I have this gnarly equality which Mathematica’s Reduce says it doesn’t have the chops to handle: $\left.k\in \mathbb{Z}\land \left\lceil \frac{k-2}{2}\right\rceil \left(\left\lceil \frac{k-2}{2}\right\rceil -((k-2) \bmod 2)+2\right)+k+1=\left\lceil \frac{k}{2}\right\rceil \left(\left\lceil \frac{k}{2}\right\rceil -(k \bmod 2)+2\right)\right]$, so I simplified to this: $k\in \mathbb{Z}\land (k \bmod 2)+k=2 \left\lceil \frac{k}{2}\right\rceil$, which Mathematica reduces to: $c_1\in \mathbb{Z}\land k=c_1$. So, can I say that […]

I can’t get eq (1) and eq (2) Question: Prove that if m and n are positive integers, and x is a real number, then: ceiling((ceiling(x)+n)/m) = ceiling((x+n)/m) Answer: Let us define the real number x as the sum of an integer ‘a’ and a positive real number ‘b’ which is lesser than 1. Therefore, […]

Let $m>2$ be an integer, $x_0 = m$ and $x_n = {(x_{n-1})}^2 – 2$ for $n > 0.$ Prove that $x_n=\lceil\tau(n) \rceil$, where $\tau(n) = α^{2^n}$ and $\alpha >1$ satisfies $\alpha + \frac{1}{\alpha} = m$. This is the problem. The only thing I can think of is that it could have something to do with […]

How do I find $f^{-1}(\{2,4\})$? Are there infinite number of answers?

Via induction I need to prove an expression is true. the expression is: $n \leq 2^k \longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot 2^k-1$ for all $n,k \in \mathbb{Z^+}$ $a_n$ is a recursive function where $a_1 = 3$ $a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1$ I am stuck at […]

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