radical 3 is irrational

Question 1

radical 3 is irrational - step 1 “let us”

Answer 1

let us assume that is rational so that we may write radical 3
= a/b

Question 2

radical 3 is irrational - step 2 “for a”

Answer 2

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

Question 3

radical 3 is irrational - step 3 “if b”

Answer 3

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

Question 4

radical 3 is irrational - step 4 “where we”

Answer 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)

Question 5

radical 3 is irrational - step 5 “the left”

Answer 5

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.