# Can we?

• Mar 13th 2008, 08:15 AM
Altair
Can we?
Can we simplify the following fraction as :-

$\displaystyle (\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0$

$\displaystyle (x^2 + y^2 +2) * (\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0 * (x^2 + y^2 +2)$
$\displaystyle x^2 + 3xy + 2 = 0$

and then further simplification....

A friend of mine was contradicting this approach and was saying that if the denominator contains variables then you simply can't do it like this. His solution however was that :-

the solution would be that :-

Either, $\displaystyle x^2 + 3xy + 2 = 0$ or $\displaystyle x^2 + y^2 +2 = infinity$

What is the real solution ?
• Mar 13th 2008, 08:24 AM
Moo
Hello,

This can be interesting... But i think the second method is strange. We look for the x and y that verify the equation. No x and y will make the denominator infinite (i mean no determined x nor y)...
• Mar 13th 2008, 08:35 AM
Altair
I also told him that his method looks weird and that intuitively one would just multiply both sides and get the result.
• Mar 13th 2008, 08:49 AM
Moo
The problem is mainly in the fact that 1/infinite is more a convention than an exact result...

Am not sure, but i see the thing like this (Giggle)
• Mar 13th 2008, 09:03 AM
Plato
Quote:

Originally Posted by Altair
Can we simplify the following fraction as :-
$\displaystyle {\color{red}}(\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0$
$\displaystyle x^2 + 3xy + 2 = 0$

The above is certainly correct.
A fraction is zero if and only if its numerator is zero.
One of infinitely many solution is (1,-1).
• Mar 13th 2008, 09:35 AM
Altair
Another thing I told him was that according to his concept of either $\displaystyle {x^2 + 3xy + 2} =0$ or $\displaystyle \frac{1}{x^2 + y^2 +2} = 0$ is not feasible here because it can only be done in case of factors.
• Mar 13th 2008, 09:46 AM
Plato
Quote:

Originally Posted by Altair
Another thing I told him was that according to his concept of either $\displaystyle {x^2 + 3xy + 2} =0$ or $\displaystyle \frac{1}{x^2 + y^2 +2} = 0$ is not feasible here because it can only be done in case of factors.

Well of course.
In $\displaystyle \frac{1}{x^2 + y^2 +2} = 0$ the numerator is not zero, just as I said.
• Mar 13th 2008, 09:55 AM
Altair
What if the denominator tends to infinity ? In this case if both x and y are infinity ?
• Mar 13th 2008, 10:25 AM
Plato
Quote:

Originally Posted by Altair
What if the denominator tends to infinity ? In this case if both x and y are infinity ?

That is a nonsense statement. Infinity is not a number!
Infinity is concept. Some wags even say it is a place where mathematicians hid their ignorance.
• Mar 18th 2008, 01:18 AM
Black Carrot
It's not nonsense, it's a perfectly reasonable question about limits. If you let either x or y approach either positive or negative infinity, the denominator will approach positive infinity. How that effects the overall fraction depends more specifically on how you choose x and y.

On the original question, your solution does leave a step unstated, I'm not sure whether you skipped it or not. In cancelling out the denominator, you make the assumption that it isn't zero. In the real numbers, it can't be, since it has to be at least 2, but on similar problems that could cause mistakes. If there was any chance it could be zero, it would be better to deal with that case explicitly.
• Mar 18th 2008, 01:25 AM
Black Carrot
Going back over the limit question more carefully, I see that there's no well-defined value of the fraction when both x and y tend to infinity. Following two different paths, you can get two different answers. Setting x=y and taking the limit as x->oo gives 2, but setting x=3y and taking the same limit gives 9/5.