# A tricky but silly doubt regarding the solutions of $x^2/(y-1)^2=1$

Motivation :

I have been confused with some degree 2 equation. I suddenly came across a simple equation and couldn't get the quintessence behind that. I have an equation $$\dfrac{x^2}{(y-1)^2}=1 \tag{1}$$ and I was looking for its solutions. It was asked by some kid ( of $9$nth standard ) to me. I did some manipulation and got $$x^2=(y-1)^2 \tag{2}$$ finally. And one can see that $(0,1)$ satisfies the Equation $[2]$ well. But I was happy, and within small time, I realized that the same solution set can't satisfy the equation $[1]$ . If you substitute $(0,1)$ in $[1]$ you get $\dfrac{0}{0}=1$ which is wrong.

The answer that convinced me finally :

We can see the same equation as this $x^2. \dfrac{1}{(y-1)^2}=1$ . We know that the set of integers form a ring. So the product of two numbers is one if one number is the inverse of other number. '$1$' present on the R.H.S is the identity element. So the product of the entity with its inverse always gives us the identity.

So when $x$ is $0$, the $0$ doesn't have an inverse in the integers. So the case is to be emitted.

Still persisting questions :

But the thing that makes me surprise is that the Wolfram Alpha gives me this solution .

In the picture you can clearly see that they both intersect at $(0,1)$ . But what is that confusion ? We omitted that solution, but in fact $(0,1)$ is the intersection of the two lines.

Questions that are to be answered by learned people :

• What is the value of term $\dfrac{0}{0}$ ? Isn't it $1$ ?

• Why the solution pair $(0,1)$ satisfies $x^2=(y-1)^2$ but not $\dfrac{x^2}{(y-1)^2}=1$ ? We know that both of them are manifestations of each other in a simple manner.

• If we need to omit that solution, why do the lines intersect at $(0,1)$ ?

Thank you everyone for giving your time.

• Omit, not emit. 0/0 is undefined in $\Bbb R$ and $\Bbb C$. People and machines often ignore removable singularities (by which I mean fill in the missing point without discussion).
– anon
Commented Aug 13, 2012 at 6:15
• If it's "tricky", then that precludes it being "silly", no? Commented Aug 13, 2012 at 7:38
• @RahulNarain : Thank you for your edit.
– IDOK
Commented Aug 13, 2012 at 8:26
• @anon : Thank you, I always have trouble with homophones.
– IDOK
Commented Aug 13, 2012 at 8:27

The equations $$x^2=(y-1)^2\tag{1}$$ and $$\frac{x^2}{(y-1)^2}=1\tag{2}$$ do not have the same solution set. Every solution of $(2)$ is a solution of $(1)$, but $\langle 0,1\rangle$ is a solution of $(1)$ that is not a solution of $(2)$, because $\frac00$ is undefined.

The reason is that $(1)$ does not imply $(2)$. Note first that $(2)$ does imply $(1)$, because you can multiply both sides of $(2)$ by $(y-1)^2$ to get $(1)$. In order to derive $(2)$ from $(1)$, however, you must divide both sides of $(1)$ by $(y-1)^2$, and this is permissible if and only if $(y-1)^2\ne 0$. Thus, $(1)$ and $(2)$ are equivalent if and only if $(y-1)^2\ne 0$. As long as $(y-1)^2\ne 0$, $(1)$ and $(2)$ have exactly the same solutions, but a solution of $(1)$ with $(y-1)^2=0$ need not be (and in fact isn’t) a solution of $(2)$.

As far as the graphs go, the solution of $(1)$ is the union of the straight lines $y=x+1$ and $y=-x+1$. The solution of $(2)$ consists of every point on these two straight lines except their point of intersection.

• Thank you. It was useful and good. +1.
– IDOK
Commented Aug 13, 2012 at 8:28
• @Iyengar: You’re welcome; I’m glad it helped. Commented Aug 13, 2012 at 8:31

Wolfram Alpha sometimes does that. It's a little bit finicky. You can see that if we ask it for the domain of $z=\frac{x^2}{(y-1)^2}$, it does indeed have us omit the case where $y=0$. I've observed this behaviour lots—for instance if you ask it to plot $y=\frac{(x-1)^2}{x-1}$, you get a similar result: it fills in the hole which should exist at $x=1$. However when you ask it for the domain of that function, it gives the correct domain. I think that this is evidence of a tradeoff between having something which interprets natural language, and having something that's precise.

• It was really good one. +1
– IDOK
Commented Aug 13, 2012 at 8:30

You cannot multiply both side of an equation with $0$. When you multiply your equation by $(y-1)^2$, you are assuring that you are not taking $y=1$ as a valid solution.

• Thank you, it was nice .+1.
– IDOK
Commented Aug 13, 2012 at 8:30

Two points which might help.

• $0/0$ is undefined
• you are not allowed to multiply an equation by 0 (if you do so then you increase the set of solutions)

So [1] is equivalent to [2] iff $y\neq 1$.

The solutions to your equation [1] thus read $$x =\pm (y-1), \qquad y\neq1.$$

• Thank you for your information. +1.
– IDOK
Commented Aug 13, 2012 at 8:29