Algebra to show Induction for $\lceil {x\over 2} \rceil \mapsto \lceil{(x+1)\over 2}\rceil $

Question

$5^{\lceil \frac x 2 \rceil}$ (Which Operation?) x = $5^{\lceil \frac x2\ + 1 ) \rceil}$

$x$ can be any value that can help equate this for induction.

I'm confused on how the algebra would work to get from $5^{\lceil \frac x 2 \rceil}$ to change it to $(x+1)\over 2$ in the ceiling function itself.

Answer 1

The wording of the question is still unclear and the notation inconsistent, but I'll take a chance that you are really asking about $5^{\lceil(x+1)/2\rceil}$ and not $5^{\lceil x+(1/2)\rceil}$ on the right-hand side of the equation.

Although you cannot prove a general fact just by looking at a few examples, sometimes it helps to look at examples anyway, especially if you do not know exactly what fact you are trying to prove. A few examples can give a hint about a general pattern.

The ceiling function $\left\lceil \frac x2 \right\rceil$ increases by $1$ whenever $x$ increases by $2$, that is, $$\left\lceil \frac 12 \right\rceil = 1, \quad \left\lceil \frac 22 \right\rceil = 1, \quad \left\lceil \frac 32 \right\rceil = 2, \quad \left\lceil \frac 42 \right\rceil = 2, \quad \left\lceil \frac 52 \right\rceil = 3,$$ and so forth. So a few examples of the known parts of your equation (the $5^n$ pieces on the left-hand and right-hand sides) with $x=1,2,3,4,5,6$ gives this:

\begin{align} 5^{\lceil 1/2 \rceil} &= 5 & 5^{\lceil(1+1)/2\rceil} &= 5 \\ 5^{\lceil 2/2 \rceil} &= 5 & 5^{\lceil(2+1)/2\rceil} &= 25 \\ 5^{\lceil 3/2 \rceil} &= 25 & 5^{\lceil(3+1)/2\rceil} &= 25 \\ 5^{\lceil 4/2 \rceil} &= 25 & 5^{\lceil(4+1)/2\rceil} &= 125 \\ 5^{\lceil 5/2 \rceil} &= 125 & 5^{\lceil(5+1)/2\rceil} &= 125 \\ 5^{\lceil 6/2 \rceil} &= 125 & 5^{\lceil(6+1)/2\rceil} &= 625 \end{align}

You may notice a pattern. Actually you should notice two patterns, one for the cases where $x$ is even and a different pattern for the cases where $x$ is odd.

It is perfectly OK for a fact to have separate statements for when $x$ is even and when $x$ is odd, or for an inductive proof to have two different cases for the inductive step, one for $n$ even and one for $n$ odd.