$A=\{\frac ab | a,b \in Z^+ , \frac{a^2}{b^2}<2 \}$

Question

Show that the set $$A=\left\{\frac ab | a,b \in Z^+ , \frac{a^2}{b^2}<2 \right\}$$ has a least upper bound $L$

My try: $$\frac{a^2}{b^2}<2$$ $$\frac{a^2}{b^2}<(\sqrt2)^2$$ $$-\sqrt2<\frac{a}{b}<\sqrt2$$ But $a,b >0$ so $\frac ab >0$

Thus, we get $$0<\frac ab<\sqrt2$$ which implies $$\sup A = \sqrt 2 = L$$ So, $L$ exists. Hence proved.

I wonder is my proof correct or not ? Or is there any other specific way to solve it using definitions in the real analysis

Answer 1

The least upper bound $L$ exists by definition of the real numbers, which have as their defining axiom that every set which has some upper bound actually has a least upper bound.

This is, in fact, how one defines $\sqrt{2}$ - so strictly speaking your proof is incorrect, since you're invoking the existence of some real number called $\sqrt{2}$ which is by definition exactly the least upper bound $L$ you're trying to construct.

A proof should go as follows: the set $A$ is a set of rational numbers and hence is a subset of $\mathbb{R}$ (by definition of $\mathbb{R}$). It has an upper bound - for example, 2 is an upper bound, since either $\lvert a/b\rvert \leq 1$, or $a/b\leq a^2/b^2 <2$ by definition of the set.

Since $A$ has some upper bound, and is a set of real numbers, by the axiom of completeness of the real numbers there exists a least upper bound (usually denoted $\sqrt{2}$).

Answer 2

It suffices to show that $A \subset \mathbb{R}$, $A\not=\emptyset$, is bounded above.

Assume $A$ is not bounded above.

For every $N \in \mathbb{Z^+}$ there are $m_N,n_N$ s.t.

$(m_N/n_N) >N;$

But then for $N \ge 2:$

$(m_N/n_N)^ 2 >N^2 >4$,

a contradiction. Hence the set $A$ is bounded above, it has a supremum.

Answer 3

In your attempt you only prove that $L = \sqrt 2$ is an upper bound, but it remains to be shown that $\sqrt 2$ is indeed the least upper bound.

Let's go at follows: assume there exists a lesser upper bound $L_\epsilon = \sqrt 2 - \epsilon$ for $\epsilon>0$. We must prove that there exists an element $x\in A$ such that $x>L_\epsilon$.

Let $x = \frac{a}{\sqrt 2 a -\epsilon/2}$, and ask that $a> 1/2$. It can be checked that $x\in A$. In order to make sure that $x>L_\epsilon$, observe that $$(\sqrt 2 - \epsilon)a = \sqrt2 a -\epsilon a> \sqrt 2 a - \epsilon/2, $$ so $x>L_\epsilon$, as we wanted. Thus, $L_\epsilon$ is not a lower bound no matter how small $\epsilon$ is.