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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 . Image

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.

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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 Aug 13 '12 at 6:15
    
If it's "tricky", then that precludes it being "silly", no? –  J. M. Aug 13 '12 at 7:38
    
@RahulNarain : Thank you for your edit. –  Iyengar Aug 13 '12 at 8:26
    
@anon : Thank you, I always have trouble with homophones. –  Iyengar Aug 13 '12 at 8:27
    
@J.M. : Yes sir. But tricky for me, silly for you ( learned people ) , is the meaning I wanted to express. –  Iyengar Aug 13 '12 at 8:37
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4 Answers

up vote 7 down vote accepted

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.

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Thank you. It was useful and good. +1. –  Iyengar Aug 13 '12 at 8:28
    
@Iyengar: You’re welcome; I’m glad it helped. –  Brian M. Scott Aug 13 '12 at 8:31
    
It was really useful. I never thought it that way. Your explanation was clear and clean. It reflects your experience and grip in mathematics. Thank you again. –  Iyengar Aug 13 '12 at 8:32
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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.

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It was really good one. +1 –  Iyengar Aug 13 '12 at 8:30
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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.

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Thank you, it was nice .+1. –  Iyengar Aug 13 '12 at 8:30
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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 $$ x =\pm (y-1), \qquad y\neq1.$$

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Thank you for your information. +1. –  Iyengar Aug 13 '12 at 8:29
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