1. ## Mathematical Induction

Can anyone verify whether the following proof is correct or incorrect? If incorrect please explain where and why I went wrong?

Use mathematical induction to show that given a set of n+1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set.

S = { $\displaystyle x_i$ such that i = 1,2,...,n+1, $\displaystyle x_i$ $\displaystyle \le$ 2n}

Show that $\displaystyle \exists$ $\displaystyle x_i$ such that $\displaystyle x_i$ | $\displaystyle x_j$ i $\displaystyle \not=$ j

Basis Step: P(1): case 1: S={1,2} 1 | 2 so P(1) is true
case 2: S={1,1} 1 | 1 so P(1) is true

Inductive Step: P(k) $\displaystyle \Rightarrow$ P(k+1)

Given S = { $\displaystyle x_i$ such that i = 1,2,...,k+1, $\displaystyle x_i$ $\displaystyle \le$ 2k}
$\displaystyle \exists$ $\displaystyle x_i$ such that $\displaystyle x_i$ | $\displaystyle x_j$ i $\displaystyle \not=$ j

Prove that $\displaystyle \exists$ $\displaystyle x_s$ such that $\displaystyle x_s$ | $\displaystyle x_t$, s $\displaystyle \not=$ t
where $\displaystyle x_s$ $\displaystyle \epsilon$ { $\displaystyle x_i$ such that i = 1,2,...,k+1,(k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)}

Note that { $\displaystyle x_i$ such that i = 1,2,...,k+1,(k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)} = S $\displaystyle \cup$ { $\displaystyle x_i$ such that i = (k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)}

Choose $\displaystyle x_s$ = $\displaystyle x_i$ that divides $\displaystyle x_j$ and $\displaystyle x_t$ = $\displaystyle x_j$ from the given part.

This proves P(k+1) is true, so the original statement is true by M.I.

2. Originally Posted by oldguynewstudent
Can anyone verify whether the following proof is correct or incorrect? If incorrect please explain where and why I went wrong?

Use mathematical induction to show that given a set of n+1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set.

S = { $\displaystyle x_i$ such that i = 1,2,...,n+1, $\displaystyle x_i$ $\displaystyle \le$ 2n}

Show that $\displaystyle \exists$ $\displaystyle x_i$ such that $\displaystyle x_i$ | $\displaystyle x_j$ i $\displaystyle \not=$ j

Basis Step: P(1): case 1: S={1,2} 1 | 2 so P(1) is true
case 2: S={1,1} 1 | 1 so P(1) is true

Inductive Step: P(k) $\displaystyle \Rightarrow$ P(k+1)

Given S = { $\displaystyle x_i$ such that i = 1,2,...,k+1, $\displaystyle x_i$ $\displaystyle \le$ 2k}
$\displaystyle \exists$ $\displaystyle x_i$ such that $\displaystyle x_i$ | $\displaystyle x_j$ i $\displaystyle \not=$ j

Prove that $\displaystyle \exists$ $\displaystyle x_s$ such that $\displaystyle x_s$ | $\displaystyle x_t$, s $\displaystyle \not=$ t
where $\displaystyle x_s$ $\displaystyle \epsilon$ { $\displaystyle x_i$ such that i = 1,2,...,k+1,(k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)}

Note that { $\displaystyle x_i$ such that i = 1,2,...,k+1,(k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)} = S $\displaystyle \cup$ { $\displaystyle x_i$ such that i = (k+1)+1 $\displaystyle x_i$ $\displaystyle \le$ 2(k+1)}

Choose $\displaystyle x_s$ = $\displaystyle x_i$ that divides $\displaystyle x_j$ and $\displaystyle x_t$ = $\displaystyle x_j$ from the given part.

This proves P(k+1) is true, so the original statement is true by M.I.
It is not wrong per say, but the reasoning is a little off. For the P(k+1) case you want to make sure that you have a set of elements that are no greater than 2k, and have k elements. Here are the three possible cases.
Case 1:
Now if more than one element is 2k+1 or 2k+2, then P(k+1) is true as any number is divisible by itself.
Case 2:
If only one element or no element is greater than 2k then we can apply the induction hypothesis (by looking at the first k+1).
Case 3:
Precisely one element is 2k+1 and one 2k+2, in which case we have k elements that are less than or equal to 2k (we need k+1).
If we have 2 in our set, then 2|(2k+2) and we are done.
Suppose that 2 is not in our set and notice that for $\displaystyle a\neq 2$, a|(2k+2) iff a|(k+1), as 2 is prime. Also $\displaystyle k+1\leq 2k$. So replace the element 2k+2 by k+1, and you get k+1 elements that are less than 2k. Apply the inductive hypothesis to conclude this case.

3. If you also wish to prove this not by induction, then you may do it this way:

Let $\displaystyle a_1,a_2,...,a_{n+1}$ be our numbers. Let us define a sequence $\displaystyle \{b_k\}_{k=1}^{n+1}$ as $\displaystyle b_i = a_i (mod \ n) \ \forall 1\leq i \leq n+1$.

Now, since there are only $\displaystyle n$ possible values for $\displaystyle b_i$ , from the pidgeonhole principle we get that at least two elements in that sequence must equal each other (since there are $\displaystyle n+1$ elements for $\displaystyle n$ possible values), and from that, one of them must divide the other.

4. ## I finally got it

Originally Posted by Focus
It is not wrong per say, but the reasoning is a little off. For the P(k+1) case you want to make sure that you have a set of elements that are no greater than 2k, and have k elements. Here are the three possible cases.
Case 1:
Now if more than one element is 2k+1 or 2k+2, then P(k+1) is true as any number is divisible by itself.
Case 2:
If only one element or no element is greater than 2k then we can apply the induction hypothesis (by looking at the first k+1).
Case 3:
Precisely one element is 2k+1 and one 2k+2, in which case we have k elements that are less than or equal to 2k (we need k+1).
If we have 2 in our set, then 2|(2k+2) and we are done.
Suppose that 2 is not in our set and notice that for $\displaystyle a\neq 2$, a|(2k+2) iff a|(k+1), as 2 is prime. Also $\displaystyle k+1\leq 2k$. So replace the element 2k+2 by k+1, and you get k+1 elements that are less than 2k. Apply the inductive hypothesis to conclude this case.
Thanks so much for this! I've been going over it off and on for the past couple of hours and I've finally lit the led in my head!

I've also been struggling with a stubborn home-made Macgyvered electrolysis experiment I cooked up for an inner city Chemistry class I'm observing for my sec ed class. It's been 35 years since I did chemistry and it shows. The chem teacher's specialty is Biology so I've been able to teach a couple of classes. I figured out my error there also, so things are looking up!

5. Originally Posted by oldguynewstudent
Thanks so much for this! I've been going over it off and on for the past couple of hours and I've finally lit the led in my head!

I've also been struggling with a stubborn home-made Macgyvered electrolysis experiment I cooked up for an inner city Chemistry class I'm observing for my sec ed class. It's been 35 years since I did chemistry and it shows. The chem teacher's specialty is Biology so I've been able to teach a couple of classes. I figured out my error there also, so things are looking up!
Glad to hear that. Good luck with your studies. Hopefully you can spend the next 35 years doing maths .