Real Analysis: Proving mn=1 Implies m & n = ±1

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SUMMARY

The discussion confirms that if m and n are integers such that mn = 1, then the only valid integer pairs are (1, 1) and (-1, -1). The reasoning is based on the properties of integers and fractions, demonstrating that for mn to equal 1, both m and n must be either 1 or -1. Any other integer values for m or n lead to non-integer results, thus invalidating those pairs.

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johnjuwax
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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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johnjuwax said:
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.
 
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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