How we could both suffer from the same delusions about the level of incompetence of each other is even beyond me. Ten plus ten is twenty. Ten minus ten is zero. Twenty times zero is zero. I will ask you this, out of all numbers on the real number line, which ones can you factor out a zero? Would one of those numbers be 10? Is zero times something 10? Would that number only be zero? If zero times something cannot be 10, then you cannot factor out a zero from 10. You would have to have 0 x NUMBER = 10, in order for zero to be a factor. Maybe we should go over the factors of 10? 1 x 10 = 10 2 x 5 = 10 I think that is all. Nowhere of these two factors of 10 is there a zero! So why on Earth would you insist that zero is a factor of 10???????????????
If you ever was able to realize that by factoring out a zero, the only number that has a factor of zero is zero. Then when you found the limit and canceled (a - b), then you would have zero times one on both sides. So then the equation only says that ( 0 + 0 ) 1 = 0 ( 1 ), since a = 0 and b = 0.
So if you had ( a + b ) ( a - b ) from factoring \( ( a^{2} - b^{2} ) \) and a = b, then ( a - b ) = 0, the only number that has a factor of zero is zero, so then ( a + b ) = 0, since a = b, then a = 0, and b = 0. That is the only number that can be added to itself that gives the value of zero.
You seem to be missing the crucial step of reducing all expressions to their simplest terms before canceling factors, in which case, it is clear that no factors need be canceled (divided out).
Well, I solved it before posting it: (0*P)+6 = 5 0*P = 5-6 = -1 Therefore the answer is -1 or 0 What do you want me to respond to since they said that the answer is undefined? So I was just watching the replies that were coming in.
What's wrong with this "proof" that 2 = 1? What's wrong is that there are two equations that say different things, namely a = b, and a + b = b. Start with the second, if it's true then a must be zero. Going back to the first, if a = b and a is zero by the second equation, then b must also be zero. And as several people have pointed out the 4th step in the "proof" is invalid if a = b, because then a - b = 0. Or (a - b)/(a - b) is fine as long as a ≠ b.
Excellent. You do agree that (10+10)(10-10) = 0 I'm sure you'll also agree that 10(10-10) = 0, right? And yet you say that (10+10)(10-10) does not equal 10(10-10)? This is the simple arithmetic you got wrong. I've agreed with this already. Several times. Zero is a factor of zero only, nothing else. I insist no such thing.
You're getting the logic mixed up, Layman. Let's try working this through in the wordy style you seem to favour. The only number that has a factor of zero is zero. Let number = factor1 x factor2 If (factor2) = 0, then (number) has a factor of zero. Since the only number that has a factor of zero is zero, then number is zero. This says nothing about (factor1), which does not have a factor zero, so it can be anything. Agreed? Now, apply that logic to \( ( a^{2} - b^{2} ) = ( a + b ) ( a - b )\): \(\array{cl$ ( a^{2} - b^{2} ) &= \ ( a + b ) ( a - b ) \\ (a-b) &= \ 0 \\ ( a^{2} - b^{2} ) &\mbox{ has a factor of zero, therefore...} \\ ( a^{2} - b^{2}) &= \ 0 }\) This says nothing about (a+b), which can be anything.
( a + b ) cannot be anything that is the whole point I have been trying to get at here. If ( a + b ) has had a zero factorored out of it, then the value of ( a + b ) can only be zero. If ( a + b ) = 0, then the only value they can be is zero. Or take a look at the other side of the equation in the exrpession \( ( ab - b^{2} ) \). You factor out ( a - b ), so then you are left with b. ( a - b ) = 0, so then you have factored out a zero from the expression. So then you said ( 0 ) ( b ) = 0, what is the value of "b"? b = 0/0, so then you say the solution of "b" be can be all reals, but 0/0 is not all reals, it can only be 0, 1, and undefined, like we said earlier with the limits. Never while dealing with limits should 0/0 be all reals. In this case it is actually zero. Those are the only three possible answers not the whole number line.
Your point is wrong. (a+b) has not had a zero factored out. \((a^2-b^2)\) has had a zero factored out (therefore \((a^2-b^2)=0\)). Right, we have factored out a zero from \(ab-b^2\), therefore \(ab-b^2=0\) We have not factored anything out of b. Obviously, b can be anything. No, I said and showed that limits approaching 0/0 can be anything. Remember this? \(\lim_{a\to 0}\frac{2a}{a} = 2\) That limit approaches 0/0, and is equal to 2. Also, we still haven't yet agreed on the simple arithmetic: Do you agree that 20 x 0 = 10 x 0 ? Do you agree that (10+10)(10-10) = 10(10-10) ?
I think that is a rule in limits, that 0/0 can only be 0, 1, or undefinied. The only reason the answer to that problem is 2, is because in that cause a/a = 1. I don't think you can assume that zero times anything is zero when dealing with limits. Say it was not zero but some variable, it could just as easily become 0/0. So then any case you would think it is all reals, it is actually zero or undefined. That is just one of the rules that makes it work out, if you don't go by that rule then you would end up getting 2=1 still... I really just don't think you are allowed to say that any number times zero is zero. They say this in arithmatic but I don't think it works in limits. If that was true then instead 0/0 could be 0, 1, undefined, or all reals. All reals just isn't a solution in these types of problems. They clearly state that it is only 0, 1, and undefined when introducing the subject. Your adding solutions that do not agree with the mathmatics.
Firstly, Do you agree that 20 x 0 = 10 x 0 ? Do you agree that (10+10)(10-10) = 10(10-10) ? Obviously, you're wrong. Right. But the fact remains that the limit approaches 0/0, and equals 2. That's actually beside the point, because the problem in question doesn't deal with limits. You're making it too complicated, confusing yourself, and getting simple maths wrong as a result. "( 0 ) ( b ) = 0, what is the value of 'b'?" Obviously, b can be anything. You don't need to know about limits to see that. 0 x 0 = 0 0 x 1 = 0 0 x 2 = 0 0 x 101.5 = 0 0 x (3i-1) = 0 Simple arithmetic. Don't be misled by making it more complicated than it is. A limit that approaches 0/0 could be anything. Not just 0,1, undefined. Not just all reals. You can easily construct complex limits that approach 0/0, and evaluate to complex numbers. You could probably do the same for vectors, matrices, and tensors. Your intuition that a limit approaching 0/0 can only be 0,1, or undefined is obviously wrong. You effectively said so yourself: \(\lim_{a\to 0} \frac{2a}{a} = 2\) Who is "they"?
Yes, 0 = 0. I do not agree that equation could be found by factoring the expressions \( a^{2} - b^{2} \) and \( ab - b^{2} \). Would you care to elaborate more on why you are wrong? I think the answer to that question is yes. That is because the limit of a/a = 1. 10/10 = 1 9/9=1 8/8=1 7/7=1 6/6=1 5/5=1 4/4=1 3/3 =1 2/2=1 1/1=1 The closer and closer you get to one the answer will always be one. So then one times 2 is 2. It has nothing to do with limits as x approuches 0/0 that it is something other than 1, 0 or undefined. You need a refresher course in limits. Your making it too simple so then you are getting the wrong answer by not using the rules provided in introductory courses in limits. I don't care about your simple arithmetic. I am using more advanced concepts of limits to approuch the problem. Your making me start to sound like alphanumeric. If this goes on much longer I don't know what I would think of myself. No it cannot be anything. The anything is undefined. Intuition has nothing to how I solved this problem, if you ever took a course in limits then you would know how to solve this problem. They, would be math instructors that teach limits. Are you suffering from a concussion? a/a =1 then 2(1)=2 that doesn't prove that the limit of a/a as "a" approuches zero is anything other than 1, 0, or undefined!
Hooray, now we're getting somewhere! You might like to go back and reconsider what you wrote in post 80. Remember this? The rest of your post is repititious, insulting, or irrelevant. So, I think we've reached an impasse. Go back to your class and ask your maths instructors whether a limit approaching 0/0 can be anything other than 0, 1, or undefined, and let us know what you discover. Or ask rpenner or AlphaNumeric. They are both maths instructors.
You see it as factoring out a zero from zero since each side of the equation is equal to zero, that is why I said ( a + b ) = 0 because it actually isn't a number that your factoring from. I just picked 10 out of the air. May be my mistake, but I think the logic holds that any number that has zero as a factor is zero, even though your not allowed to say that any number times zero is zero, because they could be manipulated in equations to say that b = 0/0 I actually thought I proved this in post #103. I don't think they have said much about this because they know I was laying down rules that has already been established in limits.
The simplest way possible I can think of to explain it is that if you had zero then you couldn't factor zero into a zero and a whole number. You can't say that 0 = ( 10 ) 0. Your not allowed to factor a zero into zero times any whole number. If you do this then the equations break down.
Layman, do you really want to insist that "You can't say that 0 = 10 x 0"? No, the equations break down if you set 0/0 = 1 But, I'm done. You clearly aren't listening to me, so go ask a maths teacher. Or anyone who passed primary school.
Yes, because then you could say that 0/0 = 10, and 0/0 is not 10 it is only 0, 1, or undefined. So then you have division by zero, and the equation breaks down. Factoring is a form of division, don't forget that kiddo.
Factoring is NOT a form of division. Factoring of a number is just writing the same number in a different representation. According to something that sounds vaguely important to learn, the Fundamental theorem of arithmetic, all integers greater than zero have a unique representation in terms of prime factors (unique, that is, up to ordering). If R is the class of all finite non-descending sequences of primes and M operates on finite non-descending sequences of primes to give their product, it follows that M is a bijection between R and the positive integers. \( \vdash R = \left{ s \; | \; \exists n \in \mathbb{N}_0 \left( s : (1\ldots n) \longrightarrow \mathbb{P} \; \textrm{and} \; \forall j, k \in (1 \ldots n ) \left( j \lt k \, \rightarrow \, s(j) \leq s(k) \right) \right) \right} \\ \vdash M = \left{ (f,z) \; | \; f \in R \; \textrm{and} \; z = \prod_{a \in \textrm{rng} f} a \right} \\ \vdash M : R \longrightarrow _{\tiny \textrm{1-to-1}} ^{\tiny \textrm{onto}} \mathbb{N} \) Proof: http://us.metamath.org/mpegif/1arith.html Thus the expression \(5^7 \times 13^5 \times 17^3 \times 29\) is a number -- written in a format which makes multiplication and division easier, while \(4132868184453125\) is the same number in a format which makes multiplication and division by 10 easy and \(EAED25C09B005_{\tiny 16}\) is the same number in a form which makes multiplication and division by 16 easy and \(359125^2 ( 19^2 + 178^2 )\) is the same number in a format which makes it easy to see that it can be written as the sum of two squares in at least one way. Factoring an expression does not change any property of the expression other than it's appearance. Thus two different representations may be fairly connected with an equals sign. \((a - b)(a + b)^2 = (a - b)(a^2 + 2ab + b^2) = a^3 + a^2 b - a b^2 - b^3 = (a^2 - b^2)(a+b)\) is an equivalence of representations for the same polynomial in a and b. \(3 \times 7^2 = 3 \times 49 = 147 = 21 \times 7\) is an equivalence of representations for the same number. \(0 \times 2^2 = 0 \times 4 = 0 = 0 \times 2\) is an equivalence of representations for the same number. Nowhere is there any division going on.
