Proofs involving the addition of natural numbers
Okay, kiddo, let's talk about adding numbers!
When we add two natural numbers together, we start with one number and then we add more of something else to it. For example, if we add 2 and 3 together, we start with 2 and then add 3 more to get a total of 5.
Now, sometimes we need to prove that adding two numbers together works in a certain way. To do this, we use something called a proof.
A proof is like a big puzzle where we have to figure out the pieces in order to show that something is true. And when we're proving things about adding numbers, we usually use something called mathematical induction.
Induction is like building a tall tower. We start with a small block and then we add more blocks, building higher and higher. When we do mathematical induction, we start with a small number and then we show that if something is true for that small number, then it's also true for the next bigger number, and so on.
For example, let's say we want to prove that if we add up the first n natural numbers (so 1+2+3+...+n), we get a certain formula. We can use induction to do this kind of proof.
First, we check if the formula works for the smallest possible n, which is usually 1. In this case, 1 is the smallest natural number, so we check if the formula works when n=1. If it does, we move on to the next step.
Next, we assume that the formula works for some arbitrary number k, which means that 1+2+3+...+k is equal to some formula. Then we want to prove that the formula also works for the next number, which is k+1.
To do this, we take the formula for 1+2+3+...+k and add (k+1) to both sides. That gives us (1+2+3+...+k)+(k+1). But we know that 1+2+3+...+k is equal to our formula (because we assumed that), so we can substitute that in:
formula for 1+2+3+...+k + (k+1) = formula for 1+2+3+...+k+(k+1)
Then we simplify the left side using our formula:
(1+2+3+...+k) + (k+1) = [k(k+1)/2] + (k+1)
And we can simplify that further:
[k(k+1)/2] + (k+1) = [(k+1)(k+2)/2]
So we've shown that the formula works for k+1, and since we started with n=1 and showed it works for n+1, we know that the formula works for all natural numbers!
That's the basic idea of using proof by mathematical induction to show that something is true about adding natural numbers. We start with a small example, assume that something is true for some number, and then show that it must also be true for the next number. By doing this over and over again, we can prove that something is true for all natural numbers.
When we add two natural numbers together, we start with one number and then we add more of something else to it. For example, if we add 2 and 3 together, we start with 2 and then add 3 more to get a total of 5.
Now, sometimes we need to prove that adding two numbers together works in a certain way. To do this, we use something called a proof.
A proof is like a big puzzle where we have to figure out the pieces in order to show that something is true. And when we're proving things about adding numbers, we usually use something called mathematical induction.
Induction is like building a tall tower. We start with a small block and then we add more blocks, building higher and higher. When we do mathematical induction, we start with a small number and then we show that if something is true for that small number, then it's also true for the next bigger number, and so on.
For example, let's say we want to prove that if we add up the first n natural numbers (so 1+2+3+...+n), we get a certain formula. We can use induction to do this kind of proof.
First, we check if the formula works for the smallest possible n, which is usually 1. In this case, 1 is the smallest natural number, so we check if the formula works when n=1. If it does, we move on to the next step.
Next, we assume that the formula works for some arbitrary number k, which means that 1+2+3+...+k is equal to some formula. Then we want to prove that the formula also works for the next number, which is k+1.
To do this, we take the formula for 1+2+3+...+k and add (k+1) to both sides. That gives us (1+2+3+...+k)+(k+1). But we know that 1+2+3+...+k is equal to our formula (because we assumed that), so we can substitute that in:
formula for 1+2+3+...+k + (k+1) = formula for 1+2+3+...+k+(k+1)
Then we simplify the left side using our formula:
(1+2+3+...+k) + (k+1) = [k(k+1)/2] + (k+1)
And we can simplify that further:
[k(k+1)/2] + (k+1) = [(k+1)(k+2)/2]
So we've shown that the formula works for k+1, and since we started with n=1 and showed it works for n+1, we know that the formula works for all natural numbers!
That's the basic idea of using proof by mathematical induction to show that something is true about adding natural numbers. We start with a small example, assume that something is true for some number, and then show that it must also be true for the next number. By doing this over and over again, we can prove that something is true for all natural numbers.
Related topics others have asked about: