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I have an equation to solve for y:
$$\frac{y^2}{y}=1$$ Normally, I would cancel out one $y$ and get $y=1$ as a single solution.
But If I think of it as quadratic equation
$$y^2=y$$ $$y^2-y=0$$ $$y(y-1)=0$$ $$y=0 \space \text{or} \space y=1$$ to have two solutions.
But when I put $y=0$ in original equation, I get $\frac{0}{0}$, so is $y=0$ a solution or not ?
If yes, then I get $\frac{0}{0}$.
If no, then how come this quadratic equation has $1$ solution ?

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up vote 3 down vote accepted

This equation has two solutions, $y=0$ and $y=1$:

$$y^2=y \tag{1}$$

This equation has one solution, $y=1$:

$$\frac{y^2}{y} = 1 \tag{2}$$

The reason that it has one solution is that either of the operations "Cancel a $y$ from the top and bottom of the fraction" and "Multiply both sides of the equation by $y$" are only valid when $y\neq 0$. In particular, if you want to manipulate (2) to look like (1), you first have to assume that $y\neq 0$, which rules out one of the solutions of the quadratic equation.

A more sophisticated answer realises that when we are asked to "solve" an equation, what we are really doing is looking for a root of a particular function (i.e. a value of the argument at which evaluating the function gives zero). In the case of (1), the function is $f(y)=y^2-y$, and in the case of (2) the function is $f(y)=y^2/y-1$.

Now, functions have to have a domain. In the case of (1) the domain is $\mathbb{R}$ (the real numbers) which includes both 0 and 1. In the case of (2), the domain is $\mathbb{R}-\{0\}$ (the real numbers without 0) which doesn't include 0.

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Thank you very much for the answer. – Happy Mittal Jun 28 '12 at 8:17
But can't I write $y^2=y$ as $\frac{y^2}{y}=1$ without any constraint ? – Happy Mittal Jun 28 '12 at 9:05
When you divide by $y$, you have to make the assumption that $y\neq 0$. – Chris Taylor Jun 28 '12 at 9:07
OK. Thanks a lot. – Happy Mittal Jun 28 '12 at 9:12

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