For instance how do we know that something weird doesn’t happen at some insanely large number?

In: 3

Some maths doesn’t change as the numbers get bigger. So as an example, for a function f(x) – meaning we put in any number x and get a result f(x) – if we can show that f(x+1) is bigger than f(x), then we know that f(x) goes to infinity as x gets larger.

When you start to tie the maths to physical objects, that’s when things get complicated.

I’ll try and give a simple example:

Let’s start at the number 2. Then we add 2 and get to 4. Then we add 2 again and get to 6, then to 8,10,12 etc. Then we add 2 again and again and again. We can do that “to infinity”.

Every single number we will ever get to will be even. I am very sure nothing “weird” will happen at “some insanely large number”.

Other sequences that go to infinity like this one have similar, although probably more complicated, rules. And from these rules we can infer all kinds of patterns and other rules, even though we can practically only observe a minuscule fraction of infinity.

In mathematics something goes to infinity if you can choose any number you like, and it goes bigger than that.

So like y = x squared. It goes to +infinity because if you pick a number, like 10000, I can pick a number like 101 and 101 squared is bigger than 10000. So then you say 1,000,000 and I say 1,001. then you say 999,999,999,999,999 and I say 999,999,999 squared is bigger. No matter what you pick there’s a number that’s bigger when you square it.

It doesn’t go to -infinity though, because if we do the opposite, it doesn’t work. If you say 5 I say 2 squared is lower than 5. But if you say -1 I’m stuck because you can’t square anything and get a number lower than -1. (we’re not doing imaginary numbers)

First of all, we need to have a good definition of “going to infinity”. It’s usually the following :

A sequence of numbers goes to infinity if for any threshold, I can find a rank (depending on that threshold) such that the sequence is larger than that threshold after that rank.

So usually, the recipe to prove that something goes to infinity is to give an explicit way to find a rank that satisfy the property when you are given an arbitrary threshold.

Let’s give two examples :

Say you have the sequence 1, 3/2 , 2, 5/2, 3, 7/2, … where you add 1/2 to the previous number at each step. It should be clear that this sequence is going to infinity, but let’s write the formal proof.

Fix an arbitrary threshold M (for convenience, let’s say that M is an integer). All you have to tell me is that there is a rank after which all the values of the sequence are larger than M. (note that it does not have to be the best rank, just one that works). For example, you can convince yourself pretty easily that after rank 2M, the values are all larger than M. That’s your proof. Nothing weird happens because you know the rules

Now, if you want a less trivial example, look at the sequence 1, 1+1/2 , 1+1/2+1/3 , 1+1/2+1/3+1/4 , … where you add 1/n to the previous number at each step where n is the rank in the sequence. It’s far less clear that this goes to infinity (and can be quite counter-intuitive as even after a billion terms, the sequence is still around 21 and is increasing veeeeery slowly).

But let’s write a proof anyway ! Fix a threshold M. Can you find a rank such that all values after that rank are bigger than M ? Not easy, but let’s convince ourselves that we can using the previous easy sequence and looking at the first terms :

The sequence starts like the previous one 1 , 3/2, so that’s a good start, but then it grows more slowly. But is there a term in our new sequence that is larger than 2 (which is the next term in our first easy sequence) ? Yes, because 1/3+1/4 > 1/4+1/4 = 1/2. So the 4th term is larger than 2.

Let’s get to the next level : is there a term in our new sequence that is larger than 5/2 ? Yes, because 1/5+1/6+1/7+1/8> 1/8+1/8+1/8+1/8 = 1/2. So the 8th term of the sequence is larger than 5/2.

Let’s get to the next level : is there a term in the sequence larger than 3 ? Yes because 1/9+…+1/16> 1/16+…+1/16=1/2. So the 16th term of the sequence is larger than 3.

So for any threshold M, you can convince yourself with that same trick that the value at rank 2*2^M of the sequence will be larger than M. Bingo, that’s a proof.

Mathematical proofs aren’t observational. We know it because we can prove it, using the definition of a limit. Yes, we can often take a good guess based on observations, but that’s not a proof.

The definition of a limit itself is pretty formal, and a bit hard to ELI5. But it is a definition, and you can prove that a certain function meets the definition.