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## Answers

The Irrationality of

__Problem__:

Prove that is an irrational number.

__Solution__:

The number, , is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that is rational so that we may write = a/b1.for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1:3 = a2/b22.or3b2 = a22a.If b is odd, then b2 is odd; in this case, a2 and a are also odd. Similarly, if b is even, then b2, a2, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may writea = 2m + 13.andb = 2n +14.where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain3(4n2 + 4n + 1) = 4m2 + 4m + 15.Upon performing some algebra, we acquire the further expression6n2 + 6n + 1 = 2(m2 + m)6.The Left Hand Side of eq. 6 is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship = a/b cannot be found. We are forced to conclude that is irrational.

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**⇒**

**SOLUTION ;-**

→

*Let us assume √3 as rational number for sometime ...*

⇒ √ 3 = a /b {

**Where a & b are some integers , co - prime numbers and b ≠ 0**}

⇒

*Now square on both the sides ,*

→ Where 3 divided's a² and also 3 divides b too. ( Let 7 be a prime number and a be a positive number ,so it divides a and a²) ⇒ Equation 1

→

*Now , Let a = 3c*

*Now , substiute this in eq - 1*

⇒

⇒

⇒

⇒

→

**Where 3 divides b² and also 3 divided b (Let 3 be a prime number and a be a positive number , so, it divides a and a² )**

⇒ So , 3 divided a and b..

⇒ So , 3 divided a and b..

**⇒ Where a and b are co- prime numbers and 7 is factor of a and b**

**→ Hence , it is contradicting .**

**∴**

__√3__is an irrational number