equality of surdic expressions (1 Viewer)

[bos1234]

a + brtx =A + brtx then a=A and b = B

Proof

Rearranging, (b-B)rtx = A -a
rtx = A-a/B-b

which contradicts rtx being irrational so b=B and a = A

could someone explain the part in bold?

thanks!

 

[SoulSearcher]

Dividing by B-b is equivalent to dividing by 0 since you made b = B, therefore B-b = 0, which is undefined.

 

[bos1234]

could u explain a bit more?

 

[AMorris]

Sorry SoulSearcher, but that was a pretty convoluted explanation. Heres the proof:

Suppose B =/ b, and that a, A, b, B are all integers (they can be rational but we just multiply by the lcm of the denominators to make them integers), then:

a + brtx = A + Brtx
a - A = rtx(B - b)
rtx = (a-A)/(B - b)

now the LHS is irrational, but the RHS is rational (for b-B =/ 0), ergo a contradiction. Thus B = b

Similarly:

Suppose a =/ A, and that a, A, b, B are all integers (they can be rational but we just multiply by the lcm of the denominators to make them integers), then:

a + brtx = A + Brtx
a - A = rtx(B - b)
1/rtx = (B - b)/(a - A)

now the LHS is irrational, but the RHS is rational (for a-A =/ 0), ergo a contradiction. Thus a = A

Proof complete. (Though there may be a faster way to do it)

 

[SoulSearcher]

My bad Morris
Ahh, I like the proof by contradiction there.

 

[bos1234]

what u mean by this step

LHS is irrational, but the RHS is rational (for b-B =/ 0), ergo a contradiction. Thus B = b

 

[watatank]

Suppose B =/ b, and that a, A, b, B are all integers (they can be rational but we just multiply by the lcm of the denominators to make them integers), then:

a + brtx = A + Brtx
a - A = rtx(B - b)
rtx = (a-A)/(B - b)

now the LHS is irrational, but the RHS is rational (for b-B =/ 0), ergo a contradiction. Thus B = b

Ok they have assumed that B =/= b at the start (as a condition so they can play around with the equation) and by proving that condition to be false they can prove the opposite is true, ie B = b.

So they have assumed that rtx = (a-A)/(B - b) will hold true for any if B =/= b. The LHS is a surd, so its irrational. Therefore the RHS must be irrational too for this assumption to be correct.

the RHS is a rational number (recall rational numbers must have form p/q, where p,q are integers) because you have said that if B =/= b. I mean you can see that much, as long as the denominator doesnt equal zero its a rational number as you can put it over any non zero denominator. WHich you shouldnt work due to your assumption, you can't have an irrational number (LHS) equalling a rational number (RHS).

This means your assumption was false. The opposite must be true, ie B = b

Same deal with the A, a and you can prove that when you put them both in the denominator, that A =a for the statement to hold true.

Also a tip: Make assumptions only when you are doing proof by contradiction to proving the opposite to be true, you can't assume a statement is true and then prove it to be true.

Sorry for the long winded explanation hopefully you don't fall asleep and I hope it helps your maths! :wave: