First, √ 2 is irrational... Since if √ 2 = p/q with gcd (p,q)=1
then p²=2q² so p² is even, hence so is p, hence p² is a multiple of 4, therefore q² is even,
so q is even too and lcd (p,q)!=1... OK ?

Here is another proof that √ 2 is irrational (Tom Apostol ; thx to The Universe of Discourse) :

If √ 2 is rational, one can rescale the isosceles right triangle with hypothenuse √ 2 into a isosceles right triangle with integer sides. Let ABC be the smallest isosceles right triangle with integer sides q and p.

The circle centered at C of radius q intersects [BC] at E, where BE = p-q is integer.

The tangent to the circle at E intersects [AB] at F. BEF is right isoceles in E since EBF=BFE=45°. (FE) and (FA) are tangential to the same circle so FA equals FE, i.e. BE, and hence is integer, implying that BF = q-FA is integer.

BEF is a isosceles right triangle with integer sides, contradicting that ABC is the smallest one.

To be continued below...

Well... with x = √ 2 and y = √ 2 we have x^y = √ 2 ^ √ 2... Is this number rational ?

If it is the case we're done, else we try x = √ 2 ^ √ 2 and y = √ 2.

Hence x^y = 2 ! qed... (note : the Gelfond-Schneider tells us that the second case is the correct one)

Here is another proof that √ 2 is irrational (Tom Apostol ; thx to The Universe of Discourse) :

If √ 2 is rational, one can rescale the isosceles right triangle with hypothenuse √ 2 into a isosceles right triangle with integer sides. Let ABC be the smallest isosceles right triangle with integer sides q and p.

The circle centered at C of radius q intersects [BC] at E, where BE = p-q is integer.

The tangent to the circle at E intersects [AB] at F. BEF is right isoceles in E since EBF=BFE=45°. (FE) and (FA) are tangential to the same circle so FA equals FE, i.e. BE, and hence is integer, implying that BF = q-FA is integer.

BEF is a isosceles right triangle with integer sides, contradicting that ABC is the smallest one.

To be continued below...

Well... with x = √ 2 and y = √ 2 we have x^y = √ 2 ^ √ 2... Is this number rational ?

If it is the case we're done, else we try x = √ 2 ^ √ 2 and y = √ 2.

Hence x^y = 2 ! qed... (note : the Gelfond-Schneider tells us that the second case is the correct one)

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