Irrational identity?

  • Thread starter basil
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  • #1
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How do you deduce that

a1 - [itex]\sqrt{N}[/itex] b1 = a2 - [itex]\sqrt{N}[/itex]b2

to be

a1=a2 and b1=b2?
 

Answers and Replies

  • #2
724
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Matching coefficients?
 
  • #3
543
1
You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6
 
  • #4
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You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6

In this case, they are only equal because N is a perfect square.

If a + b√N = c + d√N, then
a - c = (d - b)√N

Assuming a, b, c, d are integers, then (a - c) is an integer, thus for equality, (d - b)√N must be an integer. This is only true if a = c and b = d, or if √N is an integer, making N a perfect square.
 
Last edited:
  • #5
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Hey guys,

I see now...Thanks! Appreciate it.
 
  • #6
disregardthat
Science Advisor
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Assuming a, b, c, d are integers, then (a - c) is an integer, thus for equality, (d - b)√N must be an integer. This is only true if a = c and b = d, or if √N is an integer, making N a perfect square.

You skipped a step. sqrt(N) could have been rational, with (d-b) divisible by the denominator. And so this problem is reduced to proving that sqrt(N) cannot be rational for square-free N.
 

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