What does it mean to solve an equation?

This question might be more philosophical than mathematical.

In school we are taught how to solve equations such as x21=0 or sin(x)1=0. Solutions to these equations are quite simple. For example x=1 and x=1 are the solutions to the first equation. One could say that solving equation f(x)=0 is same as finding the values of x that satisfy the equation. To me this answer doesn’t really tell what does it mean to solve a equation, because the meaning of the verb to find is ambiguous. If someone says that the solutions of the equation are all the numbers in the set {yR|f(y)=0}
has he or she solved the equation? I don’t think he/she has.

Answer

Interesting question. I’d say that solving f(x)=0 amounts to

  • exhibiting the set S = {xf(x)=0}, typically by enumerating its members, or giving a sequence whose elements are all the members of S

  • demonstrating that the listed or enumerated items are exactly equal to S.

So if I say that the solutions of sinx=0 are nπ,n=0,±1,±2,, I’ve given a purported solution set S. I now need to show that for each element t of S, we actually have sint=0, and that no other values of t satisfy sint=0.

The method by which I arrive at the set S is not really germane, despite the active verb “solve”; the solution might come from algebraic or geometric manipulations, or it might come to me in a dream. But the second part — the demonstration that the purported solution set is the actual solution set — that must follow the rules of logic and mathematics.

This is, however, mostly opinion about common mathematical speech, rather than a fact about mathematics.

PS: For infinite solution sets that are not countable, Christian Blatter’s answer starts to get at a good description, although it doesn’t take into account things like “the solution set is all irrationals,” where a parametrization of the set may be very hard to come up with. Roughly speaking, as the solution sets get more complicated, exhibiting the set gets more and more complicated. No big surprise there…

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Source : Link , Question Author : 12345678 , Answer Author : John Hughes

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