Question

prove that root 3 is irrational  

Answer 1

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.

Answer 2

root 3 is irrational 1.7320508 and it is nonreccuring