Kainui
  • Kainui
If I know that a and b are integers, how would I prove that a=b in the equation ab=a+b?
Mathematics
schrodinger
  • schrodinger
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cwrw238
  • cwrw238
i dont understand that question - the only values i can think of to fot satisfy it is if a and b wre both = 2
Kainui
  • Kainui
Or 0, I suppose I'm just curious how to come to this conclusion logically rather than just guessing.
cwrw238
  • cwrw238
oh ok

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anonymous
  • anonymous
see if this helps or not\[b=\frac{a}{a-1}=1+\frac{1}{a-1}\]from here show that\[a=b=2\]
anonymous
  • anonymous
and or\[a=b=0\]
RadEn
  • RadEn
ab-a=b a(b-1)=b a=b/(b-1) a would be integer, satisfied if numerator (b) is 0 or denominator (b-1) is 1
Kainui
  • Kainui
Ah thank you very much I see. By making it in terms of fractions we eliminate the impossible solutions by seeing fairly obviously which values will give non-integer answers. Cool.
RadEn
  • RadEn
thanks for medal, my teacher @mukushla :)
anonymous
  • anonymous
Oh...man :) ur very welcome my friend :)
RadEn
  • RadEn
:)
Kainui
  • Kainui
If you're having fun, can we take this a couple steps further and look at how to show abc=a+b+c as all integers can only allow 0 or 1,2,3 as answers (in 3! ways). Since the extra variables are involved it gets a little trickier. Then I noticed I could extend these rules to n number of variables so that abcde=a+b+c+d+e would have answers of 0 or 1,1,1,2,5. But it might possibly have more. Anyone interested in playing around with this with me for fun?
anonymous
  • anonymous
sure.. i'll look at this later...
Kainui
  • Kainui
Haha alright. If you know any websites that talk about this or what this is called if it has a name, that would be extrodinarily helpful.

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