Answer
P(1) is true because
√2 > 1 ≥ 1/b for all positive integers b.
Work Step by Step
--Let P(n) be the statement
that there is no positive integer b such that √2 = n/b.
--Basis step:
P(1) is true because
√2 > 1 ≥ 1/b for all positive integers b.
-- Inductive step:
- Assume that P(j) is true for all j ≤ k,
-where k is an arbitrary positive integer; we prove that
-P(k+1) is true by contradiction.
Assume that
-√2 = (k+1)/b
for some positive integer b.
Then 2b^2 = (k +1)^2, so (k +1)^2 is even,
and hence, k + 1 is even.
So write k + 1 = 2t for some positive integer t ,
-whence 2b^2 = 4t^2 and b^2 = 2t^2.
-By the same reasoning as before, b is even, so b = 2s for some
positive integer s.
Then
-√2 = (k + 1)/b = (2t)/(2s) = t/s.
--But t ≤ k, so this contradicts the inductive
