Dividing a number by zero

I would say that it is mathmatical proof that there is something wrong with claiming a=b. I never seen any algebra books that claimed that $$a^{2} = ab$$ from then on it is just nonsense.

When you substitute equations into other equations, then it makes the variable have to agree with all of those equations. So then by saying that a=b, and then factoring out a zero from both sides, the solution to "a" and "b" would then have to be zero. It is the only number that you could factor out a zero and not change its value. Everything multiplied by zero is zero, so if it had a zero multiplied in it and you factored that out, then that number would have to be zero.

I think this is why they say "a" does not equal "b", and they cannot be equal to zero. The algebra just wouldn't work out anymore, as shown. I think this is a good example of that. It is never said that "a" can equal "b" in any fundemental math principals.
It's simple.

A = B (FINE)

Then multiply both sides by A, so we get A^2 = AB (FINE)

Subtract B^2 from both sides, so we A^2 - B^2 = AB - B^2 (FINE)

Divide both sides by (A-B), so we get A + B = B (NOT FINE!!)

If a = b, then a[sup]2[/sup] - b[sup]2[/sup] = 0. The rest of emil's equations skip over this; the equation a + b = b is true if a and b are both 0.
Right, zero is the only number that can satisfy all of the equations. So then just by saying that a=b, and then putting them into a quadradic formula, makes the value of "a" and "b" become zero. So then in quadradic formula, "a" cannot equal "b" and they cannot be zero, zero has the ability to make equations like a+b=b become true when any other number wouldn't be able to satisfy the equation.

It's simple.

A = B (FINE)

Then multiply both sides by A, so we get A^2 = AB (FINE)

Subtract B^2 from both sides, so we A^2 - B^2 = AB - B^2 (FINE)

Divide both sides by (A-B), so we get A + B = B (NOT FINE!!)
That is not fine. Since, a=b then you cannot put it into a quadradic formula. If you did then $$a^{2} - b^{2}$$ would equal zero, and then $$ab - b^{2}$$ would equal zero. So then at that time you just have 0 = 0. You then factor a zero from both sides and divide by zero. If you wanted to find the solution to "a" and "b", you could have stopped right when you found $$a^{2} - b^{2} = ab - b^{2}$$, the solution is 0 = 0.

That is not fine. Since, a=b then you cannot put it into a quadradic formula. If you did then $$a^{2} - b^{2}$$ would equal zero, and then $$ab - b^{2}$$ would equal zero. So then at that time you just have 0 = 0. You then factor a zero from both sides and divide by zero. If you wanted to find the solution to "a" and "b", you could have stopped right when you found $$a^{2} - b^{2} = ab - b^{2}$$, the solution is 0 = 0.
...or 1=1, or 2=2, or 3=3, or -100=-100. It's almost like there exists a solution for every time the first number equals the second number!

The problem resides in the step that I pointed out. I don't know what you're talking about regarding the quadratic formula, etc.

...or 1=1, or 2=2, or 3=3, or -100=-100. It's almost like there exists a solution for every time the first number equals the second number!

The problem resides in the step that I pointed out. I don't know what you're talking about regarding the quadratic formula, etc.
I may have said quadratic formula by mistake, the point is that in the equations $$a^{2} - b^{2}$$ and $$ab - b^{2}$$, if "a" and "b" are equal to each other then they could be no other value other than zero. That is why "a" and "b" cannot be equal to each other in this kind of situation.

Think about it, one squared minus one squared would be zero. One times one minus one squared would be zero. Zero doesn't play by the same rules as other numbers, so then it could only lead to wrong answers. A variable that is a zero would not act the same as any other variable. Both sides of the equation would already be nothing, so anything more than that would just be working on nothing even more still. It would just be a lot to do about nothing. I think that is why they put those mathmatical rules into place, if a = b, then $$a^{2} - b^{2} = 0$$ and $$ab - b^{2} = 0$$. Then neither of them could really say anything meaningful about something else.

Those are not "equations" those are "expressions."

$$a^2 - b^2 \; = \; ab-b^2$$ is an equation which is true if a = b OR if a = 0. Either condition is sufficient.

Those are not "equations" those are "expressions."

$$a^2 - b^2 \; = \; ab-b^2$$ is an equation which is true if a = b OR if a = 0. Either condition is sufficient.
Ah yes, that was the word I was looking for, it has really been a long time since I heard that one. I don't know what you mean by, "either condition is sufficient". I would think that it is the only values that could satisfy those conditions, but zero and a=b is said to not be valid in either of these expressions.

Prof., you are wrong. There is no need for "a" to equal zero. "a" can equal any number. It is only in the (a-b)/(a-b) step that things become undefined.

Prof., you are wrong. There is no need for "a" to equal zero. "a" can equal any number. It is only in the (a-b)/(a-b) step that things become undefined.
"a" can equal any number, but like I said, zero and a=b are the only conditions where the equations would be valid, and those expressions are no longer valid when a=0 and b=0 and a=b.

They say any variables can be any number, but they can only be certain numbers in certain equations in order for them to be true. So all I did really is think what numbers the equations are true for, so then you can know what the value of "a" and "b" actually are even without constants in this case. The equations simply do not allow "a" and "b" to be any number when they are put together in this fashion. They could have been any number when it was said $$a^{2} = ab$$ and a = b, but then when it was put into the next expressions, the values of "a" and "b" could only be zero. The expression would limit the values of "a" and "b" since $$a^{2} - b^{2} = 0$$ and $$ab - b^{2} = 0$$, so then from then on "a" and "b" would have to be zero. They could no longer be any number once these expressions where set equal to each other.

Remember the difference between an expression and an equation.
This is an expression:
$$a^2 + b^2$$
This expression can be evaluated for all values of a and b.

This is an equation. It equates two expressions:
$$a^2 - b^2 \; = \; 0$$
It is true for some values of a and b. In this case, for any value of a, there are two values of b that make the equation true.

They could have been any number when it was said $$a^{2} = ab$$ and a = b, but then when it was put into the next expressions, the values of "a" and "b" could only be zero.
No. The next equation is:
$$a^2 - b^2 \; = \; ab-b^2$$
In this equation, a and b can be anything, as long as a=b.
(considered in isolation, the equation is also true for all values of b when a=0, but it's already given that a=b.

Layman said:
The expression would limit the values of "a" and "b" since $$a^2 - b^2 \; = \; 0$$ and $$ab-b^2 \; = \; 0$$, so then from then on "a" and "b" would have to be zero.
The two expressions both evaluate to zero, but that does not imply that a=0 or b=0.

The two expressions both evaluate to zero, but that does not imply that a=0 or b=0.
It does if you factor it, you would be saying that zero times whatever is left is equal to zero. You could only take a zero out of zero, so then it couldn't be any number at that point. You wouldn't factor out a zero from one and say that it is still one. If you did anything with it from that point "a" and "b" would have to equal zero. The entire expression is equal to zero, so then any other operation would force it to become zero from that condition, since the expression itself is equal to zero.

Otherwise it would be like saying that you could factor zero out of a whole number, that is something that you cannot do.

Layman,
One of the factors of the expressions on both sides of the equation is (a-b), which we know is zero.
Right?

And I'm sure you'll agree that if a=10 and b=10 (for example), then (a-b) is still zero.
Right?

And this will make both sides of the equation equal to zero.
Right?

I'm not sure exactly where your confusion is, but perhaps you'll see if you plug some numbers in.

Plug in a=10 and b=10 into $$a^2-b^2 = ab-b^2$$.
You'll end up with 0=0, which I'm sure you'll agree is a valid equation.

I missed this from yesterday.
$$\lim_{a \to 0} \frac{0}{a} = 0$$

$$\lim_{a \to 0} \frac{a}{0} = undefined$$

$$\lim_{a \to 0} \frac{2a}{a} = 2$$

In the first example, I think 0 divided by any number is 0, so then the answer to that one would be zero. This seems to show that the zero comes from a different variable than "a", so then it would seem that zero's coming from different sources cannot be canceled.
That's correct. This is why it's important to be clear what the limit expression is.
You can't just say "The limit of 0/0" because it doesn't tell you where those zeros came from.

In the secound example, any number divided by zero is undefined, so then I don't think it would have an answer since the limit would never say "a" is actually 0. That does seem odd that you would get a different answer in this limit even though it seems to really be saying that it is searching for the limit of the same division problem $$\frac {0}{0}$$. In this case it is undefined and in the other it is 1. But since this example is undefined, I think the real solution for $$\frac {0}{0}$$ would be 1. The problem has only been rearranged so that it is undefined. In the example $$\lim_{a \to 0} \frac {a}{a}$$ you never actually have to divide by zero.
The point is that different limit expressions that approach 0/0 have different answers. There is no single answer to "The limit of 0/0".

In the third, any number divided by itself would be one, so every answer would then be 2. It is as simple as canceling out the "a" in the equation even if "a" approuches zero. You shouldn't allow yourself to think that just because "a" approaches zero, that it would give you a different answer rather than just canceling them out, you can cancel $$\frac {0}{0}$$ in limits and that should always be 1
No, this example show that approaching a limit of 0/0 does not always cancel to 1. You need to know how the numerator and denominator approach zero. "The limit of 0/0" is just not clearly defined enough to give it a definite value. You need to spell out the limit expression in full. No shortcuts.

What happens if you divide infinity by zero?

What happens if you divide infinity by zero?
:spank: Division by zero is forbidden!

it appears if you try to divide infinity by 0 it will still be infinity since there's nothing ( 0 ) to divide by

What happens if you divide infinity by zero?

$$\frac{0}{8}=0 \frac{8}{0}=8 \frac{0}{0}=R \frac{8}{8}=R (0)(8)=R$$

where R is some real number. I may be wrong though.

I used 8 in place of ∞.

Ha, pretty funny.

I'm not sure exactly where your confusion is, but perhaps you'll see if you plug some numbers in.

Plug in a=10 and b=10 into $$a^2-b^2 = ab-b^2$$.
You'll end up with 0=0, which I'm sure you'll agree is a valid equation.
I see where I got you confused now. So then it would follow that, $$(10 + 10 ) (10 - 10 ) = 10 ( 10 - 10 )$$ , by doing this would have factored out a zero out of each side of the equation. It would be the same as saying $$(10 + 10) 0 = 10 ( 0 )$$, but you cannot factor out a zero from a 10! So then "a" and "b" can no longer be 10, because the only time this would be true is when "a" and "b" is equal to zero. So then you would have $$(0 + 0 ) (0 - 0 ) = 0 ( 0 - 0 )$$, the value of "a" and "b" being zero is the only number that would be true by factoring out a zero.

The point here I am trying to make as that the mathmatical rules that say "a" is not equal to "b", and "a" and "b" are not equal to zero are in the right place where you would end up having these expressions. Not sooner, or later, when you have those expression they are no longer valid. Any type of multiplication or division or factoring would force them to become zero, and they wouldn't work for any other number other than zero.