Question

Prove that √2+√3 is irrational

Answer 1

similar question
Two symmetrical proofs by contradiction can be rendered:

Assumption: 2–√3–√23 is rational, which, by definition, means that there exist two integers mm and nn such that:

2–√3–√=mn(1)(1)23=mn

and:

gcd(m,n)=1(2)(2)gcd(m,n)=1

Square both sides of (1):

23=m2n2(3)(3)23=m2n2

Multiply both sides of (3) by 3n23n2:

2⋅n2=3⋅m2(4)(4)2⋅n2=3⋅m2

Proof a, multiples of 33 contradiction:

From (4) it follows that the number 2⋅n22⋅n2 is divisible by 33. Since 22 is prime and hence is not divisible by 33, it is n2n2that must be divisible by 33, which means that nn must be divisible by 33*, which means that it is representable as:

n=3r(5)(5)n=3r

for some integer rr.

Put (5) into (4):

2⋅9⋅r2=3⋅m22⋅9⋅r2=3⋅m2

3⋅2⋅r2=m2(6)(6)3⋅2⋅r2=m2

Now, from (6), it follows that m2m2 is divisible by 33 and hence mm is divisible by 33*, which means that, since nn and mmboth have a common factor, 33, nn and mm are not in lowest terms:

gcd(m,n)≠1(7)(7)gcd(m,n)≠1

which is a contradiction with (2). Therefore, our original assumption that the number in question is rational is false and hence that number is irrational.

Proof b, multiples of 22 contradiction :

Pretty much the same line of reasoning: from (4) it follows that m2m2 must be even, which, in turn, means that mmmust be even*, which means that it is representable as:

m=2k(8)(8)m=2k

for some integer kk.

Put (8) into (4):

2⋅n2=3⋅4⋅k22⋅n2=3⋅4⋅k2

n2=2⋅3⋅k2n2=2⋅3⋅k2

Now it follows that n2n2 must be even and, hence, so must be nn - both mm and nn being even implies that they are not in their lowest terms - contradiction, etc.

*This type of statements can be proved by using the fact that a direct statement and its converse are logically equivalent. For example: we need to prove the following statement pp:

IF n2n2 is even THEN nn must be even

The converse of pp is:

IF nn is not even (odd) THEN n2n2 is not divisible by 22

Proof: if nn is not divisible by 22 then its prime factorization does not contain 22 and hence the square of nn, which simply doubles the power of each prime factor, does not contain 22 also. Another way: if nn is odd then n=2r+1n=2r+1for some integer rr and n2=(2r+1)2=2(2r2+2r)+1n2=(2r+1)2=2(2r2+2r)+1 which is still odd, qed.