Inequalities in the real numbers

[mcastillo356]

TL;DR

It's so a basic question that I feel shame.
I'm still not sure about this solved reasoning that involves inequalities

How do we get $\epsilon(2p+\epsilon)<\epsilon(2p+1)<2-p^2$ from $0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$?

Answer: As we have $\epsilon<1$, we've got $2p+\epsilon<2p+1$; therefore, $\epsilon(2p+\epsilon)<\epsilon(2p+1)$;
-as we have $\epsilon<\dfrac{2-p^2}{2p+1}$, we also have $\epsilon(2p+1)<2-p^2$. Is the red-coloured argue what I don't understand

Greetings, appreciated PF

 

[fresh_42]

If $\varepsilon < 1$ then $x+\varepsilon < x+1$. Set $x=2p$. Then $2p+\varepsilon <2p+1.$ Now if $a<b$ then $a\cdot \varepsilon < b\cdot \varepsilon$ because $\varepsilon >0$. Now set $a=2p+\varepsilon$ and $b=2p+1$ to get $(2p+\varepsilon )\cdot \varepsilon < (2p+1)\cdot \varepsilon$.

 

[mcastillo356]

Fantastic, I never imagined it could be explained in so an easy way.
Thanks, fresh_42!

 

[mcastillo356]

Hi, PF, there is something more about this thread: I want to contextualize the question, and, by the way, ask another two questions. I beg you a pardon if this means to change the thread's place.

It's about Dedekind Cuts, concretely $\sqrt 2 = \{r\in \mathbb Q\mid r<0\}\cup \{r\in \mathbb Q\mid r\geq 0,\ r^2<2\}$

And the task is to prove $\alpha=\sqrt 2$ is a cut, without mentioning $\sqrt 2$

1- As $1^2<2<2^2$, it satisfies $1\in\alpha$ and $2\in \mathbb Q\setminus\alpha$; so the first property of Dedekind Cuts is satisfied (for example, the proof that $\alpha\neq \mathbb Q$ is this: $2\notin \alpha$, because it doesn't satisfy neither $2<0$ nor $2\geq 0$ and $2^2<2$).

2-If $p\in\alpha$ and $q<p$, or $q<0$, in which case $p\in \alpha$, by definition, or $0\leq q<p$; hence $p^2<q^2<2$. Therefore, $q\in \alpha$.
Why is that $p^2<q^2<2$?

3- If $p\in\alpha$, alternatively $p\leq 0$, in which case $p<1\in\alpha$, or $p>0$, in which case $p^2<2$. Let's see what must require a rational number $\epsilon>0$ so that $p+\epsilon\in \alpha$:

$(p+\epsilon)^2=p^2+2p\epsilon+\epsilon^2<2$

This is equivalent to $\epsilon(2p+\epsilon)<2-p^2$. If we take $0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$ then
$2(p+\epsilon)<\epsilon(2p+1)<2-p^2$, so $p<q=p+\epsilon\in \alpha$

Why this implication?

Greetings! Hope not to have treated too badly the english.

 

[PeroK]

3- If $p\in\alpha$, alternatively $p\leq 0$, in which case $p<1\in\alpha$, or $p>0$, in which case $p^2<2$. Let's see what must require a rational number $\epsilon>0$ so that $p+\epsilon\in \alpha$:

$(p+\epsilon)^2=p^2+2p\epsilon+\epsilon^2<2$

This is equivalent to $\epsilon(2p+\epsilon)<2-p^2$. If we take $0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$ then
$2(p+\epsilon)<\epsilon(2p+1)<2-p^2$, so $p<q=p+\epsilon\in \alpha$

Why this implication?

Greetings! Hope not to have treated too badly the english.

I don't understand the bit in red at the end. You have:

$0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$

Now you have to show that, with these bounds on $\epsilon$ we have $(p + \epsilon)^2 < 2$. I.e. $p + \epsilon \in \alpha$.

 

[mcastillo356]

Hello, PeroK
The target consists in suppose $p\in \alpha$, this is, we have $p$ that satisfies $p<0$ or $p\geq 0$ and $p``^2<2$ (not mentioning $\sqrt 2$), and I must find a $q>p$ that satisfies the same (whithout references to $\sqrt 2$.
All I find in the proof is a dance of $\epsilon$, $p$, 0, 1, and 2.
Let's see, I think I understand:
$2(p+\epsilon)<(p+\epsilon)^2<2$: $2(p+\epsilon)=q\in \alpha$

 

[PeroK]

Hello, PeroK
The target consists in suppose $p\in \alpha$, this is, we have $p$ that satisfies $p<0$ or $p\geq 0$ and $p``^2<2$ (not mentioning $\sqrt 2$), and I must find a $q>p$ that satisfies the same (whithout references to $\sqrt 2$.
All I find in the proof is a dance of $\epsilon$, $p$, 0, 1, and 2.
Let's see, I think I understand:
$2(p+\epsilon)<(p+\epsilon)^2<2$: $2(p+\epsilon)=q\in \alpha$

I don't understand that at all.

You have to show that:

$0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1} \ \ \Rightarrow \ \ (p + \epsilon)^2 < 2$

That should be quite simple.

 

[PeroK]

PS you could check it out for $p = \frac 7 5$.

 

[mcastillo356]

It works, but, how can I prove it algebraically?

 

[PeroK]

It works, but, how can I prove it algebraically?

Using algebra! Starting with $(p + \epsilon)^2 = p^2 + 2p\epsilon + \epsilon^2$

 

[mcastillo356]

Ok, processing...

 

[Mark44]

@mcastillo356, I revised your thread title slightly, changing "inequations" to "inequalities," which is what we call them in English. Inconsistent, but that's English for you.

 

[mcastillo356]

You have to show that:

$0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1} \ \ \Rightarrow \ \ (p + \epsilon)^2 < 2$

That should be quite simple.

Using algebra! Starting with $(p + \epsilon)^2 = p^2 + 2p\epsilon + \epsilon^2$

I know only to prove this:
Dedekind Cuts, third property:
III If $p\in\alpha$ then $p<r$ for some $r\in \alpha$
For example, the cut $\sqrt 2 = \{r\in \mathbb Q\mid r\leq 0\}\cup \{r\in \mathbb Q\mid r> 0,\ r^2<2\}$. If $p\in\alpha$, alternatively $p\leq 0$, in which case $p<1\in\alpha$, or $p>0$, in which case $p^2<2$. Let's see what must require a rational number $\epsilon>0$ so that $p+\epsilon\in \alpha$:

$(p+\epsilon)^2=p^2+2p\epsilon+\epsilon^2<2$

This is equivalent to $\epsilon(2p+\epsilon)<2-p^2$. If we take $0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$ then we've found the proper $\epsilon$.
Problem solved?

$2(p+\epsilon)<\epsilon(2p+1)<2-p^2$ would be another way to express the two inequalities that bound $\epsilon$

 

[PeroK]

If we take $0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}$ then we've found the proper $\epsilon$.
Problem solved?

I would still like to see a direct calculation that starts with that $\epsilon$ and shows that $(p+\epsilon)^2 < 2$.

 

[mcastillo356]

I will try. I'll answer tomorrow

 

[mcastillo356]

Hello, PeroK, PF
$0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}\Rightarrow{(p+\epsilon)^2<2}$
$0<\epsilon<\dfrac{2-p^2}{2p+1}$
$\Rightarrow{0<\epsilon(2p+1)<2-p^2}$
$\Rightarrow{p^2<2}$
Dedekind Cuts, third property
$\therefore{(p+\epsilon)^2<2}$

?

 

[PeroK]

Hello, PeroK, PF
$0<\epsilon<1$ and $\epsilon<\dfrac{2-p^2}{2p+1}\Rightarrow{(p+\epsilon)^2<2}$
$0<\epsilon<\dfrac{2-p^2}{2p+1}$
$\Rightarrow{0<\epsilon(2p+1)<2-p^2}$
$\Rightarrow{p^2<2}$
Dedekind Cuts, third property
$\therefore{(p+\epsilon)^2<2}$

?

We already knew that $p^2 < 2$. It was $(p+ \epsilon)^2 < 2$ that must be shown.

I gave you the first step previously (which you ignored!): $$(p+\epsilon)^2 = p^2 + 2p\epsilon + \epsilon^2 = p^2 + \epsilon(2p + \epsilon)$$ The second step is: $$p^2 + \epsilon(2p + \epsilon) < p^2 + \epsilon(2p + 1) \ \ (\text{as} \ \ \epsilon <1)$$ Just two steps left for you now!

 

[mcastillo356]

Hi, PeroK, PF

$$(p+\epsilon)^2=p^2+2p\epsilon+\epsilon^2=p^2+\epsilon(2p+\epsilon)$$

$$\epsilon<1\Rightarrow p^2+\epsilon(2p+\epsilon)<p^2+\epsilon(2p+1)$$

$$\epsilon<\dfrac{2-p^2}{2p+1}\Rightarrow \epsilon(2p+1)<2-p^2$$

$$p^2+\epsilon(2p+\epsilon)<p^2+\epsilon(2p+1)<2-p^2+p^2$$

$$\Rightarrow (p+\epsilon)^2<2\Rightarrow (p+\epsilon)\in \alpha$$

?

 

[PeroK]