**Theorem**:

**The square root of 2 is irrational,**

**Proof**

Assume for the sake of contradiction that . Hence holds for some

*a*and*b*that are coprime.
This implies that . Rewriting this gives .

Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., . Since 2 is prime, we must have that .

So we may substitute

*a*with , and we have that .
Dividing both sides with 2 yields , and using similar arguments as above, we conclude that . However, we assumed that such that that

*a*and*b*were coprime, and have now found that and ; a contradiction.
Therefore, the assumption was false, and cannot be written as a rational number. Hence, it is irrational.

**Another Proof**

The following reductio ad absurdum argument is less well-known. It uses the additional information √2 > 1.

- Assume that √2 is a rational number. This would mean that there exist integers
*m*and*n*with*n*≠ 0 such that*m*/*n*= √2. - Then √2 can also be written as an irreducible fraction
*m*/*n*with*positive*integers, because √2 > 0. - Then , because .
- Since √2 > 1, it follows that
*m*>*n*, which in turn implies that*m*> 2*n*–*m*. - So the fraction
*m*/*n*for √2, which according to (2) is already in lowest terms, is represented by (3) in strictly lower terms. This is a contradiction, so the assumption that √2 is rational must be false.

Similarly, assume an isosceles right triangle whose leg and hypotenuse have respective integer lengths

*n*and*m*. By the Pythagorean theorem, the ratio*m*/*n*equals √2. It is possible to construct by a classic compass and straightedge construction a smaller isosceles right triangle whose leg and hypotenuse have respective lengths*m*−*n*and 2*n*−*m*. That construction proves the irrationality of √2 by the kind of method that was employed by ancient Greek geometers.
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