# Lost Solutions

By Susam Pal on 05 Sep 2010

## Algebraic Equation

Let me start this post with a simple equation: $x(x - 2) = 3x.$

Let us first solve it in a flawed manner.

Dividing both sides by $$x$$ we get $x - 2 = 3.$

Adding $$2$$ to both sides we get $x = 5.$

Indeed, $$x = 5$$ is a solution of this equation. If we substitute $$x$$ with $$5$$ in the equation we get $$15$$ on both sides, so it confirms that this solution holds good.

However, $$x = 0$$ is also a solution of this equation. If we substitute $$x$$ with $$0$$, we get $$0$$ on both sides. But we did not get this solution when we solved the equation in the flawed manner above. Why did we lose this solution?

We lost this solution in the step where we divided both sides by $$x$$. In doing so, we made the wrong assumption that $$x \ne 0$$. If we want to consider the possibility of $$x = 0$$ as a solution, then we cannot divide both sides of the equation by $$x$$. Division by $$0$$ is undefined in arithmetic. This is why the typical way to solve such an equation is to bring all terms involving $$x$$ to one side of the equation and then solve it. For example, \begin{align*} x(x - 2) = 3x & \iff x(x - 2) - 3x = 0 \\ & \iff x^2 - 2x - 3x = 0 \\ & \iff x(x - 5) = 0 \\ & \iff x \in \{ 0, 5 \}. \end{align*}

## Mathematical Fallacy

Often fallacies or spurious proofs of contradictions is formed by sneaking such a division by $$0$$ into the proof. Here is one I could think of.

Let $$a = b$$. Then \begin{align*} a + b = 2a & \iff a + b - 2b = 2a - 2b \\ & \iff a - b = 2(a - b). \end{align*}

Dividing both sides by $$a - b$$, we get $1 = 2.$

We arrived at this contradiction because we incorrectly divided both sides of the equation by $$a - b$$ even though $$a - b = 0$$.

## Differential Equation

We need to be careful about division by zero while solving differential equations too.

Consider this differential equation: $y' = x^2 y^2.$

Dividing both sides by $$y^2$$, the variables separate and we get $y^{-2} y' = x^2.$

Now solving this becomes a routine task. By solving this differential equation, we get $y^{-1} = \frac{x^3}{3} + c$ where $$c$$ is the constant of integration. Taking the negative reciprocal of both sides, we get $y = \frac{-3}{x^3 + 3c} = \frac{-3}{x^3 + c'}.$ where $$c' = 3c$$.

This is indeed a valid solution of the differential equation. However, we have lost a solution. The lost solution is $$y = 0$$. We can verify that with $$y = 0$$, we get $$0$$ on both sides of the differential equation. We lost this solution by dividing both sides of the differential equation by $$y^2$$.

Whenever we need to divide both sides of a differential equation by a function to separate the variables, we also need to perform an additional step of verifying whether the zero function is a solution of the equation. In this case $$y = 0$$ is indeed a solution of the differential equation.

In general, when we need to divide both sides of the equation by a function of $$y$$, say $$f(y)$$, to separate the variables, we need to perform an additional step of verifying whether $$f(y) = 0$$ is a solution of the differential equation.