Students who are beginning to learn proofs (and some seasoned pros) occasionally commit the error of assuming what they're trying to prove. My question involves assuming at the onset that a solution to a problem exists, then using that existence to get the solution you assumed existed. Two examples:

  1. A common technique for proving that any function $f$ can be written as the sum of an even and an odd function is to assume such functions $O$ and $E$ exist, then note \begin{align*} f(x) &= E(x) + O(x) \\ f(-x) &= E(x) - O(x)\end{align*} and solve for $O(x)$ and $E(x)$.

  2. A nested root such as $\sqrt{2+\sqrt{\vphantom{2^5}3}}$ can be written as $\sqrt{1/2} + \sqrt{3/2}$ by first assuming there exist $A$ and $B$ such that $\sqrt A + \sqrt B$ is equal to the original radical, then solving for $A$ and $B$.

In general, when and how is it legal to assume the solution to a problem exists a priori? Or do the ends justify the means? Or is there some higher structure where the existence of solutions to the two problems above, and others like them, are proven, but I haven't gotten there yet (similar to telling Calc II students that a partial fraction decomposition exists, now find it)?

  • 2
    $\begingroup$ In both of those examples you show what the solution must be provided that it exists, and then you verify that this actually is a solution. In some other examples that I’ve seen the steps are interchangeable: you can show in either order that a solution must exist (e.g., that a certain sequence is convergent) and what it must be if it exists (e.g., what the limit is). $\endgroup$ – Brian M. Scott Dec 8 '14 at 4:20
  • $\begingroup$ @BrianM.Scott Proving a solution exists and then finding the solution I get. The gray area is pointing at a problem and saying "If the solution exists then it would have this property, which leads to that, and ta-da! Here's the solution I assumed existed." I'm comfortable solving things in this manner, I'm just curious about justification. $\endgroup$ – Jon Dec 8 '14 at 5:20
  • $\begingroup$ But that isn't a grey area: it's simply incomplete. It only becomes a complete solution when you either prove that the tentative solution that you've obtained actually is a solution, or prove independently that there is a solution. $\endgroup$ – Brian M. Scott Dec 8 '14 at 6:36
  • $\begingroup$ Your point being, assuming the solution exists will not alter whether or not a solution exists? I guess I can buy that. $\endgroup$ – Jon Dec 8 '14 at 16:49
  • $\begingroup$ That's true, but it's not my point. My point is that you seem to be misunderstanding the arguments in question. In all of them one does prove that a solution exists, either independently of finding it, or by verifying that one's tentative solution actually is a solution. $\endgroup$ – Brian M. Scott Dec 8 '14 at 18:10

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