Deducing Irrational Identity - a1=a2 & b1=b2?

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Discussion Overview

The discussion revolves around the conditions under which the equation a1 - √N b1 = a2 - √N b2 implies that a1 = a2 and b1 = b2. Participants explore the implications of matching coefficients and the role of perfect squares in this deduction.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions how to deduce that a1 - √N b1 = a2 - √N b2 leads to a1 = a2 and b1 = b2.
  • Another suggests that matching coefficients could be a method to approach the problem.
  • Some participants argue that the deduction cannot be made universally, providing a counterexample with specific values for N, a1, b1, a2, and b2.
  • A participant elaborates that equality holds only if N is a perfect square, explaining that if a + b√N = c + d√N, then certain integer conditions must be met for equality.
  • Another participant points out a missing step in the reasoning, suggesting that √N could be rational under certain conditions, leading to a discussion about square-free N.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the deduction's validity, with multiple competing views and counterexamples presented throughout the discussion.

Contextual Notes

Limitations include assumptions about the nature of N and the integers involved, as well as the implications of rationality for √N in the context of square-free numbers.

basil
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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?
 
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Matching coefficients?
 
You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6
 
daveb said:
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:
Hey guys,

I see now...Thanks! Appreciate it.
 
Unit said:
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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