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If I know that a and b are integers, how would I prove that a=b in the equation ab=a+b?
 one year ago
 one year ago
If I know that a and b are integers, how would I prove that a=b in the equation ab=a+b?
 one year ago
 one year ago

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cwrw238Best ResponseYou've already chosen the best response.0
i dont understand that question  the only values i can think of to fot satisfy it is if a and b wre both = 2
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
Or 0, I suppose I'm just curious how to come to this conclusion logically rather than just guessing.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
see if this helps or not\[b=\frac{a}{a1}=1+\frac{1}{a1}\]from here show that\[a=b=2\]
 one year ago

RadEnBest ResponseYou've already chosen the best response.1
aba=b a(b1)=b a=b/(b1) a would be integer, satisfied if numerator (b) is 0 or denominator (b1) is 1
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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 noninteger answers. Cool.
 one year ago

RadEnBest ResponseYou've already chosen the best response.1
thanks for medal, my teacher @mukushla :)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
Oh...man :) ur very welcome my friend :)
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
sure.. i'll look at this later...
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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.
 one year ago
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