# Inequalities in the real numbers

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TL;DR Summary
It's so a basic question that I feel shame.
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

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

• fresh_42
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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.

Homework Helper
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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
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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##

Homework Helper
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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.

• mcastillo356
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PS you could check it out for ##p = \frac 7 5##.

• mcastillo356
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It works, but, how can I prove it algebraically?

Homework Helper
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It works, but, how can I prove it algebraically?
Using algebra! Starting with ##(p + \epsilon)^2 = p^2 + 2p\epsilon + \epsilon^2##

• mcastillo356
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Ok, processing...

Mentor
@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
Gold Member
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##

Homework Helper
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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
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I will try. I'll answer tomorrow

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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}## ?

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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
Gold Member
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 • 