The given statement can be written in the form of ‘if then’ is given below

If a and b are real numbers such that a^{2} = b^{2}, then a = b

Let p: a and b are real numbers such that a^{2} = b^{2}

q: a = b

The given statement has to be proved false.

To show this, two real numbers, a and b, with a^{2} = b^{2 }are required such that a ≠ b

Let us consider a = 1 and b = – 1

a^{2} = (1)^{2 }= 1

and

b^{2} = (-1)^{2 }= 1

Hence, a^{2} = b^{2}

However, a ≠ b

Therefore, it can be concluded that the given statement is false.

Answered by Sudhanshu | 11 months agoDetermine whether the argument used to check the validity of the following statement is correct: p: “If x^{2} is irrational, then x is rational.” The statement is true because the number x^{2} = π^{2} is irrational, therefore x = π is irrational.

Which of the following statements are true and which are false? In each case give a valid reason for saying so

**(i)** p: Each radius of a circle is a chord of the circle.

**(ii) **q: The centre of a circle bisect each chord of the circle.

**(iii)** r: Circle is a particular case of an ellipse.

**(iv)** s: If x and y are integers such that x > y, then – x < – y.

**(v)** t: \( \sqrt{11}\) is a rational number.

By giving a counter example, show that the following statement is not true. p: “If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.”

Show that the following statement is true “The integer n is even if and only if n^{2} is even”

Show that the following statement is true by the method of the contrapositive p: “If x is an integer and x^{2} is odd, then x is also odd.”