Real Analysis Proof

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  • #1
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Prove or disprove that if m and n are integers such that mn = 1 then either m= 1 & n = 1 or else m = -1 & n = -1.
 

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  • #2
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Prove or disprove that if m and n are integers such that mn = 1 then either m= 1 & n = 1 or else m = -1 & n = -1.
If mn=1, then m/1=1/n. But for n an integrer, then m is a fraction. The only way both are integers is if n=1 so that m=1, or if n=-1 so that m=-1.
 
  • #3
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It follows from the fact that, if m>1, then 0<n<1, and if m<-1, then -1<n<0. Therefore we have three options left to check, m=-1, 0, 1.
 

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