Assuming a Solution Exists

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)?

• 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). – Brian M. Scott Dec 8 '14 at 4:20
• @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. – Jon Dec 8 '14 at 5:20
• 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. – Brian M. Scott Dec 8 '14 at 6:36
• Your point being, assuming the solution exists will not alter whether or not a solution exists? I guess I can buy that. – Jon Dec 8 '14 at 16:49
• 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. – Brian M. Scott Dec 8 '14 at 18:10