Discrete Mathematics Question Prove or Disprove that 19 is a

Discrete Mathematics

Question: Prove or Disprove that 19 is an irrational number.

Solution

Answer:

The number, 19, 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 19 is rational so that we may write

19 = a/b ----------( 1 ) for a and b are any two integers.

We must then show that no two such integers can be found.

We begin by squaring both sides of eq. 1:

19=a²/b² ---------( 2 )

or a²=19b² ------------(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

(2m+1)²=19(2n+1)² ----------------( 5 )

(4m²+1+4m)=19(4n²+1+4n)

Upon performing some algebra, we acquire the further expression

4m²+1+4m=76n²+19+76n

4m²+4m=76n²+18+76n

on dividing by 2 , we get

2(m²+m)=38(n²+n)+9 ----------( 6 )

The Right Hand Side of eq. 6 is an odd integer. The Left 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 19 = a/b cannot be found. We are forced to conclude that 19 is irrational.