‘the “CPU” (head) of the TM only sees one symbol (bit) per cycle. It’s hardly possible to have a smaller bandwidth.’

is incorrect. There is no requirement that a turing machine operate on bits. The only requirement is that the alphabet contains a finite number of symbols. Thus I can define as many symbols as I want to have as much “bandwidth” as I want per step. But it seems saying “infinitely fast CPU and infinite bandwidth” was just confusing people, so I took it out.

In Turing machine, infinitely fast CPU does not make sense, and infinite bandwidth is plain wrong: the “CPU” (head) of the TM only sees one symbol (bit) per cycle. It’s hardly possible to have a smaller bandwidth.

In problem, the example given is not a decision problem, it’s a meta-question on the algorithm. This makes no sense. I’d advise not to use sorting as an example, as it is not easy to come up with a convincing decision problem associated with sorting. Shortest path in a graph seems better. Say “instead of returning the shortest path between A and B, tell whether they are at distance at most k”.

Your definition of reduction is wrong. It is generally used to prove that it is NP-hard. (By reducing from another NP-hard problem). When you do that, you don’t assume that P2 is solvable in polynomial time at all. You show that you can use the solution to P2 as a solution from P1 in polynomial time, say TR(n). Then if P2 can be solved in time T2(n), P1 can be solved in time T2(n) + TR(n). So if P2 is in P, P1 too. There’s no proof by contradiction going on here at all.

NP-hard is inexact: there’s no hypothesis that we can’t prove point 1 of NP-completeness. In particular, NP-complete problems are NP-hard.

Folks, once you have read this post, read the wp article before you go and try to impress your friends (especially around reputables universities). You might save yourselves some embarassment.