Question

show that1/√3 is irrational

Answer 1

Hey there,

The number,√3, 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 √3 is rational so that we may write

√3 = a/b

 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/b²  or 3b² = a² 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 write a = 2m + 1

and

b = 2n +1

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 obtain

3(4n² + 4n + 1) = 4m² + 4m + 1 

Upon performing some algebra, we acquire the further expression

6n² + 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 √3 = a/b cannot be found. We are forced to conclude that √3  is irrational.

 Hope this helps!