# Substituting an equation into itself, why such erratic behavior?

Until now, I thought that substituting an equation into itself would $always$ yield $0=0$. What I mean by this is for example if I have $3x+4y=5$, If I substitute $y=\dfrac {5-3x}{4}$, I will eventually end up with $0=0$. However, consider the equation $\large{\sqrt {x+1}+\sqrt{x+2}=1}$ . If we multiply by the conjugate, we get $\dfrac {-1}{\sqrt{x+1}-\sqrt{x+2}}=1$, or $\large{\sqrt{x+2}-\sqrt{x+1}=1}$. Now we can set this equation equal to the original, so $\sqrt{x+2}-\sqrt{x+1}=\sqrt {x+1}+\sqrt{x+2}$ , and you get $0=2 \sqrt{x+1}$ which simplifies to $x=-1$ , which is actually a valid solution to the original! So how come I am not getting $0=0$ , but I am actaully getting useful information out of this? Is there something inherently wrong with this? Thanks.

## 4 Answers

You didn't actually substitute anything (namely a solution for $x$) into the original equation; if you would do that the $x$ would disappear. Instead you combined the equation with a modified form of itself to obtain a new equation that is implied by the original one; the new equation may or may not have retained all information from the original one. As it happens the new equation has a unique solution and it also solves the original equation; this shows the new equation implies the original one, and you did not in fact loose any information.

If you consider the operation of just adding a multiple of an equation to itself, you can see what can happen: in most cases you get something equivalent to the original equation, but if the multiple happened to be by a factor$~-1$ then you are left with $0=0$ and you have lost (in this case all) information contained in the equation.

An equation will simplify to the tautology $0=0$ only if the equation is true for all values of its variables. This happens in your first case, because you're basically going:

This equation will be true if and only if $y$ is equal to this, and no matter what $x$ is. So now, if $y$ is equal to that, will the equation be true?

And the answer is of course: yes, no matter what $x$ is.

The question is therefore: why does your second manipulation not result in something that is true for every value of $x$?

You start with

$$\sqrt {x+1}+\sqrt{x+2}=1$$

Step 1: You multiply by the conjugate to obtain

$$(\sqrt {x+1}-\sqrt{x+2})(\sqrt {x+1}+\sqrt{x+2})=\sqrt {x+1}-\sqrt{x+2}$$ $$(x+1)-(x+2)=\sqrt {x+1}-\sqrt{x+2}$$ $$-1=\sqrt {x+1}-\sqrt{x+2}$$ $$1=\sqrt{x+2}-\sqrt {x+1}$$

Observing that you can set this equal to the original expression, you do so (Step 2), and are surprised to obtain something that is not a tautology, ie, not true for all $x$.

But watch more closely what happens in step 1. Why did we end up with that $1$ on the left hand side? Merely because the $2$ and the $1$ in the original expression happened to produce a 1. This was a coincidence. Had the original equation been:

$$\sqrt {x+5}+\sqrt{x+2}=1$$

That wouldn't have happened, and you wouldn't have been able to set the result equal to the original expression. In other words, step 2 was not something that would have worked for all $x$. It only worked for this particular case, for $x$ solving this particular equation. Therefore everything deduced from step 2 are equations equivalent to the original, non-tautologous, equation. In particular, you even happened to get an equivalent equation that made one of the solutions obvious.

Another way of seeing this is that your solution involved finding that any radical expression satisfying your equation must be equal to its conjugate. You then exploited that fact, which is not at all true in general, to get the solution.

• This is lovely! Do you happen to know whether the statement "an equation simplifies to $0 = 0$ only if it's true for all variable values" can be strengthened to an if and only if? It seems obvious that it can, but I'm not sure how I'd go about proving it... Nov 9 '15 at 7:37
• @Vectornaut It depends what you mean by "simplify". If "simplfy" means "apply distributivity, commutativity, associativity, and add or multiply both sides by the same thing", then probably not. You couldn't simplify $e^{a+b}=e^ae^b$ that way, for instance. Nov 9 '15 at 8:07

I was going to comment but it got too long.

In your second procedure, you did not substitute the inverse into the original. (I see what you mean by expecting $0=0$ here, in the sense that $f(f^{-1}(x))=x$ so that one can subtract and get $0=0$)

Instead you carried out a valid re-write of the left side by multiplying top and bottom by the conjugate expression, which just means left side was multiplied by 1, not changing it. Assuming the quantity given by the conjugate is not zero (OK in your case) this step is perfectly valid, and since the original equation has not changed the resulting transformed equation should have exactly the same solution(s) as the original.

So by this kind of process you should always expect solutions of the transformed equation, if the original one has solutions.

• The step of multiplying with the conjugate is always valid, even if the conjugate should be$~0$; only in the latter case the multiple formed does not imply the original equation. However since one is combining that multiple with the original here, it is irrelevant whether the multiple implies the original; it is more interesting whether the combination of the two still implies the original (it does here). Jun 22 '13 at 7:55
• @MarcvanLeeuwen No, it is not "multiplying by the conjugate" that was done in the OP, but rather multiplying one side by the ratio of the conjugate over itself. This cannot of course be done when the conjugate is zero. One can multiply both sides of an equation by a zero conjugate, but that just produces $0=0$. In this case clearly the new equation holds for all $x$ and doesn't imply the original one (unless it began as $0=0$). Jun 22 '13 at 16:58

Notice the first equation has two variables, whereas your second equation has one variable. The performed substitution was not an authentic swap of variables; the equation obtained is essentially $$\dfrac {\sqrt{x+1}-\sqrt{x+2}}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac {{(\sqrt{x+1}-\sqrt{x+2})}^2}{1}$$

for which we get $\dfrac {1}{1}=1$, thus obtain 0=0.

As a bonus, if in the process you divided something by an expression that evaluates to zero, a no solution situation may result. Allegedly Einstein has done that :)