Question

Show that 3√2 is irrational.​

Answer 1

Step-by-step explanation:

Proof

Let us assume that 3+√2 is a rational number.

A rational number can be written in the form of p/q.

3+√2=p/q

√2=p/q-3

√2=(p-3q)/q

p,q are integers then (p-3q)/q is a rational number.

But this contradicts the fact that √2 is an irrational number.

So,our assumption is false.

Therefore,3+√2 is an irrational number.

Hence proved.

Answer 2

Answer:

Hint- Here, we will assume the opposite and will prove that the assumption contradicts.

Step-by-step explanation:

Let us suppose that 32–√32 is a rational number.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because aa and bb are two co-prime positive integers and 3 is also an integer.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because aa and bb are two co-prime positive integers and 3 is also an integer.Also, the LHS of the equation (1) i.e., 2–√2 is an irrational number.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because aa and bb are two co-prime positive integers and 3 is also an integer.Also, the LHS of the equation (1) i.e., 2–√2 is an irrational number.Therefore, equation (1) is contradicting since LHS is irrational and RHS is rational.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because aa and bb are two co-prime positive integers and 3 is also an integer.Also, the LHS of the equation (1) i.e., 2–√2 is an irrational number.Therefore, equation (1) is contradicting since LHS is irrational and RHS is rational.So, our assumption is not correct.

Let us suppose that 32–√32 is a rational number.As we know that any rational number can be represented in the form of abab where aa and bb are two co-prime positive integers.∴32–√=ab⇒2–√=a3b →(1)∴32=ab⇒2=a3b →(1)Now if we observe the RHS of equation (1) carefully, we can say it is always a rational number because aa and bb are two co-prime positive integers and 3 is also an integer.Also, the LHS of the equation (1) i.e., 2–√2 is an irrational number.Therefore, equation (1) is contradicting since LHS is irrational and RHS is rational.So, our assumption is not correct.Hence, 32–√32 is irrational.