**Q.1: Prove that p: “If a is real such that a ^{3}+ 4a = 0, then a is 0″ is true **

**(i). by direct method**

**(ii). by method of contradiction**

**(iii). by method of contra positive**

**Sol:**

**p: “if a is real such that a ^{3}+ 4a = 0, then a is 0”**

**Let q: a is real such that a ^{3}+ 4a = 0 r, a is 0.**

**(i).** To show that “p” is true, we take that “q” must be true and then prove that “r” is true.

Therefore, assume statement “q” be true.

a^{3}+ 4a= 0 a (a^{2} + 4) = 0

=> a = 0 or a^{2 }+ 4 = 0

However, since a is real, so it is O.

Hence, “r” is true.

**Thus, the statement is true.**

**(ii). ** To show “p” is true using contradiction, we take that “p” isn’t true.

Let x be a real number such that a^{3}+ 4a = 0 and let x is not 0.

Therefore, a^{3}+ 4a = 0 x (a^{2} + 4) = 0 a = 0 or a^{2} + 4 = 0 a = 0 or a^{2}= — 4

However, ‘a’ is real. Thus, a = 0, which is contradiction as we assume that a is not 0.

**Hence, the statement “p” is true.**

**(iii). ** To prove “p” to be true by using contrapositive method, let r is false and show that q is false.

Here, “r” is false states that its requirement of the negation for statement r.

This obtains the following statement.

~ r. x is not 0.

It is seen (a^{2} + 4) cannot be negative so it will be positive.

a ≠ 0 states that product of a positive number with “a” is not zero.

Let us assume the product of a with (a^{2} + 4).

a (a^{2}+ 4) ≠ 0

a^{3} + 4a ≠ 0

This proves that “q” isn’t true.

Hence, it is proved that ~ r => ~ q

**Therefore, the statement “p” is true.**

**Q.2: Prove the statement “For real numbers b and a, b ^{2} = a^{2} implies that b = a” isn’t true. Give a counter example.**

**Sol:**

Using “if- then” the given statement may be written as follows.

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

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

q: b = a

To prove: given statement is false. For this we need to prove that p. then ~q. To prove this, we need two numbers b and a with b^{2} = a^{2} such that b ≠ a (the numbers must be real numbers)

Let b = (-1) and a = 1 b^{2} = (-1)^{2} = 1 and a^{2} = (1)^{2} = 1

Therefore b^{2} = a^{2}

However, b ≠ a

**Thus, it is proved that given statement is false.**

**Q3: By using contrapositive method prove the following statement is true**

**p: if a is an integer and a ^{2} is even, then a is even.**

**Sol:**

**p:** if a is an integer and a^{2} is even, then a is even.

**Assume q:** a is an integer with “a^{2}” even. r. a is even

**To show:** By using contrapositive method p is true, q is false.

r is false [we assume]

let a is odd number

To show that ‘q’ is false, we have to prove that ‘a’ isn’t an integer or a^{2} isn’t even

‘a’ isn’t even implies that a^{2} is not even

**Therefore, ‘q’ is false. Thus, given statement i.e p is true.**