Infinity Question?

Question

First you would agree that 1/∞=0 am I correct?
so then if you do (1/∞)∞ would it be 1 or would it be 0 for the answer. If you multiply a number by the recipical you get 1 which they are i suppose. but if you multiply a number my 0 which 1/∞ is then you get 0.

Answer 1

(1/∞)∞ is essentially ∞ / ∞
This is an indeterminate form.
Actually 1*∞ is an indeterminate form.
So you can't say what it is. In some cases it may be zero, in other cases it could be 1.

Here are some links:
http://mathworld.wolfram.com/Indeterminate.html
http://en.wikipedia.org/wiki/Indeterminate_form

Answer 2

This is a good question but infinity is not a number. It cannot be algebraically manipulated. Think in terms of limits. If you have an equation that goes to 0 faster than the number it is being multiplied by it will be 0.
(1/n) goes to 0 slower than 1/(n^2)

Answer 3

First you would agree that 1/∞=0 am I correct?

That's the problem right there.

Based on that equation, 1/0=∞ would also have to be true. Let's look at that equation.

lim 1/x Does not exist because
x->0

lim 1/x =∞
x->0+

lim 1/x =-∞
x->0

So, there is no general limit. If you haven't learned about limits yet, here's a simpler explanation.

lim 1/x =∞
x->0+

This means that as the value of x approaches 0 from the right (0.1, 0.01, 0.001...), f(x) approaches infinity. There are no problems here.

lim 1/x =-∞
x->0-

Here's the problem. As x approaches infinity from the left (-0.1, -0.01, -0.001...) f(x) approaches negative infinity.

So, they approach two different values from the left and the right, so you can't make the assumption that 1/0 = ∞.

For a visual, graph 1/x to see why the value of 1/0 is left undefined.

Answer 4

Well, i disagree with you. The statement 1/∞ is not = 0.
In order to understand infinity you need to consider convergence and divergence. For example:
lim 1/x = 0
x -> ∞
The latter statement says that as x approaches infinity, then 1/x will become zero. Now:
lim (1/x)^x = 0 because the inner statement 1/x = 0
x -> ∞
and finally:
lim (1/x)(x/1) = lim (1) = 1
x -> ∞ x -> ∞

Remember, infinity is not a value. When you learn L'Hôpital's rule, if you haven't done so, you'll understand that the way you approach a variable to infinity is very important.

Answer 5

don't think of infinity as a number that would follow all the rules of an integer. look at infinity /infinity for example.... you would think that if you divide infinity by infintiy it would be 1..... incorrect.
I will skip talking about indeterminite values (you will learn exaclty what that means once you get into some more advance calculus) instead I will give you an example showing you that 2 infinitys are not always the same.

Cosider 2 sets of numbers both with infinite numbers in the set. The first set is the set of all integers ... we will call that set A. The seconed is the set of all even integers .. call that set B.

Now even though both set A and Set B have infinite numbers in their set they are not equal. If we look at 1 to 1 correspondence of parts of the set we can see that as we are increasing in set A, we will be incresing values in set B much faster.
A: ( 1,2,3,4,5......)
B : (2,4,6,8......)
So if we look at the nth integer in set A it will Corresponde to the Nth integer in Set B but their values will be VERY different. So dividing these 2 infinities would never equal 1.
...That is an easy example of why you can't use infinity as an integer and expect the same results.
.... For those math gurus out there please excuse my incorrect use of set symbols as I just didn't feel like finding the right keyboard keys to make it all look good.

Answer 6

NOTICE: I am not a mathametician, however , My mentor is a math teacher who has more degrees than a thermometer ( Hello, Dr. Michael Payne ! ) who broke down the mysteries of the universe to me and I will try to impart the wisdom as he did to me.

From what I know infinity as a mathematical concept its quantitative value is greater than any and all numbers. However, if one was to multiply or divide any number by zero it will be zero. And since infinity as a mathematical concept can be increased ( infinity + 1 is greater than infinity ). It would follow as well that it would be subject to same mathematical rules, hence it can be decreased as well. So therefore I would answer your question as zero because the same rules would apply in this case as well

Answer 7

"infinity" is not a real number; therefore, arithmetic with it is not possible. Nevertheless, it is possible to define infinite "numbers" which have their own form of arithmetic. There are different sizes of infinity, and a whole rich theory involving them, with some famous unsolved problems. Another context in which arithmetic with infinity occurs is in the "non-standard real numbers". But in all of these theories, it is never true that 1/infinity = 0

Answer 8

The reason why the logic you are using does not work is you are using inf. as a number. You can not use inf as a number and the 1/inf. does not equal 0. It is the limit 1/x as x approaches inf that = 0.
Hope this explains a little bit. But your cleaver your way of thinking is legit if inf was just another number which it is not*.

Answer 9

The reciprical doesn't work because you can't guarantee it'll be the same number on top and bottom (infinity is not a variable or a definite number) I'd assume the answer would be zero because 1/(infinity) is definitly approaching zero (although it will never reach it). Anything multipled by zero is zero.

Answer 10

you can't manipulate them like .. you have to know "how fast" things are going to zero or infinity to know the composite effect.

what you are doing is like saying 0*1 = 0 * 2 therefore 1 = 2

Answer 11

I think i can help you Doc. Yes, 1/x will shrink to zero when x blows up, and (1/x)x goes to one as x blows up because we can cancel the x's before they get out of hand. Now suppose we have (1/x)(2x). As before 1/x shrinks and 2x blows up because x blows up. But this time when we cancel we get
the answer two, so really it depends on the rates that these things change.