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Help: Understanding infinity

Hello, i have a few questions regarding infinity, which i'd like to discuss with you all.

• Is $$\frac{1}{0} = \infty$$
• Is $$\frac{1}{\infty} = 0$$
• Is $$\tan90^\circ = \pm\infty$$
• Is $$\log_x0 = -\infty$$

All my teachers tell me that the above expressions are undefined, and so is $$\infty$$ (undefined), and we cannot compare the two.

I understand that we cannot determine the exact value of the above expressions, but what is the problem of terming them as $$\infty$$?

On graphing these, we do notice that they approach infinity, so why say that they are undefined? You may say that infinity is a concept, and not an actual number, but then isn't everything a concept?

If the number 1 is defined, then what is its definition? 1? (circular proof) Every number is a concept, and I don't see how certain operations on them cannot lead to infinity, (another concept).

Note by Harshit Kapur
5 years, 2 months ago

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Harshit,

There are numerous questions that you raise in this single discussion. In general, when you're asking for clarification, keeping things to a single question / query (and opening several discussions) could help ensure that you get proper replies to all of them.

I can classify the numerous questions that you raised in the following 4 broad categories:

1) What is infinity? Can we compare infinities? Can we do any arithmetic with infinities? What do those equations mean, if anything?

2) What concepts can be represented by another concept? For example, can the concept of 1 be represented by the number line? Can the concept of infinity be represented by the number line? Can the concept of pigeonhole principle be represented by the number line?

3) What is the definition of 1? How about 2, 3? How about 0? Is there a way to define it that isn't circular?

4) Why are some operations defined on some numbers, but not on others?

For the purposes of this discussion, let's keep it mainly to 1. If you're interested in the others, please open up another discussion topic for each of them.

Staff - 5 years, 2 months ago

1) I will (partially) answer the first part. The is significant theory involved, that I'm not going into detail about. If you are interested, you can read up on set theory.

The concept of infinity is best expressed using set theory. Without having to assume the existence of 0 or 1, we know that there is the empty set $$\emptyset$$, Furthermore, we can build more sets by taking the union of their elements, or taking the set of the set. We know that there is the set of the empty set $$\{ \emptyset \}$$. We know that there is the set of the union of them empty set and the previous set. $$\{ \emptyset, \{ \emptyset \} \}$$. So on and so forth.

We say that 2 representations are equal, if there is a bijection between them. For example, $$\{ \emptyset \}$$ and $$\{ \{ \emptyset \} \}$$are equivalent. (Prove the following statement using the transitive property of bijections.) These representations form an equivalence relation, which means that it doesn't matter which choice of representations we use.

This allows us to build up a concept of counting. We can let the a set of $$n$$ elements represent the integer $$n$$. In the first paragraph, we gave representations of 0, 1 and 2. Verify for yourself that taking the union of 2 sets is equal to taking the sum of the integers that they represent. What does taking the set of a set do to the integer? I.e. What is $$\{ (any set) \}$$ equal to?

Any set that can be built up in this way (taking unions, and sets), is called a finite set. Any (nonnegative) integer that is represented in this way is finite.

An infinite set is a set that is not finite. An infinite number is a number represented by an infinite set. It is not obvious that an infinite set exists, since it is defined "not something".

You should show the following:

1. The set of natural numbers is an infinite set. Hint: To show this, show that it is not equivalent to any finite set. Hence, it is "not finite set".

2. The set of natural numbers is equivalent to the set of rational numbers

3. The set of natural numbers is not equivalent to the set of real numbers.

Staff - 5 years, 1 month ago

1 Denote the set of natural numbers by N. Fix a natural number n. Denote the set whose n elements are indexed xi, i = 1 to n, by Xn. Consider the map F: Xn => N, F(xi) = i. Then n + 1 (and all higher natural numbers) cannot be in the range of F because Xn has only the n elements xi, i = 1 to n. That is, F cannot be onto. But then, N cannot be equivalent to any finite set because if the order of that finite set was n, the only maps from it into N are of the form F(p), where p is a map into Xn, mapping different elements of the arbitrarily chosen finite set onto different xi. Therefore, N is an infinite set.

2 Any rational number can be expressed in the form (a - b) / s, where a, b, and s are natural numbers. So if we could prove that some subset of N is equivalent to the set of ordered triples N^3, that N is equivalent to the set of rationals Q would follow. Denote the set of natural numbers divisible by no primes higher than 5 by V. Consider the map v: V => N^3, v(2^a 3^b 5^s) = (a, b, s). Because prime factorization in N (hence in V) is unique, v is one to one. We therefore have the cycle N => Q => N^3 => V => N, where S1 => S2 means S1 is equivalent to a subset of S2. As a side note, this logic easily extends to the fact that N is equivalent to the set of algebraic numbers A, which by definition is all complex numbers that satisfy polynomials of finite degree n, with coefficients in Q and with nth-power coefficient 1, except this time replace V with N itself (and use the fact that N is equivalent to Q).

3 Cantor's diagonal argument is our main tool for proving that N cannot be equivalent to set of real numbers R. It's easier if we replaced R with E, where the members of E are precisely the real numbers between 0 and 1 inclusive. Consider the map r: N => E, defined as follows:

r(0) = 0.a(0,0) a(0,1) a(0,2) a(0,3) ...

r(1) = 0.a(1,0) a(1,1) a(1,2) a(1,3) ...

r(2) = 0.a(2,0) a(2,1) a(2,2) a(2,3) ...

r(3) = 0.a(3,0) a(3,1) a(3,2) a(3,3) ...

and so on. Now consider x = 0.e0 e1 e2 e3 ... where

ei = a(i,i) + 5, 0 ≤ i ≤ 4,

ei = a(i,i) - 5, 5 ≤ i ≤ 9.

Now x cannot be in the range of r because if it was, each of the ei would equal the respective a(j,i) for some fixed j, and in particular ej would equal a(j,j) which it can't. Ergo, r cannot be onto. We can therefore conclude that N is not equivalent to E (and hence, to R).

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I have a doubt

0^n=0 Where n is any real number n^0=1

Where n is any real number

Then what is 0^0 ?
is it 0 or 1?

- 2 years, 1 month ago

There's actually a wiki on it: What is $$0^0$$.You can check that out.

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I have a comment on one over zero: $$\frac{1}{0}=\infty$$ Multiply both sides by zero to remove the fraction bit. You get 1=0

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We can show that any number is equal to any other number by setting $$\frac{1}{0}=\infty$$

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hi we dont no wht infinity is so we say that till now it is not defined. whn we divide any no. by very small no. surely quotient will be>>> tht means infinity\not defined.is answer

- 5 years, 2 months ago

IT IS UNMEASURABLE

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Post this as a note. It will reach to more people that way.

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