Question

Prove using contradiction that the square root of any irrational number is irrational.

Answer 1

Let r be irrational number
Claim : sqrt( r ) is irrational:

Proof : suppose sqrt(r) is rational

then sqrt (r) = p/q
then r = (p/q)^2


do you see the contradiction now?

Answer 2

not really :(

Answer 3

okay, thanks

Answer 4

(i cleaned the proof a bit)


Suppose you are given an unspecified irrational number, which we will label r.

Claim : sqrt( r ) is irrational:

Proof: (by contradiction):
Assume the negation of the claim; i.e. assume sqrt( r ) is not irrational.
This is the same thing as saying sqrt(r) is rational.
Then by definition of a rational number, sqrt (r) = p/q for some integers p, q, where q is non-zero.

We can square both sides of sqrt(r) = p/q. (You can justify this is a valid step using field axioms on the real numbers).

sqrt(r)^2 = (p/q)^2
r = p^2/q^2

But now r is a ratio of integers, therefore r is a rational number.
But we were given that r is irrational before we even did the proof. Therefore r is both rational and not rational.
Contradiction .

Conclusion: Since assuming the negated version of the claim led to a contradiction, the negated claim is false. That is, it is false that sqrt(r) is not irrational. Hence it is true that sqrt(r) is irrational.