Ordinary Induction

Question

Define a function $f:\mathbb{Z}\to\mathbb{Z}$ by $f(x)=2x^2-8x+1$. Use ordinary induction to prove that $f(x)\geq 0$ for all integers $x\geq 4$.

A bit perplexed on the IS. We solve for P(K+1) right? Can someone help me prove this?

Answer 1

All you need to do is show that:

1) $f(4) \geq 0$ (base case)

2) $f(x+1) \geq f(x)$ for $x \geq 4$ (inductive argument)

For the latter, simply compute the expression $g(x)=f(x+1)-f(x)$ and show that $g(x) \geq 0$ for $x \geq 4$.

Answer 2

Let $f(x)=2x^2-8x+1=2(x-2)^2-7.$ To show $f(x)\geq 0$ for all $x\geq4$, $x\in\mathbb{Z}$, we must show it is at least true for the base case $x=4$: $$f(4)=2(4)-7=1\geq0$$ Now we assume the statement holds for some $x=n$ where $n>4,$ that is, assume $f(n)\geq0, $ we shall show $f(n+1)\geq 0$:$$f(n+1)=2(n-1)^2-7=2n^2-4n+2-7=2n^2-4n +(4n-4n)+1+1-7=(2n^2-8n+1)+(4n-6)$$ Thus $f(n+1)=(2n^2-8n+1)+(4n-6)=f(n)+(4n-6)$

Since $n>4$, $4n-6\in\mathbb{Z}^+$ so we have that since $f(n)\geq0$ by our induction hypothesis, $$f(n+1)=f(n)+(\text{element of $\mathbb{Z}^+$)=$\text{element of $\mathbb{Z}^+$}$}$$

Thus since the statement is true for $f(n+1)$, it is true for all $n\geq4.$ Conclude by induction that this statement is true.