1/infinity =?
It would be my guess that $\displaystyle \frac{1}{\infty}$ will be extremely close to 0. Some would argue that is IS 0, but I don't think that will ever be possible, no matter how large your number.
$\displaystyle \frac{1}{2}=0.5$
$\displaystyle \frac{1}{20}=0.05$
$\displaystyle \frac{1}{200}=0.005$
$\displaystyle \frac{1}{2000}=0.0005$
$\displaystyle \frac{1}{20000}=0.00005$
As you can see, as the denominator increases in size, the answer decreases but never seems to actually get to 0 - the 0 can always recur one decimal place further.
Hi Mukilab,
Good question. See here: Math Forum - Ask Dr. Math
1/(infinity) is definitely not equal to 0. lets see why
suppose
1/(infinity)=0 then
1=(infinity)x0
but product of 0 and any other number is 0 therefor
1=0 which is not possible ever
therefor the supposition taken is wrong hence 1/(infinity) is never equal to 0.
now the question arise what is 1/(infinity)
the value 1/x will get smaller and smaller if x is increased
since infinity is the biggest value for x then 1/x should be the smallest value, and the smallest value is near to 0 (as we can not take exact zero as answer).
one more thing...since infinity is indeterminate therefor 1/x will also be indeterminate (as one can not give the smallest value in numerical form)
Infinity is a bit like "Ixtlan".
You can keep moving in the general direction without ever really getting "there"
wherever "there" may be. Infinity is more a concept of "ever-increasing", without stop.
Just as, if "something" keeps decreasing, without ever stopping, it never makes it to zero. (You can imagine becoming billions of times smaller than an electron).
If we say
$\displaystyle \infty+1=\infty$
then
$\displaystyle \Rightarrow\ \frac{\infty}{\infty}+\frac{1}{\infty}=\frac{\inft y}{\infty}$
$\displaystyle \Rightarrow\ 1+\frac{1}{\infty}=1$
and we can compare that to $\displaystyle 1+0=1$
We cannot define infinity as a value, since whatever value we may consider,
we may consider a larger one. The ratio being set to zero is more a reflection of the "relative ratio" of the "values", one insignificant compared to the other.
Hence, we consider values approaching what we call infinity,
but the calculations of the limits are based on a constant being insignificant
compared to a value that never stops increasing,
that is why constants divided by infinity are allowed to be considered zero,
when deciding on a value for a limit.
What we are really evaluating usually, is a value being "approached", as though we "could" reach it.
THanks, I understand now.
What I do no understand, however is how when you need 1 more person in a hotel with infinite rooms but infinitely full with infinite people you only need to move people up a room. However if you do this with let's say several friends then you need room 1-2, then 2-4, 4-6. Why?
In a "hypothetical" situation,
infinitely many people, infinitely many rooms,
it's being considered that there are "always" rooms available,
since neither the number of people nor the number of rooms have
been numerically quantified.
In other words, they are both countless with always "more" available.
The numbers of either are not being "defined".
If the hotel is full and you want to accomodate another person,
we need another room.
These quantities are "defineable" and can be dealt with "exactly".
If you are dealing with the real number system, $\displaystyle 1/\infty$ isn't equal to anything because $\displaystyle \infty$ is not a real number.
There are a variety of ways of "extending" the real number system to include something called "infinity". Whether you can divide by infinity and, if so, what the result is, depends on how the real number system is extended.
Infinity is not a number but a concept, used in what we call limits, with which we can prove that $\displaystyle \lim_{x \to \infty} \frac{1}{x} = 0$, that is, as $\displaystyle x$ approaches infinity, $\displaystyle \frac{1}{x}$ approaches zero (but never reaches zero because $\displaystyle \frac{1}{x} \neq 0 \ \forall x \in \mathbb{R} - \{0\}$ (we don't include zero since division by zero is undefined). Proof : if $\displaystyle \frac{1}{x} = 0$ for some real $\displaystyle x$ different from zero, it implies that $\displaystyle 1 = 0$ which is necessarily wrong.
Exactly,
The problem is, if we write $\displaystyle \frac{1}{x}=0\ \Rightarrow\ 1=x(0)$
however, in the real number system, zero of any value is zero.
Hence $\displaystyle x(0)=0$
$\displaystyle x(0)=1$ is a contradiction in the reals.
There is no real value x, for which $\displaystyle \frac{1}{x}=0$