.25$ should equal $125$% when multiplying $1.25$ by any number.  Therefore $125$% of $0.75$ should equal $1$ if $0.75$ is also $75$% of $1$.   Am I missing something super obvious here or what?!  Please help. Thank you!
114767
First I have checked the search option but found nothing relevant to my problem and also level of math. I just started learning the language of mathematics, on my own and I have trouble understanding why  $  \frac{1}{\infty } \approx 0 $ and  why $\frac{1}{0} = {\infty}$?  I want to know how exactly do I get to these answers. It might sound strange but it's hard for me as a beginner to picture things like this on my own.
68549
Let $z, w$ be distinct complex numbers. Show that if $|z| = 1$ or $|w| = 1$, then $$\frac{|z − w|}{|1 − z^*w|} = 1$$ [Hint: Note that $|a|^2 = aa^*$.]  Hey guys, couldn't get my thinking cap on for this question. Any helpful input? Will appreciate it!
36572
I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:  $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$  I would appreciate your help.
26202
I just prove $$S_1=\{0.4,0.8\}$$ $$S_2=\{0.04,0.08,0.44,0.48,0.84,0.88\}$$ ... So I can calculate the number of elements in $S_n = 2 \times 3^{n-1}$  I just prove it is countably infinite.
53559
I'm having problem with showing that: $$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$  I would need some help in the right direction
36423
Is $\frac{1}{\sqrt{2}}$ a fraction or not?   I just came across a Grade 8 student's class work where the teacher has given the above as an example to differentiate fractions from rational numbers. I think this is wrong fundamentally, as $\sqrt2$ is irrational while all fractions would be rational. Please advise. Thanks!
2500
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?  I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
105485
The common difference is d  Multiplying and dividing by d  $$\frac 1d [\frac{3-1}{\sqrt 3 + \sqrt 1} + \frac{5-3}{\sqrt 5 + \sqrt 3}....]$$ $$=\frac 1d [\sqrt 3 - \sqrt 1 + \sqrt 5 - \sqrt 3........\sqrt {2n-1} - \sqrt {2n-2}]$$ $$\frac 1d [\sqrt {2n-1} - \sqrt 1]$$  Since n term will be $$a_n=1+(n-1)2$$ $$=2n-1$$  Final answer  $$\frac 12 [\sqrt{2n-1}-1]$$  This isn’t the right answer however. What am I doing wrong?
113978
I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
19798
Are the floor functions of $0.999\cdots$ and 1 equal?  It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ?
56179
Problem :  Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.  My approach : Since : $(a^3+b^3)=(a+b)^3-3ab(a+b)$ $\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$ Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.