Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question.

I have been given this proof: "For any $n$ in the integers where $n>2$, show there are at least $2$ elements in $U(n)$ that satisfy $x^2=1$."

I have gone through and actually proved this, (that the numbers are $1$ and $n-1$) but i didn't' know how to prove that $n-1$ is in fact in the set $U(n)$. Is it because two consecutive numbers are always relatively prime?

share|improve this question
3  
What is $U(n)$? –  Sigur Feb 7 '13 at 0:22
2  
@Sigur: $U(n)$ is the multiplicative group of integers mod $n$ that are relatively prime to $n$. –  Brian M. Scott Feb 7 '13 at 0:23
    
Yes, that is why, and Jason Bourne has just supplied a proof. –  Brian M. Scott Feb 7 '13 at 0:24
    
$-1$ is an integer. –  jspecter Feb 7 '13 at 0:31

1 Answer 1

$n$ is coprime to $n-1$, for if $d$ divides $n$ and $d$ divides $n-1$, then $d$ divides $n-(n-1)=1$.

share|improve this answer
    
ah ! such a nice, neat, simple proof. thank you so much –  Sam Feb 7 '13 at 0:24
    
Nice, and concise! (I'd say well-done, and concise, but nice rhymes with concise!) +1 –  amWhy Feb 7 '13 at 1:11
    
$n$ is coprime to $n−2$, for if $d$ divides $n$ and $d$ divides $n−2$, then $d$ divides $n−(n−2)=2$ ? –  alancalvitti Feb 7 '13 at 2:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.