# Thread: Well ordering property and induction

1. ## Well ordering property and induction

From ROSEN, section 4.2, problem 36

The well-ordering property can be used to show that there is a unique greatest common divisor of two positive integers. Let a and b be positive inteters, and let S be the set of postive integers of the form as + bt, where s and t are intergers.

a) Show that S is nonempty.

Since a, b, s, and t are integers and the set of integers is closed under addition and multiplication, as + bt must be an integer.

My question is this: Is it implied that s and t are POSITIVE integers. If s and t were both negative integers, then S could be empty. Could someone explain what I'm missing?

b) Use the well ordering property to show that S has a smallest element c.

Since S was proved to be non-empty in a), then by the well ordering property, S must contain a smallest element since S is a set of positive integers. Call that element c.

c) Show that if d is a common divisor of a and b, then d is a divisor of c.

Since c is an element of S and all elements of S have the form as + bt, then by the theorem (that if d divides a and d divides b then d divides a + b) we show that d divides c.

d) Show that c divides a and c divides b.

Assume that c does not divide a, then a = qc + r where 0 < r < c.

I am stuck proving that r is an element of S, and I don't see how this contradicts the choice of c. I see that r is a positive integer, but how is it of the form as + bt?

Thanks

2. Originally Posted by oldguynewstudent
From ROSEN, section 4.2, problem 36

The well-ordering property can be used to show that there is a unique greatest common divisor of two positive integers. Let a and b be positive inteters, and let S be the set of postive integers of the form as + bt, where s and t are intergers.

a) Show that S is nonempty.

Since a, b, s, and t are integers and the set of integers is closed under addition and multiplication, as + bt must be an integer.

My question is this: Is it implied that s and t are POSITIVE integers. If s and t were both negative integers, then S could be empty. Could someone explain what I'm missing?

Where is that implied and why would it?? Just take $\displaystyle s=t=1\,\Longrightarrow\,0<as+bt=a+b\in\mathbb{Z}^+ \,\Longrightarrow\,S\neq\emptyset$

b) Use the well ordering property to show that S has a smallest element c.

Since S was proved to be non-empty in a), then by the well ordering property, S must contain a smallest element since S is a set of positive integers. Call that element c.

c) Show that if d is a common divisor of a and b, then d is a divisor of c.

Since c is an element of S and all elements of S have the form as + bt, then by the theorem (that if d divides a and d divides b then d divides a + b) we show that d divides c.

d) Show that c divides a and c divides b.

Assume that c does not divide a, then a = qc + r where 0 < r < c.

I am stuck proving that r is an element of S, and I don't see how this contradicts the choice of c. I see that r is a positive integer, but how is it of the form as + bt?

Thanks
Fort some $\displaystyle s\,,\,r\in\mathbb{Z}\,,\,c=ar+bt$ , so $\displaystyle r=a-qc=a-q(as+bt)=a(1-qs)+b(qt)\in S$ , since $\displaystyle r>0$ , and since also $\displaystyle r < c$ we get here a contradiction to the minimality of c in S.

Tonio

3. ## Thank you for your help so far

Originally Posted by tonio
Fort some $\displaystyle s\,,\,r\in\mathbb{Z}\,,\,c=ar+bt$ , so $\displaystyle r=a-qc=a-q(as+bt)=a(1-qs)+b(qt)\in S$ , since $\displaystyle r>0$ , and since also $\displaystyle r < c$ we get here a contradiction to the minimality of c in S.

Tonio
Tonio, thank you for all the help. It is greatly appreciated.

I forgot about part e), would you verify that my reasoning is correct?

e) Conclude from (c) and (d) that the greatest common divisor of a and b exists. Finish the proof by showing that this greatest common divisor is unique.

From part c) we have shown that all elements of S divide a and b. Also S is not empty from part a), therefore there is a common divisor of a and b.

If c is the minimum element of S then choose $\displaystyle S_1$ = {$\displaystyle s_1$ | $\displaystyle s_1$ $\displaystyle \epsilon$ S, $\displaystyle s_1$ $\displaystyle \not=$ c}

Then by the well ordering property, there is an element $\displaystyle c_1$ $\displaystyle \epsilon$ $\displaystyle S_1$ that is the smallest element of $\displaystyle S_1$. Since $\displaystyle c_1$ $\displaystyle \epsilon$ S, $\displaystyle c_1$ | a and $\displaystyle c_1$ | b.

Now choose $\displaystyle S_2$ = {$\displaystyle s_2$ | $\displaystyle s_2$ $\displaystyle \epsilon$ $\displaystyle S_1$, $\displaystyle s_2$ $\displaystyle \not=$ $\displaystyle c_1$}

Keep choosing subsets until only one element is left. That one element is gcd(a,b), call it m.

The only thing I can come up with to prove m is unique is the following:

Assume m is not unique, then there are $\displaystyle m_1$ and $\displaystyle m_2$ both elements of S that divide a and b. If $\displaystyle m_1$ is less than $\displaystyle m_2$ then $\displaystyle m_1$ could not be the gcd(a,b). If $\displaystyle m_2$ is less than $\displaystyle m_1$ then $\displaystyle m_2$ could not be gcd(a,b). Therefore $\displaystyle m_1$ = $\displaystyle m_2$. Is this correct?

Thanks again.

4. First, concerning a).
My question is this: Is it implied that s and t are POSITIVE integers. If s and t were both negative integers, then S could be empty. Could someone explain what I'm missing?
You need to show that there exists an element in $\displaystyle S$, not that $\displaystyle \{as+bt\mid s,t\in\mathbb{Z}\}$ consists of positive elements or something like that. When you prove existence, you have the right to choose an element. In this case, you can choose $\displaystyle a$, which belongs to $\displaystyle S$ with $\displaystyle s=1$, $\displaystyle t=0$. This is different from when you are proving something for all $\displaystyle s$ and $\displaystyle t$: then it is your opponent (so to speak; the one who doubts the validity of the statement) and not you who has the right to choose $\displaystyle s$ and $\displaystyle t$.

e) Conclude from (c) and (d) that the greatest common divisor of a and b exists. Finish the proof by showing that this greatest common divisor is unique.
You have already shown that it exists: it is $\displaystyle c$. Indeed, it is a common divisor by d), and every other common divisor $\displaystyle d$ divides $\displaystyle c$ by c), so every other common divisor $\displaystyle d$ does not exceed $\displaystyle c$.

From part c) we have shown that all elements of S divide a and b.
This is not correct; c) shows that only the smallest element $\displaystyle c$ of $\displaystyle S$ divides $\displaystyle a$ and $\displaystyle b$. You used the fact that $\displaystyle c$ is the smallest when you proved that the remainder of $\displaystyle a / c$ is 0, since it is smaller than $\displaystyle c$. Also, e.g., $\displaystyle a\in S$, but $\displaystyle a$ does not have to divide $\displaystyle b$.

Finish the proof by showing that this greatest common divisor is unique...
Assume m is not unique, then there are and both elements of S that divide a and b. If is less than then could not be the gcd(a,b). If is less than then could not be gcd(a,b). Therefore = . Is this correct?
You need to show not that the least element of $\displaystyle S$ is unique but that the GCD is unique. Not every common divisor of $\displaystyle a$ and $\displaystyle b$ is in $\displaystyle S$; for example, when $\displaystyle a$ and $\displaystyle b$ are both even, then every element of $\displaystyle S$ is even and therefore $\displaystyle 1\notin S$.

But the idea is basically correct. If there are two GCDs, then each of them is $\displaystyle \ge$ than the other, so they are equal.

5. Originally Posted by emakarov
First, concerning a).
You need to show that there exists an element in $\displaystyle S$, not that $\displaystyle \{as+bt\mid s,t\in\mathbb{Z}\}$ consists of positive elements or something like that. When you prove existence, you have the right to choose an element. In this case, you can choose $\displaystyle a$, which belongs to $\displaystyle S$ with $\displaystyle s=1$, $\displaystyle t=0$. This is different from when you are proving something for all $\displaystyle s$ and $\displaystyle t$: then it is your opponent (so to speak; the one who doubts the validity of the statement) and not you who has the right to choose $\displaystyle s$ and $\displaystyle t$.

You have already shown that it exists: it is $\displaystyle c$. Indeed, it is a common divisor by d), and every other common divisor $\displaystyle d$ divides $\displaystyle c$ by c), so every other common divisor $\displaystyle d$ does not exceed $\displaystyle c$.

This is not correct; c) shows that only the smallest element $\displaystyle c$ of $\displaystyle S$ divides $\displaystyle a$ and $\displaystyle b$. You used the fact that $\displaystyle c$ is the smallest when you proved that the remainder of $\displaystyle a / c$ is 0, since it is smaller than $\displaystyle c$. Also, e.g., $\displaystyle a\in S$, but $\displaystyle a$ does not have to divide $\displaystyle b$.

You need to show not that the least element of $\displaystyle S$ is unique but that the GCD is unique. Not every common divisor of $\displaystyle a$ and $\displaystyle b$ is in $\displaystyle S$; for example, when $\displaystyle a$ and $\displaystyle b$ are both even, then every element of $\displaystyle S$ is even and therefore $\displaystyle 1\notin S$.

But the idea is basically correct. If there are two GCDs, then each of them is $\displaystyle \ge$ than the other, so they are equal.
Thank you very much for the excellent advise.