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Kainui

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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  1. cwrw238
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    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
  2. Kainui
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    Or 0, I suppose I'm just curious how to come to this conclusion logically rather than just guessing.

    • one year ago
  3. cwrw238
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    oh ok

    • one year ago
  4. mukushla
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    see if this helps or not\[b=\frac{a}{a-1}=1+\frac{1}{a-1}\]from here show that\[a=b=2\]

    • one year ago
  5. mukushla
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    and or\[a=b=0\]

    • one year ago
  6. RadEn
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    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

    • one year ago
  7. Kainui
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    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.

    • one year ago
  8. RadEn
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    thanks for medal, my teacher @mukushla :)

    • one year ago
  9. mukushla
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    Oh...man :) ur very welcome my friend :)

    • one year ago
  10. RadEn
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    :)

    • one year ago
  11. Kainui
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    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
  12. mukushla
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    sure.. i'll look at this later...

    • one year ago
  13. Kainui
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    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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