The purpose of this project is to use the well-ordering principle to prove that

Question

The purpose of this project is to use the well-ordering principle to prove that certain square roots are irrational. THEOREM If p is a positive integer that is not the square of an integer, and x2 = p, then x is irrational. This will be a proof by contradiction using the well-ordering principle. I. Assume that x2 = p, p is a positive integer that is not the square of an integer, and x is a positive rational number. Define A to be the set n E J:nx is an integer). Show that A is nonempty, and let m be the smallest member of A. 2. Show that there is k EJ such that k

Explanation / Answer

1. A = {n J : nx is an integer}

Let x = r + o/q = (rq + o) / q (r,q and o are integers)

=> qx = rq + o an integer

=> qx2 = qp = (rq + o)x is an integer

Thus there is atleast one element in A which is rq + o.

Let m be the smallest member of A. Currently it is rq + o.

Thus m = rq + o.

2. Since x is not an integer,

r < x < r + 1   

Thus k = r.

3. z = mx - mk = m (x - k)

We have x > r => x > k => x - k > 0

=> mx - mk > 0

=> z > 0

Also since x < k +1,

x - k < 1

=> m(x - k) < m

=> z < m

Thus 0 < z < m.

4. z = mx - mk

=> z = m (x - k)

=> z = m (x - r)

Substituting m = rq + o and x = r + o/q

=> z = (rq + o) o/q = o (r + o/q) = rx

Thus z is an integer.

zx = rx * x = rx2 = rp another integer

=> z A.

Since z < m, z should be the smallest integer in A.

But it was given that m is the smallest integer and we have a contradiction.

Hence x is not rational.

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