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

[basil]

How do you deduce that

a₁ - $\sqrt{N}$ b₁ = a2 - $\sqrt{N}$b₂

to be

a₁=a₂ and b₁=b₂?

 

[TylerH]

Matching coefficients?

 

[daveb]

You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6

 

[Unit]

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.

 

[basil]

Hey guys,

I see now...Thanks! Appreciate it.

 

[disregardthat]

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.