Search found 194 matches
- Thu May 06, 2021 10:38 pm
- Forum: Algebra
- Topic: A question about FE
- Replies: 9
- Views: 15305
Re: A question about FE
SO I guess your function is $f(x)=\frac{1}{x}$ if $x \neq 0$, and $f(0)=0$ if $x=0$. I guess then it works. Actually the main question is does $f(xf(x))=xf(x) \Rightarrow f(x)=x\ \ \forall x \in \mathbb{R}$? :roll: Probably not why? i have just shown example of such functions but can it proven?(exc...
- Thu May 06, 2021 12:37 pm
- Forum: Algebra
- Topic: A question about FE
- Replies: 9
- Views: 15305
Re: A question about FE
I think the problem with $f(x)=\frac{1}{x}$ is that its domain is not $\mathbb{R}$, it is $\mathbb{R} \backslash \{0\}$. But the function should have a domain of the whole set of real numbers. This is what I think, but I may be wrong too. Doesn't $x=0 \Rightarrow f(0)=0$? then the domain $\mathbb{R...
- Wed May 05, 2021 10:19 pm
- Forum: Algebra
- Topic: A question about FE
- Replies: 9
- Views: 15305
Re: A question about FE
Does $f(xf(x))=xf(x) \Rightarrow f(x)=x$ $\forall x \in \mathbb{R}$??(I don't think so :| as if $f(x)=1/x$ we would get a constant value each time or may be other weird function may satify this inversive property making it constant each time but idk ) I think the problem with $f(x)=\frac{1}{x}$ is ...
- Wed May 05, 2021 2:04 pm
- Forum: Algebra
- Topic: A question about FE
- Replies: 9
- Views: 15305
A question about FE
Does $f(xf(x))=xf(x) \Rightarrow f(x)=x$ $\forall x \in \mathbb{R}$??(I don't think so as if $f(x)=1/x$ we would get a constant value each time or may be other weird function may satify this inversive property making it constant each time but idk )
- Sun May 02, 2021 3:08 pm
- Forum: Number Theory
- Topic: A Problem from Dustan
- Replies: 4
- Views: 11763
- Sun May 02, 2021 2:54 pm
- Forum: Number Theory
- Topic: A Problem from Dustan
- Replies: 4
- Views: 11763
Re: A Problem from Dustan
My SOLN(PLS CONFIRM): We will do by casework. Case 1: $p=q=r=x$ Not possible since $3 \perp 2$ Case 2: WLOG $p=q=x \neq r$ then $2x!+r!=2^s$ if $x>r$ then $r!(2k+1)=2^s$ which is clearly not possible. So, $r>x$ then we would get $x!(2+\frac{r!}{x!})=2^s$ It is clear that $x=2$ either we would get ot...
- Sun May 02, 2021 2:22 pm
- Forum: Number Theory
- Topic: A Problem from Dustan
- Replies: 4
- Views: 11763
A Problem from Dustan
Find all $p,q,r,s>1$ positive integers such that $p!+q!+r!=2^s$.
- Fri Apr 30, 2021 12:13 pm
- Forum: Junior Level
- Topic: Brain Teaser :D
- Replies: 1
- Views: 7441
Brain Teaser :D
A bear starting from the point P, walked one mile due south. Then he changed direction and walked one mile due north, and arrived exactly at the point P that he started from.What was the color of the bear?
- Thu Apr 29, 2021 8:44 am
- Forum: Test Forum
- Topic: Mathematics
- Replies: 4
- Views: 8425
Re: Mathematics
Yes! it's the distributive property of real numbers. Isn't there any proof of this? Why is this property called an axiom? It doesn't seem trivial to me. Was the multiplication between two negative real numbers or a negative real number and a positive real number defined before setting up the distri...
- Tue Apr 27, 2021 10:25 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 664601
Problem 26
$\textbf{Problem 25}$ Find all functions $f:\mathbb Z\rightarrow \mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds: \[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\](Here $\mathbb{Z}$ denotes the set of integers.) For the sake to continue the ...