3

We consider the set $$A=\{x:x \in \mathbb{Q} \text{ and } x^2<2\} $$ We wish to show that this set has no largest member since this is all the rationals less than $\sqrt{2}$. I have been given the hint consider $$r^2<2, r>0 \qquad 0<\delta<1, \delta<\frac{2-r^2}{2r+1} $$ We wish to show that $(r+\delta)^2<2$.

I have tried expanding this in several ways and haven't been successful

$$(r+\delta)^2=r^2+2r\delta+\delta^2$$

We might factor in the two following ways

$$r(r+2\delta)+\delta^2 \qquad \text{ or } \qquad r^2+\delta(2r+\delta)$$

Any suggestions?

RhythmInk
  • 3,140

3 Answers3

4

Use the second factorization: $$\begin{align} (r+\delta)^2&=r^2+\delta(2r+\delta)\\\\ &<r^2+\delta(2r+1)&&(\delta<1)\\\\ &<r^2+2-r^2&&(\delta<\tfrac{2-r^2}{2r+1})\\\\ &=2 \end{align}$$

Pedro
  • 19,965
  • 9
  • 70
  • 138
0

`first of all the choise of the right set is important.:

So let A={$`x\in Q:x>0,x^2<2$}.

But this is not the only set that it does not have a Sup in Q,there is an infinite No of others..

Now since $A$ is a non empty,$1\in A$,bounded , $x\leq 2$ for all $x\in A$ subset of $Q$ it must have a Sup. call it $r$.

Assume now that this $r$ belongs to $Q$ . Now our main efford is to show that this assumption will lead us to a contradiction

By trichotomy we have :$r<\sqrt2$ or $r>\sqrt2$ or $r=\sqrt2$

Now by using the following two theorems:

a) if $r^2<2$,then there exists a $\delta>0$ such that $(r+\delta)^2<2$ and

b) if$2<r^2$,then there exists a $\delta>0$ such that $(r-\delta)^2>2$ and $r-\delta>0$

and the definition of the Sup$r$ we will conclude that the only possibility is:

...............................$r=\sqrt2$.....................................

But we know that $\sqrt2$ does not belong to $Q$ ,hence $r$ does not belong to $Q$ and $A$ has no SUP in $Q$.

Can you fill in the rest of the details ??

chris
  • 151
-1

Assume there is $y\in \mathbb{Q}$ such that $y=sup\{x\in \mathbb{Q} : x^2<2\}$. then $y^2\leq 2$. Since the square root of $2$ is not rational $y<2$, i.e. $y\in \{x\in \mathbb{Q} : x^2<2\}$. To get a contradiction we now use the tip. Let $0< \delta \in \mathbb{Q}$ be such that $\delta <\frac{2-y^2}{2y+1}$. Then certainly $y+\delta \in \mathbb{Q}$ and furthermore $(y+\delta)^2<(y+\frac{2-y^2}{2y+1})^2=(\frac{y^2+y+2}{2y+1})^2 \leq 2$. Contradiction.

Edit: The last inequality is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$ which is true because $y^2<2$ and $y\geq 1$.

Edit 2: We want to prove that $(y^2+y+2)^2 \leq 2(2y+1)^2$ expanding both sides and cancelling yields $y^4+2y^3+2\leq 3y^2+4y$. Subtracting both sides by $3y^2+4y$ and factoring one sees that this is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$.

math635
  • 1,645
  • 11
  • 19