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# Calculus allows dividing by zero?

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thedustbustr    191
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:
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   
I may have confused some laws, its been a while since proofs, but they all exist. [edited by - thedustbustr on August 14, 2003 2:39:23 PM] [edited by - thedustbustr on August 16, 2003 9:55:39 AM]

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Hyatus    122
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"

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rypyr    252
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.

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thedustbustr    191
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.

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thedustbustr    191
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)

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cmptrgear    306
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

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Way Walker    745
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]

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Sfpiano    126
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.

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Way Walker    745
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.

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Alias2    122
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]

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v71    100
...Once again... personal math....someone should have finished his calculus course at university/ college and is still fascinated by the limit concept....

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Way Walker    745
quote:
Original post by Sfpiano
Zero divided by zero is not possible.

He never divided anything by zero, and "anything" includes zero.

quote:

The limiting value, and the definitive value are different things.

That''s precisely why he had to use limits to get an answer for something that looks like 0/0, but isn''t. For another example, look up L''Hopital''s rule.

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emilk    216
(lim x->0) 0/x = 0
(lim x->0) x/x = 1
(lim x->0) +n/x = oo (infinity)
(lim x->0) -n/x = -oo (negative infinity)

So should division by zero be 0, 1, infinity or negative infinity?
The answer is: neither, division by zero is ALWAYS undefined.

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Mastaba    761
Step 7 does not follow from applying step 2 to step 6.

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thedustbustr    191
cmptgear: Yes I get it thx. But you ignored the limit in front of it

"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)."

Why can't I say explicitly that the LHS x and the RHS x are the same variable?

Let me expand this to a real life example: Calculate the derivative of f(x)=x^2 using the definition of a derivative.
1. (limit as d approaches 0) (f(x+d)-f(x))/d2. (limit as d approaches 0) (x+d)^2 - x^2)/d3. (limit as d approaches 0) (x^2 + 2xd + d^2 - x^2)/d4. (limit as d approaches 0) (2xd + d^2)/d5. (limit as d approaches 0) 2x+06. 2x

On line 4 to 5 you divide by "almost 0" to eliminate the d in the denominator. On line 5 you add "exactly 0" to get rid of the final d. This property of d being both "almost 0" and "exactly 0" is the property I exploit in my impossible proof.

I believe I've countered every criticism so far, if I haven't, please explain which mathematical law I violated.

Dustin
edit: forgot /code

[edited by - thedustbustr on August 14, 2003 3:10:38 PM]

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Sfpiano    126
Whoops, I misunderstood what he was saying. I think I've got it now . You completely left out the limit part from 6 on. It should be:

6. (the limit as x approaches 0) (x * a) / x = (the limit as x approaches 0) (x * b) / x7.  (the limit as x approaches 0)(1 * a) = (the limit as x approaches 0)(1 * b )//law of substitution (6, 2)8.  (the limit as x approaches 0) a = (the limit as x approaches 0) b         //multiplicative identity of 1

I think, right?

[edited by - Sfpiano on August 14, 2003 3:08:02 PM]

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thedustbustr    191
sfpiano: yes oops, but the proof still works
waywalker: yep you get it
Guys please don''t flame me. FYI I''m a senior in HS this year, taking ap calc, took half semester of calc last year. So the university comments arent necessary thx

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Sfpiano    126
How does it still work exactly? When you divide by x, you''re not dividing by zero. Now that I think about it, how do steps 3 or 4 work?

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Way Walker    745
quote:
Original post by thedustbustr
"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)."

Why can''t I say explicitly that the LHS x and the RHS x are the same variable?

Because they''re not.

quote:

Let me expand this to a real life example: Calculate the derivative of f(x)=x^2 using the definition of a derivative.
1. (limit as d approaches 0) (f(x+d)-f(x))/d2. (limit as d approaches 0) (x+d)^2 - x^2)/d3. (limit as d approaches 0) (x^2 + 2xd + d^2 - x^2)/d4. (limit as d approaches 0) (2xd + d^2)/d5. (limit as d approaches 0) 2x+06. 2x

On line 4 to 5 you divide by "almost 0" to eliminate the d in the denominator.

You''ve done nothing of the sort. You''ve divided d by d to get 1.

quote:

On line 5 you add "exactly 0" to get rid of the final d.

Again, you''ve done nothing of the sort. You''ve simply used the rules stating that (limit as d approaches x)d = x and (limit as d approaches x)f(y) = f(y). Maybe it would have been clearer if you included the following steps:

(limit as d approaches 0) (2xd + d2)/d
(limit as d approaches 0) (2x + d)
(limit as d approaches 0) 2x + (limit as d approaches 0) d
2x + 0

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thedustbustr    191
Step 3 can be set equal to step 4 by the multiplicative identy of zero (anything times zero is zero) and then setting 0=0 by transitive law (a=c, b=c, therefore a=b).

In steps 3,4,5 the "exactly zero" property of x is used. In step 6, the "almost zero" property of x is used.

My point is that calculus lets you contradict thimgs (ie having x as both "almost 0" and "exactly 0" (see the trivial derivative example).

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Way Walker    745
quote:
Original post by thedustbustr
sfpiano: yes oops, but the proof still works
waywalker: yep you get it
Guys please don''t flame me. FYI I''m a senior in HS this year, taking ap calc, took half semester of calc last year. So the university comments arent necessary thx

Actually, they sort of are. At university they''re much more strict about what you can actually do in math. One of the hardest things to understand (it''s not applicable here, but it''s an example of the sort of "wishful thinking" that math doesn''t allow, and "wishful thinking" is the problem in your proof) is that

a => b

does not mean that

b => a

so, if you want to prove that

a <=> b

you really need to do two proofs. One proving

a => b

and one proving

b => a

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Sfpiano    126
How can you say the lim of x*a is zero? You have no idea what a is. Just because the lim value of x is zero doesn''t mean is holds true for anything times x.

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Guest Anonymous Poster
Hrm; just to clarify a few things. A division by 0 is not "undefined". x/0 is Infinity.

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Hyren    122
quote:

5. (the limit as x approaches 0) x * a = x * b //law of transitive (3,4)
6. (the limit as x approaches 0) (x * a) / x = (x * b) / x //divide both sides of equality by the same number

In steps 5 and 6 above, you divide each side by x (appearingly). However, what you're really doing is using the property of limits such that:

lim x->0 f(x) * lim x->0 g(x) = lim x->0 (f(x) * g(x))

However, in this case your g(x) on both sides is 1/x and lim x->0 1/x is undefined. However, this property is only valid if both limits exist (are defined) . Therefore, you cannot do this and this invalidates the proof.

It's been a while since I've taken calculus, so maybe I'm off on this. Maybe someone taking it now or having more knowledge of calc can corroborate this.

Edit: Just remember to say why lim x->0 (1/x) is undefined. Because lim x->0(-) 1/x = -infinity and lim x->0(+) = infinity, then there is no continuity of limits at that point and the limit of the function is not defined.

"Back to the code mines... ka-chink... ka-chink..."
Tachyon Digital - Down for the summer, be back in the fall.

[edited by - Hyren on August 14, 2003 3:33:02 PM]

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Way Walker    745
quote:
Original post by thedustbustr
Step 3 can be set equal to step 4 by the multiplicative identy of zero (anything times zero is zero) and then setting 0=0 by transitive law (a=c, b=c, therefore a=b).

Actually, all you need is the transitive law. You have statements saying

a = 0
b = 0

So you can simply say

a = b

The thing is,

a = (limit x approaches 0)xf
b = (limit x approaches 0)xg
(limit x approaches 0)xf = (limit x approaches 0)xg

this is all good until you try doing

(limit x approaches 0)(xf = xg)

where things break down.

quote:

In steps 3,4,5 the "exactly zero" property of x is used. In step 6, the "almost zero" property of x is used.

"the exactly zero property of x" and "the almost zero property of x" don''t exist. x is simply a variable in some function. You''re asking "What is the behavior of this function as x approaches 0?". Now, the two properties you mention might be useful in beginning to understand limits, and they might be useful when you begin to evaluate limits, but they''re not actually true.

quote:

My point is that calculus lets you contradict thimgs (ie having x as both "almost 0" and "exactly 0" (see the trivial derivative example).

Yes, see the trivial derivative example. "almost 0" and "exactly 0" can be useful as a shorthand to evaluate it, but they''re not what''s actually being done. There''s no contradiction. Another instance of a common shorthand in calculus is

(the integral from 0 to infinity)f(x)dx

which is a shorthand for

(the limit as s approaches infinity)((the integral from 0 to s)f(x)dx)

as long as you remember it''s a shorthand, nothing''s wrong.