Proof problem?

Number Theory Level 3

Let's evaluate the following proof:

Theorem: Suppose $n \in N.$ Then $4n^2 - 12n + 9 ≥ 1.$

Proof: Suppose $n \in N.$ Examine 3 cases.

Case 1: Suppose $n = 1.$ Then $4n^2 - 12n + 9 = 1,$ which is greater than or equal to $1$ .

Case 2: Suppose $n = 2.$ Then $4n^2 - 12n + 9 = 1,$ which is greater than or equal to $1.$

Case 3: Suppose $n > 3.$ Then $4n^2 - 12n + 9 = 4(n)(n - 3) + 9.$ Since $n > 3, n - 3 > 0.$

Since both $n$ and $n-3$ are greater than $0,$ we see that $4(n)(n-3) > 0,$ so $4(n)(n-3) + 9 > 9.$

Hence, $4n^2 - 12n + 9 >1$

Look at the proof, and choose the correct option.

The proof and the theorem are correct. The theorem is false and the proof is incorrect,for

n = 1.5

We have

4n^2 - 12n + 9 =0

The proof is incorrect because of arithmetic errors. The proof is correct, but the theorem is false The proof leaves out the case where

n=3,

but is otherwise correct.

1 solution

Tom Engelsman
Sep 27, 2017

Here's a much better proof via induction:

CASE I (n = 1): $4(1^2) - 12(1) + 9 = 4 -12 + 9 = 1 \ge 1 \Rightarrow TRUE;$

CASE II (n = k, (k > 1)): $4k^2 - 8k + 9 \ge 1$ ;

CASE III (n = k+1): $(4k^2 - 8k + 9) + (8k - 8) \ge 1 + (8k-8);$

or $(4k^2 + 8k + 4) - (12k + 12) + 9 \ge 1 + (8k-8);$

or $4(k+1)^2 - 12(k+1) + 9 \ge 1 + 8(k-1) \ge 1$ ;

or $\boxed{4(k+1)^2 - 12(k+1) + 9 \ge 1 \Rightarrow TRUE}.$