Prove that square root of three is an irrational number

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/b

1.

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

2.

or

3b2 = a2

2a.

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

3.

and

b = 2n +1

4.

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

5.

Upon performing some algebra, we acquire the further expression

6n2 + 6n + 1 = 2(m2 + m)

6.

The Left Hand Side of eq. 3a is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 3a. Therefore, integer values of a and b which satisfy the relationship = a/b cannot be found. We are forced to conclude that is irrational.

From:http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm