# Proving natural numbers in Pascal's Triangle

1. Jul 21, 2013

### Seydlitz

1. The problem statement, all variables and given/known data
Taken from Spivak's Calculus, Prologue Chapter, P.28

b) Notice that all numbers in Pascal's Triangle are natural numbers, use part (a) to prove by induction that $\binom{n}{k}$ is always a natural number. (Your proof by induction will be be summed up by Pascal's Triangle)

2. Relevant equations

$$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$

3. The attempt at a solution

I'm quite disoriented to be honest in applying this method of proof in completely new situation like this. Moreover I'm not that comfortable yet with manipulating binomial coefficient, though I've managed to prove (a) using the definition given and I've no problem so far in doing proof by induction to sum of series and the first problem in the chapter.

It's very clear to me that binomial coefficient will only be composed of natural numbers, I just need to build the logical framework in which to prove it. I need your help guys, what is your view to the problem. I'll be very verbose in doing this.

So following the step of the proof by induction that goes like this:
(1) 1 is in A
(2) k+1 is in A, whenever k is in A

Ok so $$\binom{1}{1}$$ is 1 according to the definition. So I assume I've completed step (1).

Now let's try step (2). I can imagine that this equation adds two number one line above, and it is in fact true.
$$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$
Maybe $$\binom{n}{k}$$ will always give natural number, and it does in fact give 1 if n=k, or if k=0. Suppose k=1, then $$\binom{n}{k-1}$$ will give 1, a natural number. Hence the same will happen to $$\binom{n+1}{k}$$ Better than that, by step (2), I have also shown that binomial coefficient will always give natural number.

Secondly I want to ask what do you think of the problem no.3 in general in this chapter? Is it trivial? Is it normal for beginning students to stumble a bit in this type of question? I'm quite happy actually to be able to finish most of the problems in the preceding chapter before this.

Thank You!

2. Jul 21, 2013

### micromass

Staff Emeritus
This is a pretty tricky problem. It requires a clever induction. Why is this problem different from a regular induction problem? Because here you have two variables $n$ and $k$ instead of one.

So, a right move in the right direction would be to say exactly which statement to prove through induction. I propose the following:

P(n): The binomial coefficient $\binom{n}{k}$ is an integer for any $k$.

So, the first thing we need to do is to prove $P(0)$. This is rather easy.

Next, assume that $\binom{n}{k}$ is an integer for any $k$. Then you need to prove that $\binom{n+1}{k}$ is an integer for any $k$. So take an arbitrary $k$, apply $(a)$ and see what you get.

3. Jul 21, 2013

### reenmachine

Don't want to disturb the conversation , if not I will just bring it to my thread.

4. Jul 21, 2013

### micromass

Staff Emeritus
I think it's probably better to do it in your own thread.

5. Jul 21, 2013

### reenmachine

No problem!

6. Jul 21, 2013

### micromass

Staff Emeritus
Once the OP solved his problem here (or doesn't return), then you can ask questions in this thread. But before that, I think it might be a "hijack" of the thread, and we might give away solutions to the problem.

7. Jul 21, 2013

### Seydlitz

Ok thanks Micromass,

So I think this fact is true. $$\binom{n}{k}$$ will give integers if $k=0$ or if $n=k$. But k must always be $n \geq k$

Applying this to (a), we already accept the second term in the right will give integer. But in order for the first term to be defined then $(n+1) \geq k$ If that's so the second term will give at least $\binom{n+1}{n+1}$. Well this is supposed to give integers no doubt.

So are we there yet? I think we have already proved n and n+1, and k indirectly with n. Though I'm still feeling, I've not grasped all.

8. Jul 21, 2013

### micromass

Staff Emeritus
Why?

Sorry, I don't get your last sentence. Why can you assume $k=n+1$ ? (you can't by the way).

Also, the induction hypothesis says that $\binom{n}{k}$ is an integer for all $k$ such that $0\leq k\leq n$. Don't forget the crucial word "all".

9. Jul 21, 2013

### Seydlitz

They all give 1 according to the raw definition that includes factorial. So it will give integer right?

I'm assuming $n+1$ because it's the max value k can be in (a). I want to show that if that's true then as you said all k below that will give integer. If I just write k, I don't know what (k-1) will give with the first term in the right side.

10. Jul 21, 2013

### micromass

Staff Emeritus
I never said that all $k$ below give integers. Why is this true?

11. Jul 21, 2013

### Seydlitz

It is all k right? If k in $\binom{n}{k}$ will give integer then so does where $n+1=k$ in (a). If all k is true then $k-1$ will be true also.

12. Jul 21, 2013

### Seydlitz

Edit: I just realized that I was false. I cannot assume n+1 = k in a. Because that will make k larger than n in the second term in the right.

I am not quite sure how to show the part you say for all k micromass.

I can execute instruction like take the square root of x and materialize that on to paper. But I am not quite sure when you say take arbitrary x. Yes x can take any value. But how do you symbolize it? Does it mean that I've to impose a condition instead in such a way for x to be arbitrary?

Last edited: Jul 21, 2013
13. Jul 21, 2013

### Seydlitz

I'm going to do the second try of the proof.

If $0 \leq k \leq n$, the binomial coefficient $\binom{n}{k}$ is defined by:
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$ if $k \neq 0,n$
$$\binom{n}{0}=\binom{n}{n}=1$$ considering that $0!=1$

Clearly the binomial coefficient will give an integer of 1 in case
$$\binom{0}{0}=\binom{n}{n}=1$$
This $\binom{0}{n}$ will only be possible if $n=0$, in that case it will also give 1.

For any $k$ in $0 \leq k \leq n$ I assume $\binom{n}{k}$ is an integer.
Then for any $k$, $\binom{n+1}{k}$ is supposed to be an integer.

We apply the formula (a), which has been proven previously in previous problem, and is assumed to be true.
$$\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$$
Then it is clear that for any $k$ in $0 \leq k \leq n$,$\binom{n+1}{k}$ is an integer and the proof is complete. █

14. Jul 22, 2013

### Theorem.

You have got the right idea here. The main part of your argument is just about spot on. The whole thing just needs a little cleaning up. First, what is your base case?
If you are proving by induction on k for $0\leq k \leq n$ then your base case should be the statement $n \choose 0$ regardless of the value of n, what does this give you? is it a natural number? you have answered these questions but in a somewhat confusing manner. After the base case your argument is a lot clearer and I understand it well.

Last edited: Jul 22, 2013
15. Jul 22, 2013

### Theorem.

I should add something: It is probably better to state that you are doing strong induction here. You need to justify why $n \choose {k-1}$ is also an integer. Your inductive hypothesis also isnt correct. You are assuming the statement holds for all n choose k. You need to say suppose $n \choose k$ is a natural number for all $k<l$ for some natural number l. Then you show that $n\choose l$ is also a natural number using the argument you used. Does this make sense to you?

16. Jul 22, 2013

### Seydlitz

Yes it is indeed missing something, I can't feel it right yet. The task is to prove by induction that $\binom{n}{k}$ is always a natural number. So the base case as suggested by Micromass, is to prove first that $\binom{n}{0}$ is a natural number. According to the definition whatever $n$ is, the value is always 1. 1 is a natural number.

Afterwards I've to prove $\binom{n+1}{k}$ is true for all $k$ in $0 \leq k \leq n$. But this seems to be the problematic part as I cannot get the necessary argument right.

Ok it's getting better, but how to justify precisely that $n \choose {k-1}$ is an integer? By going back to the definition? $$\binom{n}{k-1}=\frac{n!}{(k-1)!(n+1-k)!}=\frac{n(n-1)...(n+2-k)}{(k-1)!}$$

I still don't understand the last part entirely, I assume the hypothesis can be proven if I show $\binom{n+1}{k}$ is an integer.

Last edited: Jul 22, 2013
17. Jul 22, 2013

### Theorem.

Once you have rewritten your induction hypothesis as strong induction, the statement would actually be automatic. EXCEPT I have noticed a flaw that I initially glanced over. If you are doing induction on k, your hypothesis should read as follows (strong induction) :
Assume $n\choose l$ is a natural number for all $l<k$ for some natural number k. Then in order to prove by induction we must show it is also true that $n \choose k$ is a natural number.

you mixed up what you are doing induction on. Are you doing induction on n or k? in your base case it is definitely k, but you seems to switch to n for your induction step. Think this through and decide what works best as MICROmass suggested. I think I may have confused you because I read your induction base step and thought you were doing induction on k but really you were doing induction on n as micromass suggested. I apologize for that.

18. Jul 22, 2013

### Theorem.

To clarify:
Base Case: $0 \choose k$ is a natural number (show this as you did).
Inductive hypothesis: suppose for some natural number l $l \choose k$ is a natural number for all natural numbers k.

Inductive step: Show that then ${l+1} \choose k$ is also a natural number for any k.
You have $$\binom{l+1}{k} =\binom{l}{k-1} + \binom{l}{k}$$
Why is $l \choose {k-1}$ a natural number? HINT: your inductive hypothesis.
Why is $l \choose {k}$ a natural number? HINT: same reasoning.

SORRY for the confusion earlier, i really wasn't sure what you were applying induction to (n or k). This post should fix that. DISREGARD everything i said before!

19. Jul 22, 2013

### Seydlitz

No problem Theorem, you're helping me, I should thank you for your time instead.

And yes I'm doing induction on $n$ with arbitrary $k$. I think that is right. I just don't know how to write it down for arbitrary k. Should I just say it like this crudely:

"Okay so this is arbitrary k, believe it, as long as it is not equal or greater than $n$, then I propose $\binom{n}{k}$ to be integer, then I show afterwards using formula that $\binom{n+1}{k}$ with arbitrary k is in fact an integer. Ok I've shown you. Then the proof is complete.

Edit: I have not read post 18 when writing this.

20. Jul 22, 2013

### Theorem.

I think i clarified this in the last post I managed to sneak in before you posted this. Basically by the inductive hypothesis $\binom{l}{k-1}\in \mathbb{N}$ (we assumed this) and
also $\binom{l}{k}\in \mathbb{N}$ for the same reasoning.
Recall that the assumption was that $\binom{l}{k} \in \mathbb{N}$ for ANY k, so it works for k-1 as well as k. Then the final result is obtained because the sum of two natural numbers is a natural number! As you have suggested