A simple question but I know two conflicting rules on this:   Multiplication is stronger that dividing: $8:4\times2=8:8=1$ Dividing is the same as multiplication with the inverse: $8:4\times2=8\times4^{-1}\times2=4$   Which rule should rule?  Sorry if it is a duplicate.
8308
There is a huge debate on the internet on the value of $48\div2(9+3)$. I believe the answer $2$ as I believe it is part of the bracket operation in BEDMAS.  yields the same answer. I also believe that if $48\div2\times(9+3)$ was asked it would be $288$ which Mathway agrees with as well. However,  says it is $288$ either way. A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason) What is the answer and why?
32876
I'm currently tutor in an undergraduate course and the students are asked to find how addition and multiplication behave on the field with 4 elements: $\{0,1,x,y\}$. In the solution the teacher gave me, he uses the fact that $(\mathbb{F}_4^*, \cdot )$ is cyclic with order 3 to find that $x$ and $y$ are both generators and that $(\mathbb{F}_4,+)$ is a group of order 4 to deduce that $0=1+1+1+1=(1+1)(1+1)$ and hence, $(1+1)=0$ since a field is integral.   However, I'm not satisfied with this reasoning because this question is asked to student in first year and they juste have been introduced to fields (and this was part of an analysis course so they won't see any result soon). So I tried to find how it behaves by myself, using as few results as possible.  Multiplication is the easy part, $0\cdot a$ and $1 \cdot a$ is trivial for all $a\in \mathbb{F}_4$. Then, $x\cdot y\not = 0$ because it is invertible, and it can't be equal to $x$ or $y$ because that would imply the other element to be equal to $1$. Hence, $x\cdot y=1$. We are left to find what is $x^2$ (and similarly, $y^2$). $x^2$ cannot be $0$, nor $x$. If $x^2=1$, then $y=(x^2)y=x(xy)=x\cdot 1=x$. Contradiction. Hence $x^2=y$ and by the same reasoning, we have $y^2=x$.  It was easy with the multiplication, but I can't figure out any non-trivial result concerning addition. My intuition tells me that distributivity should be used here but I'm feeling like it's just converting a problem into another one because we don't know anything about addition.  Does anyone here have an idea of how addition can be "found" without using any advanced result (or as few results as possible)?  Thanks in advance.
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.
44996
i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much!
1132
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$?  Clearly $2^n - 1$ is an odd integer, therefore, let $n$ be an odd integer and it divides $2^n - 1$. We can write $2^n - 1 = (2-1)(2^{n-1} + \cdots \cdots + 1) = (2^{n-1} + \cdots \cdots + 1)$ From here on I don't know how to proceed?
65025
In group theory it happens very often that you want to say: "The order of $H$ divides / does not divide the order of $G$." But the length of | is then always the same:  $ |H| | |G| $ produces |H|||G|  which is not ideal at all. (The negation \nmid has just this length too.) The length of "divides" should be longer than the "cardinality length". I could use   $ |H| \big| |G|$  but then there is no good way to say "does not divide" with the same length. Also, \big| is no solution when the expression has to be used in subscripts.   I thought it was just me, but I asked other group theorists, and nobody had a solution. I hope there is one.  I am seeking a division sign which is a little longer than the usual | , allows negation (with the same length) and proportionally correct use in subscripts (also under \sum etc.). It does not look professional if the "order sign" has the same length as the "divides" sign. (And what about combinations with absolute values, norms - although personally I don't use them as often as group orders.)
41484
I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$
66635
Let $r$ be an integer greater than $2$.  Is there a simple way of showing that $2^r$ divides  $\left(\begin{array}{c}  {2}^{r-2} \\ k \end{array}\right) 2^{2k}$  but it does not divide $\left(\begin{array}{c} 2^{r-l} \\ k \end{array}\right) 2^{2k}$ for $l&gt;2$ and for all $k$ greater than $0$ and less or equal than $2^{r-2}$?  Equivalently, is there a simple way of showing that the element $5$ has order $2^{r-2}$ in the group of units of the ring $\mathbb{Z}_{2^r}$?  The two questions are connected by writing $5 = 4 + 1$ and using the binomial expansion formula to study the order of $5$.
123037
It is well known that $(n!)^2$ divides $(2n)!$.  Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$?  If yes or no, could you provide the details behind the argument?  As a quick test, I tried $n=3$ and $n=4$ which worked.  Thanks,  -Larry
12243
I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!
7610
Why multiplying fractions is equal to multiply the tops, multiply the bottoms? $$\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},$$ And why $$\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},$$ Also why $$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$ I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference.
123358
How to logically understand the multiplication of two integers ?  Eg:  3 x 4 = 12 (is understandable)  -3 x4 = -12 (is also somewhat understandable)  But , 3 x -4 = -12 (is NOT understandable)  -3 x -4 = 12 (is also NOT understandable)  Now , at this we must assume that commutative property may not be true .  What is the logical explanation for something being multiplied negative times ?
28830
Let $a,b$ be . Is there an efficient algorithm to calculate $a*b$, the nim-product of $a$ and $b$?  The following rule seems like it could be helpful:  $$ 2^{2^m} * 2^{2^n} = \begin{cases} 2^{2^m} 2^{2^n} &amp; \text{if $m \ne n$} \\ 3(2^{2^m - 1}) &amp; \text{if $m = n$} \\ \end{cases}. $$  Juxtaposition denotes ordinary ordinal multiplication here (not nim-multiplication).
13989
I'm not sure I've got this right. When proving $a^n \mid b^n \Rightarrow a \mid b$, can we do this indirectly? In short,  "Suppose $a$ does not divide $b$, this implies that $a^n$ does not divide $b^n$. But $a^n \mid b^n$, hence $a$ divides $b$."  How about $n^n \mid m^m \Rightarrow n \mid m$? Can we do this the same way?
111071
I am having trouble with (a) and (b):  I evaluated (a) and I got that sun of the first three terms does not equal $\ln(5/8)$. I got that it is equal to $\ln(3/4) + \ln(2/3) + \ln(15/16)$ which is not equal to $\ln(5/8)$.  I am also having trouble with part (b). I cannot figure out how to calculate the math to cancel out the $(k+1)^2$ and turn it into $2(n  +1)$.  Any help would be amazing.
120629
I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.  Thanks in advance
47021
I'm trying to understand each part of this completed proof that my professor did, here is my interpretation in parentheses, please advise as necessary.  Proof: Assume that $2^{1/2}$ is rational, then $2^{1/2}$ = $p/q$ for some integer $p$ and $q$ with $q \neq 0$ and $\gcd(p,q) = 1$  $$(p/q)^2 = 2 = p^2 / q^2 $$ $$\text{ (What was the purpose of squaring √2 and p/q?)}$$  $$p^2 = 2q^2$$ $$ \text{ (Is this basically just canceling out the q^2 on left side, and multiplying the right side?)}$$  2 | p^2 => (2 divides p^2 since it is a factor)  2 | p => (thus 2 divides p)  p = 2k for some integer k => (definition of even)  (2k)^2 = 4k^2 = 2q^2 => (Why are we squaring 2k, and how does 4k^2 = 2q^2?)   q^2 = 2k^2 => (Canceled out 2 on one side, and dividing on other)  2 | q^2 => (2 divides q^2)  2 | q => (thus 2 divides q)  2 | gcd(p,q) => (2 divides gcd of p and q)  2 | 1 Which is a Contradiction (no clue how I got here)