Proof that the square root of 2 is irrational. Assume is rational, i.e. it can be expressed as a rational fraction of the form , where and are two relatively prime integers. Now, since , we have , or . Since is even, must be even, and since is even, so is .

Proof that root is irrational number:

To prove that root is an irrational number, we need to show that it cannot be expressed as the quotient of two integers (p/q), where p and q are coprime integers and q is not equal to zero.

Assumption:
Let's assume that root is a rational number and can be expressed as p/q where p and q are coprime integers and q is not equal to zero.

Proof by contradiction:
We will prove this assumption false by contradiction.

Step 1:
Assume that root is a rational number and can be expressed as p/q.

Step 2:
Square both sides of the equation to get rid of the root.

(root)^2 = (p/q)^2

Step 3:
Simplify the equation.

p^2 = q^2 * root^2

Step 4:
Multiply both sides of the equation by q^2 to eliminate the root from the denominator.

p^2 * q^2 = q^4 * root^2

Step 5:
Rearrange the equation.

p^2 * q^2 = (q^2 * root)^2

Step 6:
Since p^2 * q^2 is an integer, it implies that q^2 * root is also an integer.

Step 7:
Now, let's assume that root is a rational number and can be expressed as a/b, where a and b are coprime integers and b is not equal to zero.

Step 8:
Substitute the value of root in terms of a/b into the equation from step 7.

(q^2 * (a/b))^2 = p^2 * q^2

Simplifying further, we get:

q^4 * (a^2/b^2) = p^2 * q^2

Step 9:
Multiply both sides of the equation by b^2 to eliminate the root from the denominator.

q^4 * a^2 = p^2 * b^2

Step 10:
Since q^4 * a^2 is an integer, it implies that p^2 * b^2 is also an integer.

Step 11:
This means that both p^2 and b^2 are divisible by q^4, which contradicts our assumption that p and q are coprime integers.

Conclusion:
Since our assumption leads to a contradiction, we can conclude that root cannot be expressed as a rational number. Therefore, root is an irrational number.