Math Challenge

Let an Set $\displaystyle{A=\left\{ { x }_{ 1 },{ x }_{ 2 },........,{ x }_{ n } \right\} \\ n=2k+1,\quad k\in { I }^{ + }}$ ,

One-one function(s) $\displaystyle{f:A\rightarrow A\quad \\ \left| f\left( { x }_{ 1 } \right) -{ x }_{ 1 } \right| =\left| f\left( { x }_{ 2 } \right) -{ x }_{ 2 } \right| =........=\left| f\left( { x }_{ n } \right) -{ x }_{ n } \right| }$ , then Find number of all such one-one functions ?

Help Me!

Thanks..!

Karan

#Algebra #JEE #Challenge #JEE_Advance

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

@Shashwat Shukla @Ronak Agarwal @Raghav Vaidyanathan @Mvs Saketh @Sudeep Salgia @Calvin Linsir @Brian Charlesworth sir @Krishna Sharma @Tanishq Varshney Please help me in this if you can... any help will be appreciated ! Thanks!

Karan Shekhawat - 6 years, 1 month ago

if $n$ is odd then how is $n=2 ?$ , in the set A?

Mahimn Bhatt - 6 years, 1 month ago

so what ... ? Please read it properly bro .. n is odd Means total number of terms are odd , It dosn't mean that all terms are odd.... so be careful ..

Karan Shekhawat - 6 years, 1 month ago

ok ..i am sorry.....i thought that x(n) is a general term.....did not look at the end of the set. :(

Mahimn Bhatt - 6 years, 1 month ago

@Mahimn Bhatt – it's okay bro .. please try it now.. and help me

Karan Shekhawat - 6 years, 1 month ago

@Karan Shekhawat – yeah..

Mahimn Bhatt - 6 years, 1 month ago

Are each of the $x_i$ distinct? Because then you have one solution , the identity function.

Siddhartha Srivastava - 6 years, 1 month ago

Yes It is... nothing is mention , and read question properly ... We have to Count all such possible cases , obviously f(x)=x is one among such.. But we have to find all ...

Karan Shekhawat - 6 years, 1 month ago

OK. I only see 1 solution though. I even have a proof. Do you have a counter example?

Let $|f(x_i) - x_i| = m$ . If $m =0$ , we have the identity function. So let us assume $m \not = 0$ and a function other than the identity function exists.

Then $f(x_i) - x_i = \pm m \quad \forall i \in \{ 1,2,...,n \}$ .

Divide $A$ into $P$ and $Q$ , where if $f(x_i) - x_i = m \implies x_i \in P$ else $x_i \in Q$ . Since $P$ and $Q$ are disjoint, $|P| + |Q| = |A| = odd$ .

Now, $\sum f(x_i) - x_i = |P|m + |Q|(-m) = (|P| - |Q|)m$

Now, since $|P| + |Q|$ is odd, $|P| - |Q|$ is also odd, hence not zero. Therefore, $(|P| - |Q|)m$ is not zero and $\sum f(x_i) - x_i$ is not zero.

But, $f(x)$ is clearly a permutation function, which means that $\sum f(x_i) - x_i = 0$ , contradiction.

Therefore there exists no other function other than the identity function.

Siddhartha Srivastava - 6 years, 1 month ago

@Siddhartha Srivastava – Great! That's an invariance argument.

Calvin Lin Staff - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$