Simple manipulation of x^2 - 1 = 0, getting x=0 as a solution.

[Crazy Horse 11]

The expression 0/0 is undefined. The expression

\frac{x^2 -1}{x^2 - 1}

is therefore undefined for all $x$ which make the denominator $0$.

0/0=1 may be undefined but (x^2 - 1) / (x^2 - 1) when x=(-1, 1) proves it.

P.j.S .

 

[micromass]

0/0=1 may be undefined but (x^2 - 1) / (x^2 - 1) when x=(-1, 1) proves it.

P.j.S .

If $x=1$, then

\frac{x^2-1}{x^2 -1}

is undefined. Same with $x=-1$.

 

[Crazy Horse 11]

If $x=1$, then

\frac{x^2-1}{x^2 -1}

is undefined. Same with $x=-1$.

(x^2 - 1) / (x^2 - 1) = 1 is that true or false?

 

[micromass]

(x^2 - 1) / (x^2 - 1) = 1 is that true or false?

Depends on the value of $x$.

 

[Crazy Horse 11]

Depends on the value of $x$.

Something divided by itself always = 1. x^2 - 1 is no different.

Therefore for example when x=1 then 0/0=1:

(x^2 - 1) / (x^2 - 1) = 1. This may be undefined but it is proved true by the fundamental mentioned above.

Prove that 0 / 0 = 1 is not true.

 

[micromass]

Something divided by itself always = 1.

Not true.

x^2 - 1 is no different.

Therefore for example when x=1 then 0/0=1:

(x^2 - 1) / (x^2 - 1) = 1. This may be undefined but it is proved true by the fundamental mentioned above.

Prove that 0 / 0 = 1 is not true.

It's essentially a definition. The division "x/y" is defined for all real (and complex numbers) except for y=0. We have left "x/0" undefined by definition.

The reason is simple. What would we define it as?? You could argue that

\frac{x^2 - 1}{x^2 - 1}=1

for all $x$ except $1$ and $-1$. So you could say that it makes sense to let this equality hold also for $1$ and $-1$. That way, we could define $0/0=1$.
The reason we don't do this, is because we could also argue that for $x\notin \{1,-1\}$ that

\frac{2(x^2-1)}{x^2-1}=2

So we want this to hold also for $1$ and $-1$. But that forces $0/0=2$.

In the same way, we could make a perfectly fine argument that $0/0=r$ for any number r! So there is no definition of $0/0$ which unambiguously extends the laws of arithmetic. This is why we define $0/0$ to be undefined.

 

[Crazy Horse 11]

Not true.



It's essentially a definition. The division "x/y" is defined for all real (and complex numbers) except for y=0. We have left "x/0" undefined by definition.

The reason is simple. What would we define it as?? You could argue that

\frac{x^2 - 1}{x^2 - 1}=1

for all $x$ except $1$ and $-1$. So you could say that it makes sense to let this equality hold also for $1$ and $-1$. That way, we could define $0/0=1$.
The reason we don't do this, is because we could also argue that for $x\notin \{1,-1\}$ that

\frac{2(x^2-1)}{x^2-1}=2

So we want this to hold also for $1$ and $-1$. But that forces $0/0=2$.

In the same way, we could make a perfectly fine argument that $0/0=r$ for any number r! So there is no definition of $0/0$ which unambiguously extends the laws of arithmetic. This is why we define $0/0$ to be undefined.

What is your answer for 2(x^2 - 1) / (x^2 - 1) = ?

 

[micromass]

What is your answer for 2(x^2 - 1) / (x^2 - 1) = ?

It is undefined for $x=1$ and $x=-1$. It is $2$ otherwise.

 

[Crazy Horse 11]

It is undefined for $x=1$ and $x=-1$. It is $2$ otherwise.

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0. Please show how you got the answer of 2.

 

[Fredrik]

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0. Please show how you got the answer of 2.

For all real numbers x except 0, we have x/x=1. This implies that for all real numbers x except 1 and -1, we have
$$\frac{2(x^2-1)}{(x^2-1)}=2\cdot 1=2.$$

 

[micromass]

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0.

Wut?

 

[Fredrik]

So you are trying to teach me that 2(3)/3 = 2*1?

Yes, because what you said in post #39 implies that it's not.

2(x^2 - 1) = x^4 - 2x^2 + 1

This is wrong, but now at least I understand what you're doing wrong. $2(x^2-1)$ doesn't mean what you think it does. It's just the number 2 times the result of $x^2-1$. It doesn't mean the same thing as $(x^2-1)^2$.

 

[Crazy Horse 11]

Yes, because what you said in post #39 implies that it's not.


This is wrong, but now at least I understand what you're doing wrong. $2(x^2-1)$ doesn't mean what you think it does. It's just the number 2 times the result of $x^2-1$. It doesn't mean the same thing as $(x^2-1)^2$.

Thank-you for pointing out my oversight.