Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.