When is 0 / 0 =1 ? : Solution Posted

[Guest]

When can the following be true?

0 / 0 = 1

Yes, read as zero divided by zero equals one!

No Tens, unit, magic.. blah blah! guess enuff hint for now

Finally the solution:

dBm is a unit to measure absolute Power, x dBm = 10* log(X^10e+3), where X is power in Watts.

Thus, 1 mw (milli Watt) = 10e-3 W = 10 * log(1) dBm = 0 dBm

Now, 0 (in dBm) / 0 (in dBm) = 1 (but Obvious)

So fellas, spot any fallacies?

Gee, should acknowledge some guy in SD for some funny questions on dBm scale which led to this q. Me thinks it makes cracker of a q

[Guest]

I always thought that zero divided by anything is a non-real answer, or something along that lines.

But that isn't the question, so I;ll keep my mouth shut.

[Guest]

My answer is that your taking the limit of two functions that both evaluate to zero where one is divided by the other. If I gave you a math problem that was lim(x->0) (x/x), you would see that if we evaluated it we would get 0/0. Because 0/0 is undefined, we have to use L'Hopital's Rule where you take the derivative of the top and the bottom d/dx(x) = 1, so our new equation is lim(x->0) (1/1) which clearly evaluates to 1.

PS: I liked the other guys idea with the celcius to kelvin conversion too.

[bonanova]

My answer is that your taking the limit of two functions that both evaluate to zero where one is divided by the other. If I gave you a math problem that was lim(x->0) (x/x), you would see that if we evaluated it we would get 0/0. Because 0/0 is undefined, we have to use L'Hopital's Rule where you take the derivative of the top and the bottom d/dx(x) = 1, so our new equation is lim(x->0) (1/1) which clearly evaluates to 1.

PS: I liked the other guys idea with the celcius to kelvin conversion too.

Spot on. Great answer.

The OP asks: when [under what conditions] can 0/0=1?

And you've given the best answer.

The other guesses, including decibels and any situation where 0/0 has no context, are red herrings.

Note also that 0/0 unambiguously can be ...

0 limit (x->0) of x²/x

infinity limit (x->0) of x/x²

any finite number k limit (x->0) of kx/x.

[Guest]

0 limit (x->0) of x²/x

infinity limit (x->0) of x/x²

any finite number k limit (x->0) of kx/x.

Hate to disagree with you as you agreed with my answer, but none of these are technically correct. They all do satisfy the first condition that they (by inspection) are 0/0 but when they are evaluated they come out to numbers other than 1. For x^2/x if we take the derivative of the top and the bottom, we get 2x/1 = 2x. For any function kx/x we get k/1 = k. Sorry

[Guest]

a = 1

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

(a+B.) (a-B.) = b(a-B.)

divide through by a-b

(a+B.) = b

a + b -1 = b-1

1 + 1 - 1 = 1-1

1 = 0

so

0/0 = 1

is the same as

0/1 = 0

and therefore the answer is always.

i don't think you can divide through by a-b because a-b equals 0 and is not doable.

[Guest]

0 degrees C / 0 degrees C => 273.15 degrees K/ 273.15 degrees K => 1

Don't know when it would meaningful to divide temperatures but...

I agree with you keith, if I were to ask you to find the ratio of one object's heat to another. For example, if I said find the ratio of the heat of this chair over the heat of this sofa, and both of them were at 0 C then you would have 0/0 however the answer would be 1 because the temperatures are equal. As you said, this is because of the type of scale that Celcius is. To get more technical, Celcius is an interval scale, meaning that only differences in numbers are truly meaningful and multiplication and division are not meaningful. Kelvin is a ratio scale which means that multiplication and division are meaningful, which is why we must convert to kelvin in the example I gave. This gives the explanations a bit more correctly: http://en.wikipedia.org/wiki/Ratio_scale. I believe this is the most correct answer to the riddle.

[bonanova]

Hate to disagree with you as you agreed with my answer, but none of these are technically correct. They all do satisfy the first condition that they (by inspection) are 0/0 but when they are evaluated they come out to numbers other than 1. For x^2/x if we take the derivative of the top and the bottom, we get 2x/1 = 2x. For any function kx/x we get k/1 = k. Sorry

Please explain what you're disagreeing with - after actually reading what I said.

The examples I gave show that not only are there conditions under which 0/0=1 unambiguously,

there are also conditions under which 0/0 unambiguously can equal 0, infinity. and any real number k.

You say that's technically not correct, because zero, infinity and [in general] k are all different from 1?

Give it another shot.

0 limit (x->0) of x²/x

infinity limit (x->0) of x/x²

any finite number k limit (x->0) of kx/x.

If a second reading doesn't do it for you, then go ahead and give the technical explanation of the error you see.

[Guest]

Please explain what you're disagreeing with - after actually reading what I said.

The examples I gave show that not only are there conditions under which 0/0=1 unambiguously,

there are also conditions under which 0/0 unambiguously can equal 0, infinity. and any real number k.

You say that's technically not correct, because zero, infinity and [in general] k are all different from 1?

Give it another shot.

If a second reading doesn't do it for you, then go ahead and give the technical explanation of the error you see.

Gotcha! I understand what you meant now, I thought you were saying that there were an infinite number of answers to the problem, I didn't realize you were just generalizing to say that 0/0 could be equal to any number, which of course you are absolutely correct about.

[Guest]

My answer is that your taking the limit of two functions that both evaluate to zero where one is divided by the other. If I gave you a math problem that was lim(x->0) (x/x), you would see that if we evaluated it we would get 0/0. Because 0/0 is undefined, we have to use L'Hopital's Rule where you take the derivative of the top and the bottom d/dx(x) = 1, so our new equation is lim(x->0) (1/1) which clearly evaluates to 1.

PS: I liked the other guys idea with the celcius to kelvin conversion too.

Unfortunately this does not mean that 0/0=1. It means the limit as x approaches 0 is equal to one. Basically that if 0/0 were defined, which it's not, it would equal one. Just because the function has a limit at an x doesn't mean f(x) exists. If you've dealt with removable discontinuities think of that.

Here is a link for a picture of a graph that will explain the limit thing better:

http://en.wikipedia.org/wiki/Image:Removab...scontinuity.svg

the limit as x approches 1 is 1 but f(1)=0

[akaslickster]

never. U must start with something to end with it.

[Guest]

Anytime is my guess. Seeing as to how the equation in reverse works 1 X 0 = 0 and so holds true for any number. 0/0 can equal 1,2,3,4 ect. anytime.

[Guest]

Let X=0

X^2+X=0

Differentiate it,

2X+1=0

Substitute X=0,

implies 1=0

which implies 0/0=1; 1/1=1; 0/1=1; 1/0=1

Plz correct me if i m wrong!!! because i m juz an enthusiast!!!

[Guest]

Let X=0

X^2+X=0

Differentiate it,

2X+1=0

Substitute X=0,

implies 1=0

which implies 0/0=1; 1/1=1; 0/1=1; 1/0=1

Plz correct me if i m wrong!!! because i m juz an enthusiast!!!

The problem is you set the derivative equal to zero. dy/dx=2x+1 is the equation of the derivative for y=X^2+x. When the derivative equals Zero X=-1/2, not zero. Setting the derivative equal to zero gives critical values of the function, but for your equation there are none at x=0. You need set the derivative equal to dy/dx (or f') not 0.

[Guest]

The problem is you set the derivative equal to zero. dy/dx=2x+1 is the equation of the derivative for y=X^2+x. When the derivative equals Zero X=-1/2, not zero. Setting the derivative equal to zero gives critical values of the function, but for your equation there are none at x=0. You need set the derivative equal to dy/dx (or f') not 0.

Got It!!! thankx...

[Guest]

Solution posted by skale is k.. but not pure maths!!!

Edited May 4, 2008 by roz

[Guest]

O divided by O. O being a variable. Using L'Hospital's rule, derivative of O is 1... 1 divided by 1 is 1

[unreality]

L'Hospital's rule? I think you're think of L'Hopital's rule lol

[Guest]

I think that if you just take into account that any number divided by itself is 1 then

X / X = 1

Therefore

0 / 0 = 1

and the world explodes...

[Guest]

Division by (a-B.) results in indeterminate form... but keep thinking!

i agree...you said a=b. therefore a-b=o. so you dividing anything by a-b results in an indeterminate form.

[Guest]

0! = 1

0!/0! = 1

o! = 1 but 0 != 0. i think there is a mix up with the use of o! and 0 because i consider them to be different

[Guest]

When can the following be true?

0 / 0 = 1

Yes, read as zero divided by zero equals one!

No Tens, unit, magic.. blah blah! guess enuff hint for now

Finally the solution:

dBm is a unit to measure absolute Power, x dBm = 10* log(X^10e+3), where X is power in Watts.

Thus, 1 mw (milli Watt) = 10e-3 W = 10 * log(1) dBm = 0 dBm

Now, 0 (in dBm) / 0 (in dBm) = 1 (but Obvious)

So fellas, spot any fallacies?

Gee, should acknowledge some guy in SD for some funny questions on dBm scale which led to this q. Me thinks it makes cracker of a q

ok, let me try

let a=1 and b=1

i) now, notice that the expression a+b-a+b = 0 if you considered it as (a+b)-(a+b).

ii) similarly, notice that thesame expression a+b-a+b = 2 if you considered it as a+(b-a)+b.

now, sure you agree with me that a+b-a+b = 0 and/or 2.

so, a+b-a+b/a+b-a+b = 1 if we consider the numerator as i) above and the denominator as ii) above...even though i) and ii) are parabolically focused.

[Guest]

When can the following be true?

0 / 0 = 1

Yes, read as zero divided by zero equals one!

No Tens, unit, magic.. blah blah! guess enuff hint for now

wait a minute: for mw it won't be 10e 0 but 10 e1 log of which is 1!

Finally the solution:

dBm is a unit to measure absolute Power, x dBm = 10* log(X^10e+3), where X is power in Watts.

Thus, 1 mw (milli Watt) = 10e-3 W = 10 * log(1) dBm = 0 dBm

Now, 0 (in dBm) / 0 (in dBm) = 1 (but Obvious)

So fellas, spot any fallacies?

Gee, should acknowledge some guy in SD for some funny questions on dBm scale which led to this q. Me thinks it makes cracker of a q

[Guest]

Its simple

zero/zero=?

If we divide equation with z we get

ero/ero=1

and so on finaly it is

zero/zero=1

Edited August 21, 2008 by OstapBender

[Guest]

Speaking of such problems I always argued with my high school teacher that infinity*0=1 always.

The thing is there is no real infinity nor real zero so as much as you lean towards infinity you can lean 1/infinity towards zero.

So if someone says 1873182936491 is infinite number in such system 1/1873182936491 must be 0 and 1873182936491*1/1873182936491 must be 1...

[Guest]

a = 1

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

(a+B.) (a-B.) = b(a-B.)

divide through by a-b

(a+B.) = b

a + b -1 = b-1

1 + 1 - 1 = 1-1

1 = 0

so

0/0 = 1

is the same as

0/1 = 0

and therefore the answer is always.

But earlier you divided by a-b, a=b, therefore a-b=0, You can't divide by 0

Its undefined