apologize in advance if this is a duplicate, but I found a lot questions related to this but none answering this specific question.  My logic is: let's consider division the opposite of multiplication. So if I have   x / y = z. then that means that  z * y = x  i understand that when y = 0 there can be so such z that would make the equation true for x if x is not zero. But if both x AND y = zero, then there is literally an infinite amount of numbers that can be z. e.g  1 * 0 = 0  2 * 0 = 0  3 * 0 = 0  4 * 0 = 0  So again, why is this undefined instead of infinity.   (s/n: after typing out this question i realize it may be my misconception of the true definition of infinity but alas I'm deciding to post this anyway)
34511
I came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.   $\dfrac {x}{0}$ is Impossible (  If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both $1.$ and $2.$ are divided by zero, but only $2. $ has infinite solutions so as $1.$ has none solution, how and why ?) $\dfrac {0}{0}$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?) $\dfrac {0}{x}$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.
19743
I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?   In short, what is the benefit or reason to define a function this way?  At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).  Thanks,  Moshe
54709
Suppose we have $x^2+4x+3=0$  We differentiate both sides to get $2x+4=0$.  So $x=-2$  But $(-2)^2+4(-2)+3=0$   $4-8+3=-1$ is not equal to $0$.  Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy?
82407
Suppose we have an equation $ab=0$. This equation is true when statements $a=0$ or $b=0$ are true.  If $a=0$, then $b=\frac{0}{0}$. That means $b$ could be any number for $ab=0$ to be true. If the set which groups all the numbers is the complex set, then $b$ will every number within $\mathbb{C}$, so $\forall z\in\mathbb{C}:b=\frac{0}{0}=z$. Therefore, $\frac{0}{0}$ is every number.  I know it really is not defined as number but conceptually it is every number, right?  Is this right, or am I missing something?
30704
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &amp;=&amp; xy \nonumber \\ x^2 - y^2 &amp;=&amp; xy - y^2 \nonumber \\ (x + y)(x - y) &amp;=&amp; y(x-y) \nonumber \\ x + y &amp;=&amp; y \nonumber \\ 2y &amp;=&amp; y \nonumber \\ 2 &amp;=&amp; 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &amp;=&amp; y \\ x &amp;=&amp; y - y \\ x &amp;=&amp; y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close?
72872
Is the fraction  $$\frac{1}{\frac{1}{0}}$$  undefined?   I know that division by zero is usually prohibited, but since dividing a number by a fraction yields the same result as multiplying the number by the fraction's reciprocal, you could argue that  $$\frac{1}{\frac{1}{0}} = (1)\left(\frac{0}{1}\right) = 0$$  Is that manipulation permissible in this case? Why or why not?
11548
Why does $x$ divided by zero not equal $x$? After all, $x$ is not being divided by anything.
24209
Suppose you have a ring $(C[0,1],+,\cdot,0,1)$ of continuous real valued functions on $[0,1]$, with addition defined as $(f+g)(x)=f(x)+g(x)$ and multiplication defined as $(fg)(x)=f(x)g(x)$. I'm curious what the zero divisors are.  My hunch is that the zero divisors are precisely the functions whose zero set contains an open interval. My thinking is that if $f$ is a function which is at least zero on an open interval $(a,b)$, then there exists some function which is nonzero on $(a,b)$, but zero everywhere else on $[0,1]\setminus(a,b)$. Conversely, if $f$ is not zero on any open interval, then every zero is isolated in a sense. But if $fg=0$ for some $g$, then $g$ is zero everywhere except these isolated points, but continuity would imply that it is also zero at the zeros of $f$, but then $g=0$, so $f$ is not a zero divisor.   I have a hard time stating this formally though, since I'm only studying algebra, and not analysis. Is this intuition correct, and if so, how could it be rigorously expressed?
24447
I know I'm going to use Liouville's Theorem, but my main question is why is $1/|f(z)|$ entire as well if $f$ is entire? Is this just a basic property: if $f$ is entire, then $1/f$ is entire? Thanks for the help.
84382
I know the answer is obvious: In $\mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$. My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?  Edit: I would like to prove it viewing $\mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...
84369
I do not even know if the question makes sense.  The point is rather simply. If I have the sum of all natural numbers,  $$\sum_{n\in \mathbb{N}}n$$  this is clearly "equal to infinity".  But since almost a century ago, we know that there are (a lot of) different "infinities".  So, is this sum equal to something countable or something bigger?  I tried to look for references, but couldn't find anything and, since I am not an expert in Logic, Set Theory or Foundations of Mathematics, I thought that it would be good to ask here.  Thanks in advance, Davide  PS: This question is about sum of cardinals. It is not about Calculus.
118333
I have a really hard time learning math when I can't find any justifications for the existence of infinity (in this case actual infinity).  Are there any justifications for the Axiom of Infinity?   Thank you very much.
91536
I think of the number zero as a whole number.  It can certainly be a ratio = $\frac{0}{x}, x \neq 0.$  Therefore it is rational.   But any ratio equaling zero involves zero, or is irrational, e.g.$\frac{x}{\infty}, x \neq 0$ is not a ratio of integers.  Can a rational number that is rational only when the number itself is involved still be rational?  I realize that it "should" be rational but it doesn't seem to fit into the same category with other rational numbers. (I'm glad mse has a soft question category for questions like this...).
45167
My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.  On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero.   I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms.     Note that this question arose after reading  and its first comment.
2753
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, so it seems like it should be split pretty evenly.
117703
If I have $10$ apples, but you have $5$ apples, then I have $2$ times more apples than you. But what if I have $10$ apples, but you don't have any apples?   If you look at the graph $f(x)=\frac{10}{x}$, it shows that when $x$ approaches $0$, $f(x)$ approaches infinity. So it means I have infinity times more apples than you?
76183
Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.  I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?  So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?