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

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  • #26
pwsnafu
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It doesn't matter where we start since inequality is reflexive.
I'm assuming you mean equality not inequality.

Showing a set A does not equal a set B is just as good as showing the set B does not equal the set A.
In order to show ##A \neq B## you need either show ##A \not\subset B## or ##B \not\subset A##. (Or both, but only one is necessary.)

In this situation ##A \subset B## but ##B \not\subset A##. The original post is making the assumption they should be equal. We are explaining why that is invalid.

We were done when we found both sets and found the counter example, 0, in the second equation's solution set.

So, we can conclude that (1) x(x^2-1)=0 and (2) x^2-1=0 are different polynomials with different roots. (:

Do note that because (2) is a factor of (1), we can see that (2)'s solutions form a subset less than or equal to (1)'s solutions. More generally, whatever polynomial you multiply (2) by with result in a polynomial whose solution set contains the solutions of (2). This is precisely why we factor polynomials to find their roots. (or wolfram it)
You're not disagreeing with what I'm saying.

The problem is the original poster did make the assumption that they "should have" the same roots (i.e. that his manipulations were reversible) and ended up in confusion.
 
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  • #27
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Now, in your example, you do the following
[tex] x^2 - 1 = 0~~\Rightarrow~~ x(x^2 - 1)=0[/tex]

This is certainly valid. But it is not reversible. So if we are given ##x(x^2 -1)=0##, then we cannot deduce that ##x^2 - 1 = 0##.

So in your example, you have proven that IF ##x^2 - 1 =0##, then either ##x=0## or ##x=1## or ##x=-1##. This is certainly true. But the converse is not true.
This is really weird to me when I think of the solution sets. Intuitively, I figured {0,-1,1} could imply {-1,1} since the latter is a subset, and that {-1,1} couldn't imply {0,-1,1}. But it's legal to expand the solution set by multiplying so I guess that makes sense, but what about dividing to reduce the solution sets?

What if I started with x(x^2 - 1) = 0, that can imply x^2 - 1 = 0? Then is it not reversible the other way around in this case?
 
  • #28
pwsnafu
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This is really weird to me when I think of the solution sets. Intuitively, I figured {0,-1,1} could imply {-1,1} since the latter is a subset, and that {-1,1} couldn't imply {0,-1,1}.
Your intuition was the other way around.

Edit: to elaborate, the solution sets represent OR not AND.
{-1,1} means "x = -1 OR x = 1".
{-1, 0, 1} means "x = -1 OR x = 1 OR x = 0".
So the former implies the latter.

But it's legal to expand the solution set by multiplying so I guess that makes sense, but what about dividing to reduce the solution sets?

What if I started with x(x^2 - 1) = 0, that can imply x^2 - 1 = 0?
No. x(x^2-1) = 0 implies (x^2-1=0 OR x=0). I put parenthesis there for emphasis. You cannot just throw away a possible solution.

Then is it not reversible the other way around in this case?
##x^2 - 1 = 0 \implies x(x^2-1) = 0## is indeed valid.
 
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  • #29
Yes necessarily!

The starting equation is [itex]x^2-1 =0[/itex]
Thus we know that [itex]x^2-1 =0[/itex]. So dividing by [itex]x^2-1 [/itex] is equivalent to dividing by zero.

Always!



Again from the original equation [itex]x^2-1 =0[/itex] we know that x has to be 1 or -1. So the division is never valid.
do you know what has just been proven by x^2 - 1 = 0? That 0/0=1. see?

(x^2 - 1) / (x^2 - 1) = 1
0 / 0 = 1

P.j.S .
 
  • #30
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do you know what has just been proven by x^2 - 1 = 0? That 0/0=1. see?

(x^2 - 1) / (x^2 - 1) = 1
0 / 0 = 1

P.j.S .
The expression 0/0 is undefined. The expression

[tex]\frac{x^2 -1}{x^2 - 1}[/tex]

is therefore undefined for all ##x## which make the denominator ##0##.
 
  • #31
The expression 0/0 is undefined. The expression

[tex]\frac{x^2 -1}{x^2 - 1}[/tex]

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 .
 
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  • #32
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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

[tex]\frac{x^2-1}{x^2 -1}[/tex]

is undefined. Same with ##x=-1##.
 
  • #33
If ##x=1##, then

[tex]\frac{x^2-1}{x^2 -1}[/tex]

is undefined. Same with ##x=-1##.
(x^2 - 1) / (x^2 - 1) = 1 is that true or false?
 
  • #34
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(x^2 - 1) / (x^2 - 1) = 1 is that true or false?
Depends on the value of ##x##.
 
  • #35
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.
 
  • #36
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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

[tex]\frac{x^2 - 1}{x^2 - 1}=1[/tex]

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

[tex]\frac{2(x^2-1)}{x^2-1}=2[/tex]

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.
 
  • #37
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

[tex]\frac{x^2 - 1}{x^2 - 1}=1[/tex]

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

[tex]\frac{2(x^2-1)}{x^2-1}=2[/tex]

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) = ?
 
  • #38
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3,286
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.
 
  • #39
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.
 
  • #40
Fredrik
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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
$$\frac{2(x^2-1)}{(x^2-1)}=2\cdot 1=2.$$
 
  • #41
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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?
 
  • #42
Fredrik
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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##.
 
  • #43
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.
 

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