given:
a != b
anything times 0 is 0 //multiplicative identity of 0
anything times 1 is itself //multiplicative identity of 1
proof:
1. (the limit as x approaches 0) x = 0 //def. limit
2. (the limit as x approaches 0) x / x = 1 //def. limit
3. (the limit as x approaches 0) x * a = 0
//muptiplicative identity of 0 (1)
4. (the limit as x approaches 0) x * b = 0 //ditto (1)
5. (the limit as x approaches 0) x * a =
(the limit as x approaches 0) x * b
//law of transitive (3,4)
6. (the limit as x approaches 0) (x * a) / x =
(the limit as x approaches 0) (x * b) / x
//divide both sides of equality by the same number
7. 1 * a = 1 * b //law of substitution (6, 2)
8. a = b //multiplicative identity of 1 Calculus allows dividing by zero?
Author
EDIT: replies telling me I am stupid, uneducated, bad at math, have a poor understanding of limits (even if it may be true) or telling me "look stupid you can't divide by zero everyone knows that" are not appreciated. I already received a few responses explaining the flaw in the proof; however, if you must reply, please at least read this entire initial post. Thanks
--
I realized that by using limits you can divide by zero. I used this property to construct an impossible proof. It fringes around the following true statement:
(the limit as x approaches 0) x / x equals 1.
This can't have been unnoticed, it's way too obvious and so trivial - however, it completely blows calculus apart (can't define a derivative without using limits).
I've only taken half a semester of calculus in my junior year of high school; maybe there is an explanation I haven't heard about. Anyone have any ideas?
Heres the proof if you're curious:
Not that I ever took Calculus, but it says the limit as x
''approaches'' zero. I don''t believe that includes zero itself.
-Hyatus
"da da da"
''approaches'' zero. I don''t believe that includes zero itself.
-Hyatus
"da da da"
5. (the limit as x approaches 0) x * a = x * b //law of transitive (3,4)
This is not true if a != b, it is only true to say that both sides APPROACH zero as x approaches zero, but they approach zero at different rates...
What you''ve stated is the equivalent of:
0 * 4 = 0 * 5
then divide by zero so 4 must equal 5.
This is not true if a != b, it is only true to say that both sides APPROACH zero as x approaches zero, but they approach zero at different rates...
What you''ve stated is the equivalent of:
0 * 4 = 0 * 5
then divide by zero so 4 must equal 5.
Author
a limit evaluates to the specific number it approaches if it approaches the same number from the positive direction and the negative direction. x/x is undefined at 0; (limit as x approaches 0) x/x is 1.
Author
rypyr: I know. If the proof didn''t use limits, it wouldnt work because 0/0 is undefined (not 1). But, as I said, this is a true statement:
(the limit as x approaches 0) x / x equals 1
so the proof is legit. No mathematical laws are violated (other than the first postulate, a != b)
(the limit as x approaches 0) x / x equals 1
so the proof is legit. No mathematical laws are violated (other than the first postulate, a != b)
Your reasoning is incorrect, 0/0 is not equal to 1. If anything it would be undefined or infinity as it is impossible to divide any number by 0, including 0 itself. Now 1x10^-1200/1x10^-1200 will equal 1 but even though 1x10^-1200 is an extremely small number it is still not 0.
if you dont believe me you can reason it out this way.
0 is divisble by 0 how many times? 0
or 0 is divisible by 0 how many times? 1, 2, 3, 4....doesnt matter what number because they will all end up 0. This is why division by 0 is undefined because the answer can be any number.
"I may not agree with what you say but I will defend to the death your right to say it."
--Voltaire
if you dont believe me you can reason it out this way.
0 is divisble by 0 how many times? 0
or 0 is divisible by 0 how many times? 1, 2, 3, 4....doesnt matter what number because they will all end up 0. This is why division by 0 is undefined because the answer can be any number.
"I may not agree with what you say but I will defend to the death your right to say it."
--Voltaire
quote:Original post by thedustbustr
3. (the limit as x approaches 0) x * a = 0
//muptiplicative identity of 0 (1)
4. (the limit as x approaches 0) x * b = 0 //ditto (1)
5. (the limit as x approaches 0) x * a = x * b
//law of transitive (3,4)
I think you skipped a step in there. You'd have to have done something like
3. (the limit as x approaches 0) x * a = 0
4. (the limit as x approaches 0) x * b = 0
5a. ((the limit as x approaches 0) x * a) = ((the limit as x approaches 0) x * b)
5b. (the limit as x approaches 0) (x * a = x * b)
and this is where it breaks down. You can't go from 5a to 5b. The problem is you've used the same symbol (namely, x) for two different variables (the x on the LHS and the x on the RHS).
[edited by - Way Walker on August 14, 2003 2:51:54 PM]
Zero divided by zero is not possible. I fail to see how you arrived at that conclusion here, but the point is that it doesn''t work. The limiting value, and the definitive value are different things.
quote:Original post by cmptrgear
Your reasoning is incorrect, 0/0 is not equal to 1. If anything it would be undefined or infinity as it is impossible to divide any number by 0, including 0 itself. Now 1x10^-1200/1x10^-1200 will equal 1 but even though 1x10^-1200 is an extremely small number it is still not 0.
if you dont believe me you can reason it out this way.
0 is divisble by 0 how many times? 0
or 0 is divisible by 0 how many times? 1, 2, 3, 4....doesnt matter what number because they will all end up 0. This is why division by 0 is undefined because the answer can be any number.
Hence the limits. Even if f(a) is undefined, ((limit as x approaches a) f(x)) might be defined. Similarly, it''s perfectly possible for f(a) != ((limit as x approaches a) f(x)) to be true. This is essentially how you get integrals from Reimann sums.
I'm just re-iterating what I think everyone else is saying, but why not, right? :-)
Your proof has a faulty assumption moving from step 6 to 7:
6. (the limit as x approaches 0) (x * a) / x = (x * b) / x //divide both sides of equality by the same number
7. 1 * a = 1 * b //law of substitution (6, 2)
The problem is, you're dividing by x as x approaches 0 . The limit of x as it approaches 0 is 0, which you defined in step 1. So you're dividing by zero, which is undefined, not 1 (as your step 7 suggests).
[edited by - alias2 on August 14, 2003 2:59:42 PM]
Your proof has a faulty assumption moving from step 6 to 7:
6. (the limit as x approaches 0) (x * a) / x = (x * b) / x //divide both sides of equality by the same number
7. 1 * a = 1 * b //law of substitution (6, 2)
The problem is, you're dividing by x as x approaches 0 . The limit of x as it approaches 0 is 0, which you defined in step 1. So you're dividing by zero, which is undefined, not 1 (as your step 7 suggests).
[edited by - alias2 on August 14, 2003 2:59:42 PM]
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