I'm trying to prove $x+2(-1)^x ≥ x-2$ for all $x \in \mathbb Z^+$

I figured that I'm supposed to use induction but I'm stuck at the induction step. I'm trying to show true for $x=k+1$.

$(x+1)+2(-1)^{x+1} ≥ (x+1)-2$

I try to get the RS from the LS

$$\begin{matrix}(x+1)+2(-1)^{x+1} &=& (x+1)+2((-1)^x)(-1) &\text{(from definition of exponents)}\\ &>& (x+1)+2(-x)(-1)&\text{(since }1^x > x)\\ &=& 3(x)+1&\text{(I want to get to }(x+1)-2)\end{matrix}$$

(can I just say $3(x)+1 ≥ (x+1)-2$?)

I'll try to manipulate both sides

$(x+1)+2(-1)^{x+1} ≥ (x+1)-2\\ (x+2((-1)^x)+1-2) ≥ (x+1)\\ x+2((-1)^x)-1 ≥ x+1\\ x+2((-1)^x)+1-1 ≥ x+1-1\\ x+2((-1)^x) ≥ x$

but now if I let $x=1$ I get $-1 ≥ 1$ so I know I've done something wrong here too plus I don't think I'm supposed to manipulate both sides.

I have checked the base case and I know that what I'm trying to prove is correct as in $x+2(-1)^x ≥ x-2$ for all $x \in\mathbb Z^+$ but I just cant figure out what I'm doing wrong.

I know I haven't formatted the question properly with exponents I will figure it out for future questions. Any help would be greatly appreciated thanks in advance.

  • $\begingroup$ Were you asked to do this with induction? Because there is a simpler proof without induction, namely the fact that $(-1)^x$ can only take two values, and regardless of which one it takes, $2(-1)^x \geq -2$. Adding $x$ to both sides, the inequality follows. $\endgroup$ – астон вілла олоф мэллбэрг Oct 9 '17 at 3:38
  • $\begingroup$ I edited your question to have proper formatting. If you click on the grey "edit" button on the bottom of your question, you can see how I did it. $\endgroup$ – Formyer Oct 9 '17 at 3:45

There is no induction needed:

It is:

$x+2\cdot (-1)^x=\begin{cases} x-2,\text{for x=2k+1}\\ x+2,\text{for x=2k}\end{cases}\geq x-2$

with $k\in\mathbb{N}$


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