To prove 1/root 2 is irrational: Let us assume that root 2 is irrational 1/ root 2 = p/q (where p and q are co-prime)q/p = root 2q = (root 2)pSquaring both sides;q2 = 2p2 - (1)By theorem:q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equation 1:2p2 = q2 = 2c2 =4c2p2 =4c2 /2 = 2c2p2/2 = c2By theorem p is also divisible by 2;But p and q are co-primeThis is a contradiction which has arisen due to our wrong assumption∴1/root 2 is irrational.

aap uski value nikalo to aap dekh sakte ho...ager vo neither tepeating nor terminating hai to its irrational...

Proof that √2 is irrational:

To prove that √2 is irrational, we will use a proof by contradiction. This means that we will assume that √2 is rational and then show that it leads to a contradiction.

Assumption: Let's assume that √2 is rational.

Definition of a rational number: A rational number can be expressed as a fraction p/q, where p and q are integers and q is not equal to 0.

Representing √2 as a rational number: Since we assumed √2 is rational, we can express it as √2 = p/q, where p and q have no common factors other than 1.

Squaring both sides: Now, let's square both sides of the equation to eliminate the square root: 2 = (p/q)^2.

Expanding the equation: This gives us 2 = p^2/q^2.

Conclusion: From the above equation, we can deduce that p^2 = 2q^2.

Implication: This implies that p^2 is even, as it is equal to 2 multiplied by q^2.

Even number property: An even number can be expressed as 2k, where k is an integer.

Expressing p^2 as an even number: Therefore, we can represent p^2 as p^2 = 2k, where k is an integer.

Consequence: This means that p is also an even number, as the square of an odd number would result in an odd number.

Expressing p as an even number: We can express p as p = 2m, where m is an integer.

Substituting the value of p: Substituting the value of p in the equation p^2 = 2q^2, we get (2m)^2 = 2q^2.

Simplifying the equation: Simplifying this equation gives us 4m^2 = 2q^2.

Dividing both sides by 2: Dividing both sides of the equation by 2, we obtain 2m^2 = q^2.

Conclusion: From the above equation, we can deduce that q^2 is even.

Implication: This implies that q is also an even number.

Contradiction: However, we assumed earlier that p and q have no common factors other than 1, but now we have found that both p and q are even numbers.

Contradiction to the assumption: This contradiction contradicts our initial assumption that √2 is rational.

Conclusion: Therefore, our assumption is false, and √2 is irrational.

Summary: By assuming that √2 is rational and following the logical steps, we arrived at a contradiction. Hence, we can conclude that √2 is irrational.