# Thread: [SOLVED] No largest element

1. ## [SOLVED] No largest element

In showing that $\displaystyle \mathbb{Q}$ does not follow the least upper bound property consider the following sets: $\displaystyle A = \{p: p^2 <2, \ p \in \mathbb{Q} \}$ and $\displaystyle B = \{p: p^2 >2, \ p \in \mathbb{Q} \}$. We want to show that for every $\displaystyle p \in A$ we can find a $\displaystyle q \in A$ such that $\displaystyle p<q$. Likewise, for every $\displaystyle p \in B$, we can find a $\displaystyle q \in B$ such that $\displaystyle q < p$. To do this, we let:

$\displaystyle q = p-\frac{p^2-2}{p+2}$ (1)

and it all works out (e.g. (i) $\displaystyle p \in A \Rightarrow p<q$ and (ii) $\displaystyle p \in B \Rightarrow q <p$). We also consider $\displaystyle q^2-2$ to show that $\displaystyle q$ is also in $\displaystyle A$ or $\displaystyle B$ respectively.

The question is, how do we come up with (1)? Was it just a guess and check process? It seems like the second term in (1) is very small. So in a sense we are considering $\displaystyle q= p- \varepsilon$. But how do we come up with $\displaystyle \varepsilon$? Is it arbitrary?

2. Originally Posted by Sampras
In showing that $\displaystyle \mathbb{Q}$ does not follow the least upper bound property consider the following sets: $\displaystyle A = \{p: p^2 <2, \ p \in \mathbb{Q} \}$ and $\displaystyle B = \{p: p^2 >2, \ p \in \mathbb{Q} \}$. We want to show that for every $\displaystyle p \in A$ we can find a $\displaystyle q \in A$ such that $\displaystyle p<q$. Likewise, for every $\displaystyle p \in B$, we can find a $\displaystyle q \in B$ such that $\displaystyle q < p$. To do this, we let:

$\displaystyle q = p-\frac{p^2-2}{p+2}$ (1)

and it all works out (e.g. (i) $\displaystyle p \in A \Rightarrow p<q$ and (ii) $\displaystyle p \in B \Rightarrow q <p$). We also consider $\displaystyle q^2-2$ to show that $\displaystyle q$ is also in $\displaystyle A$ or $\displaystyle B$ respectively.

The question is, how do we come up with (1)? Was it just a guess and check process? It seems like the second term in (1) is very small. So in a sense we are considering $\displaystyle q= p- \varepsilon$. But how do we come up with $\displaystyle \varepsilon$? Is it arbitrary?
It seems that if we want to show that sets of the form $\displaystyle A = \{p: p^2 < n, \ p \in \mathbb{Q}, \ n \geq 2 \}$ and $\displaystyle B = \{p: p^2 > n, \ p \in \mathbb{Q}, \ n \geq 2 \}$ (where $\displaystyle n \in \mathbb{N}$ and is prime) don't have largest and smallest elements respectively, we consider:

$\displaystyle q = p-\frac{p^2-n}{p+n}$ (2)

Maybe we choose the numerator so that it "controls" whether $\displaystyle q > p$ or $\displaystyle q <p$?

3. If you note that $\displaystyle q=\frac{2(p+1)}{p+2}$ it becomes easier to deal with.

I find is best to work with contradictions.

If $\displaystyle p\in A$ then show that $\displaystyle p<q$ and $\displaystyle q\in A$.
To do that suppose that $\displaystyle q^2\ge2$ and get a contradiction.

4. Originally Posted by Plato
If you note that $\displaystyle q=\frac{2(p+1)}{p+2}$ it becomes easier to deal with.

I find is best to work with contradictions.

If $\displaystyle p\in A$ then show that $\displaystyle p<q$ and $\displaystyle q\in A$.
To do that suppose that $\displaystyle q^2\ge2$ and get a contradiction.
But this is assuming that we choose $\displaystyle \frac{p^2-2}{p+2}$ as the second term in (1). Suppose we chose $\displaystyle \frac{p^2-2}{p+3}$? Then we would get

$\displaystyle q = p-\frac{p^2-2}{p+3}$ (3)

5. Originally Posted by Sampras
But this is assuming that we choose $\displaystyle \frac{p^2-2}{p+2}$ as the second term in (1).
But that is exactly what is done.
Why? Because it works. That is all you want.

You need to show that $\displaystyle A$ has no greatest term, and $\displaystyle B$ has no least term.

Clearly $\displaystyle \sqrt{2}$ belongs to neither set.

6. Originally Posted by Plato
But that is exactly what is done.
Why? Because it works. That is all you want.

You need to show that $\displaystyle A$ has no greatest term, and $\displaystyle B$ has no least term.

Clearly $\displaystyle \sqrt{2}$ belongs to neither set.
It works but the question is how would you come up with it if you didn't know that it worked? Is this through trial and error?

7. Originally Posted by Sampras
It works but the question is how would you come up with it if you didn't know that it worked? Is this through trial and error?
Yes it is trial and error up to a point.
I have seem as many as five different ways to do this very problem.
(Although I have never seen this one before).

Someone just sat down and reversed engineered the process.

8. Originally Posted by Plato
Yes it is trial and error up to a point.
I have seem as many as five different ways to do this very problem.
(Although I have never seen this one before).

Someone just sat down and reversed engineered the process.
This was in Rudin's book.