4 ms·
Let m = n. For m^n = n^m. There are infinite solutions. Am I missing something? Edit: “Distinct”
by bofaGuy 3y ago
Let m = n.
For m^n = n^m.
There are infinite solutions.
Am I missing something?
Edit: “Distinct”
- chx 3y agoint
- 1letterunixname 3y agom = n would be trivial and pointless. Also, I'm curious why he didn't specify precisely clarify "distinct" to mean "pairs of unequal values" (e.g., exclude {n,m} where n=m) and "pairs not equal to others" (e.g., {x,y} excludes {y,x}).
- ouid 3y agoThe condition is as precise as possible "pairs of distinct integers" is not equal to "distinct pairs of integers". Don't blame the writer when you fail to read what is written.
- 1letterunixname 3y agoRelax and be civilized.
- ahmedalsudani 3y agoIt's common for the obvious cases and obvious constraints to be assumed. Adding "distinct" would have made it clearer, but it was obvious that the author meant distinct from reading the title alone.
- someweirdperson 3y ago> Adding "distinct" would have made it clearer, but it was obvious that the author meant distinct from reading the title alone. It's math. No assumption is ever obvious. Only parts of some proofs are ever allowed to be called obvious.
- Dylan16807 3y agoWould you really call that a "pair of integers"? Especially when it makes the solution so trivial?
- tirpen 3y agoYes. (1,1) is a pair of integers, and so is (17,17). Why wouldn't they be?
- Dylan16807 3y agoBut if you make it a sentence, "my pair of integers is 1 and 1", it sounds a little bit off. Because "pair" for the same concept twice is often an error. Picking a mathematical formulation like you did can avoid that kind of implication, but there wasn't a template for which formulation to use and the way you wrote that makes it look like order matters which isn't right either.
- CyberDildonics 3y agoIn binary, you will probably get 1 next to another 1 all the time.
- Dylan16807 3y agoThose are digits, not integers? And that also sounds like two ordered numbers which is not what the question calls for.