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.

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.

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)

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.

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

This is both the reason why mathematics wants formal proofs instead of heuristics, intuition or induction, as well as why it sometimes is incredibly hard to really proof things.

Take f(x) = x, as it is the simplest example. “y goes to infinity” means that y gets at some point and onward larger than any finite bound. Then “f(x) goes to infinity when x goes there” is almost tautological, as it says: “whenever x gets and stays larger then any finite bound, then x gets and stays larger then any finite bound”. Or without the fluff “if A then A”, which is absolutely always true.

Now for other functions like x² , e^x , log(x) , or x+100·sin(x) the argument might be more convoluted. Lets try the last one:

“Whenever x gets and stays larger then any finite bound, then x+100·sin(x) gets and stays larger then any finite bound”.

We can note that sin(x) is between -1 and 1, therefor we know that x+100·sin(x) is at least as big as x-100. Looking as x-100 as a kind of “worst case” simplifies things:

“Whenever x gets and stays larger then any finite bound, then x-100 gets and stays larger then any finite bound”.

No more sine to deal with! If we now want x-100 to be (and stay) larger than a finite bound (lets say 1,000,000), then we guarantee that by choosing x itself to be bigger than that bound plus 100 (here: 1,000,100). This proves what we wanted!

Never did we have to check an infinite number of cases by hand, nor is there any wiggly room, despite x+100·sin(x) wiggling up and down all the time (draw a graph if you want to see this). We argued purely by reason and logic, which goes further than simple calculation.

For some way questions, showing that something grows without bound can become seriously complicated; too complicated for even modern mathematicians. Only time will show if we are able to absolutely prove such things, but the chance for failure is always there. Heck, it is even an unsolved problem if there is a way to always and definitely decide (an “algorithm”) if two elementary expressions describe the same function.

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.

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.

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)

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