They always cancel to one if they are the same variable who's limit is approuching zero. Anytime your doing a limit and you see the expression $$ \frac {a}{a} $$ and "a" is approuching zero, you can just cancel them like you would any other variable, and that will give you the correct answer for the limit. If there is another constant like zero in the equation that is not the same variable then it will be zero or undefined.
Dividing a number by zero
- Thread starter chikis
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Why am I starting to get the feeling that people just don't believe in limits? Zero is a constant. Why would you even start out with zero in an equation? If you did it would most likely be zero or undefined. You could solve for a variable and then find out that it is zero, it would then be a constant. This is really basic stuff, I can do analysis of these variables in my head.
You don't "believe" things in math. You assert axioms and use logic to derive theorems from those axioms.
Using the axioms of arithmetic, there is no definition of $$\frac{1}{0}$$ and even if you add such an axiom, the rigidity of the axioms of arithmetic forbid $$\frac{1}{0} - \frac{1}{0}$$ or $$\frac{0}{0}$$ from being a sensible expression. This you do not contest.
Using the axioms of analysis, $$\lim_{x\to 0} \frac{x}{x} = 1$$, but there is no axiom to assert that prefers $$\frac{a}{b} = \lim_{x\to 0} \frac{x + a}{x + b}$$ to $$\frac{a}{b} = \lim_{x\to 0} \frac{15x + a}{x^3 - x + b}$$, and therefore no axiom or theorem that asserts that $$\frac{0}{0}$$ is always to be interpreted as equal to $$\lim_{x\to 0} \frac{x}{x}$$. This you do not contest.
What you apparently seem to believe is that rigor is without purpose or utility in mathematics, and therefore one is free to steamroll over little problems like invalid syllogisms. But in that, you fail in your own duty to yourself and have fallen into the category of pseudomathematics. If you think I am wrong in the above two points, you must demonstrate that by constructing a system of mathematics that does allow you to assert what you believe without greatly compromising the utility of arithmetic and/or analysis.
http://www.jstor.org/stable/2589247 (USD $12.00)
Using the axioms of arithmetic, there is no definition of $$\frac{1}{0}$$ and even if you add such an axiom, the rigidity of the axioms of arithmetic forbid $$\frac{1}{0} - \frac{1}{0}$$ or $$\frac{0}{0}$$ from being a sensible expression. This you do not contest.
Using the axioms of analysis, $$\lim_{x\to 0} \frac{x}{x} = 1$$, but there is no axiom to assert that prefers $$\frac{a}{b} = \lim_{x\to 0} \frac{x + a}{x + b}$$ to $$\frac{a}{b} = \lim_{x\to 0} \frac{15x + a}{x^3 - x + b}$$, and therefore no axiom or theorem that asserts that $$\frac{0}{0}$$ is always to be interpreted as equal to $$\lim_{x\to 0} \frac{x}{x}$$. This you do not contest.
What you apparently seem to believe is that rigor is without purpose or utility in mathematics, and therefore one is free to steamroll over little problems like invalid syllogisms. But in that, you fail in your own duty to yourself and have fallen into the category of pseudomathematics. If you think I am wrong in the above two points, you must demonstrate that by constructing a system of mathematics that does allow you to assert what you believe without greatly compromising the utility of arithmetic and/or analysis.
from "Mathematics on a Distant Planet." American Math Monthly, vol 105 no 7, pp. 640–650. (1998)Richard Hamming said:
http://www.jstor.org/stable/2589247 (USD $12.00)
Pseudomathematics? Is that even a word? I was just describing the same mathmatics that is taught to poor unexpecting children all over the world! I started out by giving the example of the tangent line to a curve. In that derivation, $$\lim_{h\to 0} \frac{h}{h} = 1$$ and the tangent line to a curve is said to be correct! If you said it was zero or undefined then it would be wrong, but since you are only taking the limit as "h" approuches zero, then the equation is true. I already proved that it is real mathmatics before I even started getting into this discussion. I don't know what kind of "pseudomathematics" you are doing!
Signs point to yes. http://en.wikipedia.org/wiki/Pseudomathematics
Incorrect.
The mathematics taught to school children is that zero times any number is zero.
This can be proved from other properties of addition and multiplication.
For example, because $$\forall x\; x + 0 = x$$ and $$\forall x\; x \times 1= x$$ and $$\forall x,y,z \; z \times ( x + y ) = z \times x \, + \, z \times y$$ and $$\forall x,y,z \; x = y \leftrightarrow z + x = z + y$$ it follows that $$ x + 0 = x = x \times 1 = x \times ( 1 + 0 ) = x \times 1 + x \times 0 = x + x \times 0 \rightarrow 0 = x\times 0$$
This property of zero is why it has no finite representation in logarithms.
Multiplication by any number except zero can be "undone" by division. But since multiplication by zero throws away all information about the original number, there is nothing for arithmetic to operate on to recover the unique original number. Because $$5 \times 0 = 7 \times 0$$ does not allow you to assert $$5 = 7$$.
The topic of limits of expressions as parameters approach some value is not taught to "poor unexpecting children" but to young adults, usually in an introduction to differential calculus and illustrating the utility of L'Hôpital's Rule which is all about avoiding trying to evaluate expressions that look like they might approach a form like $$0 \times \infty$$ or $$\frac{\infty}{\infty}$$ or $$0^0$$ or $$\frac{0}{0}$$ and and using alternative expressions which have the same limiting value (which is often not one) but better behavior. But zero is not an expression that approaches a value as some parameter is changed, so your invocation of analysis is unwarranted in this case and contrary to good instruction.
Sounds like you could use another lesson in limits. If you assumed that "h" is zero and then would make all other values in the equation invalid, then the tangent line to a curve would have to be wrong. The whole purpose of finding a limit is to bring meaning to an equation when it would have no solution. So then to find a solution, you would have to take the limit. So then since "h" would be zero in the derivation of the tangent line to a curve, you can just take the limit. Then by taking the limit "h" no longer behieves as though it is zero, so then a solution can be found and then you can derive the equation for a tangent line to a curve.
I don't think you even realize that the only way I could agree with you would be to deny that properties in mathmatics that have already been established are wrong. If you don't believe that limits actually find solution to problems and that equations found this way are wrong then that is your own problem. You should make another thread in the Pseudoscience section called Pseudomathematics to talk about it. It wouldn't belong in the math section. The topic was dividing a number by zero, so then I only explained how that is done in established mathmatics. In basic mathmatics you can't deal with zero, but there are more advanced mathmatics that you should learn about that are able to deal with zero, like limits. You should try to learn about them if you never have already.
Are you seriously saying that this equation is false?
$$ (10 + 10 ) (10 - 10 ) = 10 ( 10 - 10 ) $$
It's not factoring a zero from a ten.
It's multiplying zero and ten to get zero.
The reverse would be factoring a ten and a zero from a zero.
What rules are you talking about? You're really not making much sense.
Yes, this equation is false. $$ a^{2} - b^{2} $$ should never equal $$ ab - b^{2} $$ period. If they did you would end up getting answers like 1=2.
10 - 10 = 0 so then if you take a (10 - 10 ) out of an expression, then you would have taken a zero out of the expression. That would be because 10 - 10 = 0. The only number you could take a zero out of and it still be the same number would be zero.
The ones they put in math books that say when an expression is valid or not when they introduce the expression to begin with.
Layman, you aren't qualified to give any lessons in maths.
rpenner is. You could learn a lot by paying attention to his posts, and asking about the parts that don't yet make sense to you.
It seems that you used to relying on intuition to learn and understand mathematics. That's fine up to a point, but there comes a time (and you're reaching it now) where you need to be more formal and rigorous. Otherwise, you won't be able see where your intuition is misleading you - intuition is not self-correcting; formalism is.
I suggest you look at the Formal definition of a limit. Working through the examples on that page should help a lot.
Seriously? This is simple arithmetic:
(10+10)(10-10) = 0
Right?
10 ( 10 - 10 ) = 0
Right?
What do you mean by "take a (10-10) out of an expression?", and how is that relevant to this equation:
$$ (10 + 10 ) (10 - 10 ) = 10 ( 10 - 10 ) $$
Sorry, it still doesn't make sense.
Yes, we're all agreed that $$\lim_{a\to 0}\frac{a}{a}=1$$.
Or anything at all.
Which is why the unrigorous expression "The limit of 0/0" could be anything, and should be avoided.
Unless you define the expression rigorously, you don't know whether the numerator and denominator are constants, the same variable approaching zero, or different variables approaching zero.
Those expressions are not valid when a=b, or when a=0 and b=0. So no it is not a valid equation. You would have taken zero out of twenty and zero out of ten. That is not a valid operation and is why it is not true when a=b, or when a=0 and b=0.
That is what has been factored out of both sides of the equation. A value that is equal to zero, that then gives two different values left on each side of the equation. Factoring isn't supposed to change the value of an expression. So then you would have 20=10 if you took the limit as (a - b) approached zero.
Are you just trolling me now or what? Open an algebra book, then look up the expressions on each side of the equation. It will say that the expression are not valid for a=b, or a=0 and b=0. If you assume that they have these values then you would get wrong answers like 1=2. It says this when almost any expression is introduced in algebra. You really should learn algebra.
This is simple arithmetic. There is no a or b involved.
(10+10)(10-10) = 0
10 ( 10 - 10 ) = 0
You're talking about the algebra, which is irrelevant to the simple arithmetic.
You seem to have a problem with the factorizing done in this step:
$$\begin{align}
a^2 - b^2 &= ab - b^2 \\
(a+b)(a-b) &= b(a-b)
\end{align}
$$
There's no problem. Both sides are equal to zero before and after the factorization step.
$$\begin{align}
a^2 - b^2 &= 0 \\
(a+b)(a-b) &= 0 \\
ab - b^2 &= 0 \\
b(a-b) &= 0
\end{align}$$
Or are you talking about the following step, when the zero factors are invalidly cancelled from both sides?
No, that would be doing it wrong. You are confusing the algebra example with the simple arithmetic example, and getting confused by thinking about limits without rigorously defining them.
Sorry, I don't see how this is relevant to the discussion.
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I do have a problem with it. Since you are only doing simple arithmetic, and not algebra, you would have already done steps that are not valid in algebra, so then you would also get an incorrect answer for the limit. So the limit isn't wrong, it is the steps taken before you would take the limit that would make it wrong. You set two expressions equal to each other that should never be equal to each other in algebra. If the equation itself is set up wrong then of course you would end up getting the wrong limit. So then it actually isn't the division by zero that gives the wrong answer it is using expressions incorrectly in algebra. Then that doesn't prove that canceling a variable while taking the limit is wrong.
So then since a = b then ( a - b ) = 0, and ( a + b )( 0 ) = b ( 0 ). What number could you then factor out a zero from? Any number? No, the only number you could factor out a zero is zero and have it keep the same value. Say an expression was equal to 10, would you then have 10 and then factor out a zero to have ( x ) ( 0 ) = 10? Then it would no longer be 10 it would be zero, the only number that would have the same value after factoring out a zero would be zero, ( 0 ) ( 0 ) = 0, that is the only time the factorization would retain the same value. Zero times zero is the only time factoring out a zero wouldn't change the value of the expression, so then "a" must equal zero and "b" must equal zero in this factorization. Setting a = b then forces "a" and "b" to become zero in the factorization of these expressions.
Yeah, that makes no sense.
This is arithmetic:
(10+10)(10-10) = 0
10 ( 10 - 10 ) = 0
Those two equations stand on their own, independent of any algebra. It's easy to work them out and see that they're correct.
What limit? There are no limits involved in the artihmetic or the algebra we're discussing. You're making it more complex than it needs to be, confusing yourself, and getting simple maths wrong as a result.
In our equations, we have the equivalent of (x)(0) = 0, not (x)(0) = x, so there's no problem with x=10.
The problem is not the equations and how the arithmetic works out in them, the problem is that you factored out a zero in an expression! So then the numbers in the algebra equations could not be 10, they would have to be zero. The only number you can factor out a zero from would be zero!
Say you had 10 and you wanted to factor it. You could say that it is ( 5 ) ( 2 ) = 10. Well if you factored out a zero from 10, then there wouldn't be anything you could multiply by 0 in order to get 10.
So then you would end up with this number that doesn't exist ( x ) ( 0 ) = 10. Since nothing times zero is 10, then the number you factored out a zero from could not be 10. Say you had 0, and then you factored out a 0, you would have ( 0 ) ( 0 ) = 0.
So then the number could not be 10 and could only be 0, because that is the only number that you could factor out a zero from since there would be no "x" that you could multiply by zero in order to get anything but zero.
Say you had 10 and you wanted to factor it. You could say that it is ( 5 ) ( 2 ) = 10. Well if you factored out a zero from 10, then there wouldn't be anything you could multiply by 0 in order to get 10.
So then you would end up with this number that doesn't exist ( x ) ( 0 ) = 10. Since nothing times zero is 10, then the number you factored out a zero from could not be 10. Say you had 0, and then you factored out a 0, you would have ( 0 ) ( 0 ) = 0.
So then the number could not be 10 and could only be 0, because that is the only number that you could factor out a zero from since there would be no "x" that you could multiply by zero in order to get anything but zero.
That's what we mean when we say you can't attribute a value to the arithmetic expression $$\frac{0}{0}$$ and that division by zero cannot result in a number.
If $$2 \times 3 = 6$$ means that $$\frac{6}{3} = 2$$ and $$\frac{6}{2} = 3$$.
Then the fact that both $$2 \times 0 = 0$$ and $$0 \times 3 = 0$$ means that $$\frac{0}{0}$$ is not well-defined.
If $$2 \times 3 = 6$$ means that $$\frac{6}{3} = 2$$ and $$\frac{6}{2} = 3$$.
Then the fact that both $$2 \times 0 = 0$$ and $$0 \times 3 = 0$$ means that $$\frac{0}{0}$$ is not well-defined.
Here is the apparent paradox starting from $$\lim_{a\to b}a = b$$ instead of a=b:
$$\array{lcl$
\lim_{a\to b} \ (a) &= &b
\lim_{a\to b} \ (a^2) &= &\lim_{a\to b} \ (ab) \\
\lim_{a\to b} \ (a^2 - b^2) &= &\lim_{a\to b} \ (ab - b^2) = 0\\
\lim_{a\to b} \ (a+b)(a-b) &= &\lim_{a\to b} \ b(a-b) = 0 \\
\lim_{a\to b} \ \frac{(a+b)(a-b)}{a-b} &\ne &\lim_{a\to b} \ \frac{b(a-b)}{(a-b)}
}$$
The equality breaks at this step, because it wrongly relies on limits approaching 0/0 being equal.
$$\array{lcl$
\lim_{a\to b} \ (a) &= &b
\lim_{a\to b} \ (a^2) &= &\lim_{a\to b} \ (ab) \\
\lim_{a\to b} \ (a^2 - b^2) &= &\lim_{a\to b} \ (ab - b^2) = 0\\
\lim_{a\to b} \ (a+b)(a-b) &= &\lim_{a\to b} \ b(a-b) = 0 \\
\lim_{a\to b} \ \frac{(a+b)(a-b)}{a-b} &\ne &\lim_{a\to b} \ \frac{b(a-b)}{(a-b)}
}$$
The equality breaks at this step, because it wrongly relies on limits approaching 0/0 being equal.
One problem is that you're getting simple arithmetic wrong, never mind algebra or calculus.
Look:
(10+10)(10-10) = 0
Do you disagree?
Pay attention to detail, and you'll see that we are factoring zero out of zero, not anything else. We do not have (x)(0) = 10. We have (x)(0) = 0
Look:
Let a=10, b=10
Start with zero:
a^2 - b^2 = 0
Factor out zero:
(a+b)(a-b) = 0
(20)(0) = 0