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Suppose I have an equation $f(x) = 0$, and I know the solution exists. The solution is $x = a$ (I don't know the actual solution to begin with, I just know it exists). I perform a series of $irreversible$ operations and arrive at the answer:

$f(x) = 0 \implies \cdots = \cdots \implies \cdots = \cdots \implies x = a$

which is just a big $f(x) = 0 \implies x = a$.

Since the operations are irreversible, we cannot go in the other direction and claim $ x = a \implies f(x) = 0$

Nevertheless, we know that the solution is $a$, so if we plug in $a$ into $f$ we get zero. But hold on a minute, the statement "if we plug in $a$ into $f$ we get zero" really means $x = a \implies f(x) = 0$; we just went in the other direction, which is something we said we couldn't do!

What is wrong here? Does the fact that $f(x) = a$ has a solution imply that there must be a reversible series of steps between $f(x) = a$ and $x = a$?

Edit: A condensed version

1) We know $f(x) = 0$ has a solution.

2) We can prove $f(x) = 0 \implies x = a$ using irreversible steps.

3) Now that we computed what $a$ actually is, by 1) we know that $x = a \implies f(x) = 0$ is true. Does it follow now that there exists another method to solve the equation using purely reversible steps?

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    `I know the solution exists and it is x=a` What is that supposed to mean exactly? Taken at face value, you know a priori that $x=a$ is *the* root, so the rest of the question becomes immaterial. – 2017-02-01
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    @dxiv Sorry, I mean that all I know is a root exists; its value is $a$, but I don't know that to begin with. – 2017-02-01
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    @vagoberto I can't think of such an example for equations, but this can happen for inequalities. For example, suppose you know that $x > |a| + |b|$. Using the triangle inequality, you can conclude that $x > |a+b|$. But if you start from $x > |a+b|$, you cannot go backwards and conclude $x > |a| + |b|$. – 2017-02-01
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    `all I know is a root exists` Do you know it is unique? The question says "***the*** solution", which would tend to imply that. – 2017-02-01
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    @dxiv Yes for simplicity's sake, we can assume we know there is one unique root. – 2017-02-01

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This is impossible to answer in any rigorous way without having a precise definition of the sort of "steps" you want to consider. But basically, the answer is no.

It happens all the time in math that we can separately prove $P\Rightarrow Q$ and $Q\Rightarrow P$, but there is no direct way to prove $P\Leftrightarrow Q$ in both directions at once. So there is no reason to expect that just because you can prove both $f(x)=0\Rightarrow x=a$ and $x=a\Rightarrow f(x)=0$, you can prove $f(x)=0\Leftrightarrow x=a$ by a series of reversible steps. After all, any mathematical statement at all can be encoded in equations like this by defining your function $f$ appropriately.

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    Thank you very much for your answer. I think it pretty much answers the question, but tomorrow I will need to exaime the answer in the context of the problem I was solving (I am very tired now) – 2017-02-01
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$f(x) = 0 \Longrightarrow x = a$ tells you that if $x$ happens to be a root of $f$ then $x$ must be equal to $a$. In other words, the only possibility for a root of $f$ is $a$. However, this implication does not guarantee that $a$ is in fact a root. That's why you need to plug $a$ into $f$ and see whether or not you get zero.

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    But I am already given that $f$ has a root; so if we proved $f(x) = 0 \implies x = a$ then $a$ $must$ be a root, no? – 2017-02-01
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    Well then yes. Let $\mathcal{S}$ be the logical implication "$f(x) = 0 \Longrightarrow x = a$ for any $x$ in the domain of $f$". Next suppose that $f$ has a root - say it is the element $x_0$ in the domain of $f$. By definition of what it means to be a root, it must be that $f(x_0) = 0$. Then by the logical implication $\mathcal{S}$, it follows that $x_0 = a$. Thus $a$ is a root – 2017-02-01
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You have derived a necessary condition for $f(x)=0$, namely that $x=a$. It is, however, still possible that $a$ is not actually a root of $f$, i.e. it need not be the case that $x=a \Rightarrow f(x)=0$ .

For example, $a$ might be outside of the original domain of $f$ and $f(a)$ might not even be properly defined. Imagine $f(x)=x$ where $f: [1,2] \rightarrow \mathbb{R}$ and look for $f(x)=0$.

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    But I am already given that $f$ has a root; so if we proved $f(x) = 0 \implies x = a$ then $a$ $must$ be a root, no? – 2017-02-01
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    You didn't say in the original question that you assumed the existence of a root. By making all these additional assumptions, you seem to be assuming reversibility. If there is a root (this is a separate statement from your derivation and thus not part of it) and if additionally $f(x=a)=0$, then $a$ is a root, no matter how you derived $a$. Do you have an example where a root exists (i.e. you can proof existence of a root) and you can create an irreversible solution to $f(x)=0$? – 2017-02-01
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    Also see my comment on dxiv's answer. – 2017-02-01
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    I have edited the question a few times, but I did mention in the original version that I know a root exists. To address your question, I could not find such an example; hence the last question in the original post: "Does the fact that $f(x)=a$ has a solution imply that there must be a reversible series of steps between $f(x)=a$ and $x=a$?" – 2017-02-01
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[ EDIT ] Original answers remains in the edit history, but after clarifications given in the comments, the question appears to be (paraphrased):

Assuming that $f(x)=0$ has a unique solution, and provided we prove that $f(x)=0 \implies x=a\,$, is that sufficient to prove that $a$ is the unique solution, or is it still necessary to verify that $a$ does in fact satisfy $f(a)=0\,$?

The answer to this question is "the former" i.e. $f(x)=0 \implies x=a\,$ is sufficient to prove that $a$ is the unique root.

That's because any series of irreversible (but otherwise correct) operations can introduce additional roots, but can never discard a legitimate root. Therefore if $f(x)=0 \implies x \in A=\{a,b,c,\cdots\}$ then those $a,b,c,\cdots$ values are not all guaranteed to be roots of the original equation, but all the actual roots are guaranteed to be in $A$.

In the given case, however, the potential solution set $A=\{a\}$ contains a single value and, since it's been assumed that the equation has in fact one unique root, then that root must be $a$.

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    OP commented in response to our answers that he wants to assume the existence of such a root prior to the derivation, in which case this example doesn't work. I think the assumption of the root + the derivation of a necessary condition, does in fact imply that you know $f(a)=0$ but this is because the additional premise is circumventing the one-sided derivation via a modus ponens argument. I.e. We know that "If $f$ has a root, it is $a$" and we know "$f$ has a root" (premise), thus we do in fact know $a$ is a root of $f$. – 2017-02-01
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    @TimonG. I am as confused as anybody about the finer points of the question. I just posted a request for clarification under the OP. – 2017-02-01
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    When you plug $a$ into $f(\cdot)$ and get $f(a)=0$ you get merely $f(a)=0$---not a general $f(x)=0$ for arbitrary $x$. – 2017-02-01
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    dxiv, the tagging system may be broken I believe. Just letting you know that I responded to your comment above. Please let me know if there is anything else which needs clarification. I will try to condense it here a little bit: 1) We know $f(x) = 0$ has a solution. 2) We can prove $f(x) = 0 \implies x = a$ using irreversible steps. 3) Now that we computed what $a$ actually is, by 1) we know that $x = a \implies f(x) = 0$ is true. Does it follow now that there exists another method to solve the equation using purely reversible steps? – 2017-02-01