Given the statement For every integer n 2 n2 6n 1 0 Is t

Given the statement:

For every integer n > 2, n^2 – 6n + 1 > 0.

Is the text below a correct proof of the statement? (YES or NO? Explains your choice.)

Proof: By induction

Let S[n] be the statement that n^2 – 6n + 1 > 0

We show that if S[k] is true for some integer k > 2 then S[k + 1] is also true.

Suppose that k^2 – 6k + 1 > 0 (Induction Hypothesis)

Then (k + 1)^2 – 6(k + 1) + 1 = k^2 + 2k + 1 – 6k – 6 + 1 = (k^2 – 6k + 1) + (2k – 5)

Both bracketed expressions are positive: the first by the Induction Hypothesis, and the second because k > 2. So S[k + 1] is true.

By PMI, S[n] is true for all n > 2.

There is nothing wrong in your proof.

BUT given statement is for every integer n > 2, n^2 - 6n +1 > 0

This is False.

Reason: for n = 3

3^2 - 6*3 + 1 = -8

-8 is not greater than zero.

So Given statement is wrong.

It would have been true if the statement is

for every integer n > 5, n^2 - 6n + 1 > 0

Let me know in the comments, if you have any doubt.