infimum for {x>0 : x^2>4}={x>0: x>2}

[bluefrog]

I am trying to understand the answer to a question. The question is:
I am given the following set [MATH]A= \lbrace x>0 : x^2>4 \rbrace = \lbrace x>0 : x>2 \rbrace[/MATH] and told to determine,
a) lower bound
b) Let [MATH]L[/MATH] be a lower bound s.t. [MATH]L>2[/MATH] and let [MATH]y=\frac{L+2}{2}[/MATH]. Show that [MATH]L>y>2[/MATH]c) Show that [MATH]y \in A[/MATH] and [MATH]L \leq y[/MATH]. Show this leads to contradition, in which case [MATH]L \leq 2 \Rightarrow \inf{A}=2[/MATH]
So the answer given for a) is that 2 is the lower bound, no issues with that

The answer for b) though I do not understand part of it, which is:
Since [MATH]L>2 \Rightarrow L+2>4 \text{, in which case } y=\frac{L+2}{2} >2[/MATH]. Also [MATH]y=\frac{L+2}{2} > \frac{L+L}{2}=L [/MATH]
Firstly, I do not understand "[MATH]L>2 \Rightarrow L+2>4[/MATH]" and secondly I do not understand "[MATH]y=\frac{L+2}{2} > \frac{L+L}{2}=L [/MATH]"
Could somebody expand the answer for me perhaps?

 

[pka]

I am trying to understand the answer to a question. The question is:
I am given the following set [MATH]A= \lbrace x>0 : x^2>4 \rbrace = \lbrace x>0 : x>2 \rbrace[/MATH] and told to determine,
a) lower bound
b) Let [MATH]L[/MATH] be a lower bound s.t. [MATH]L>2[/MATH] and let [MATH]y=\frac{L+2}{2}[/MATH]. Show that [MATH]L>y>2[/MATH]c) Show that [MATH]y \in A[/MATH] and [MATH]L \leq y[/MATH]. Show this leads to contradition, in which case [MATH]L \leq 2 \Rightarrow \inf{A}=2[/MATH]
So the answer given for a) is that 2 is the lower bound, no issues with that

The answer for b) though I do not understand part of it, which is:
Since [MATH]L>2 \Rightarrow L+2>4 \text{, in which case } y=\frac{L+2}{2} >2[/MATH]. Also [MATH]y=\frac{L+2}{2} > \frac{L+L}{2}=L [/MATH]
Firstly, I do not understand "[MATH]L>2 \Rightarrow L+2>4[/MATH]" and secondly I do not understand "[MATH]y=\frac{L+2}{2} > \frac{L+L}{2}=L [/MATH]"
Could somebody expand the answer for me perhaps?

It is always the case that if \(p>q\) then \(\dfrac{p+q}{2}\) is between \(p~\&~q\).
PROOF: \(p > q \Rightarrow \frac{p}{2} > \frac{q}{2} \Rightarrow \frac{p}{2} + \frac{p}{2} = \frac{{p + q}}{2} \Rightarrow p > \frac{{p + q}}{2}\)
Likewise \(p > q \Rightarrow \frac{p}{2} > \frac{q}{2} \Rightarrow \frac{p}{2} + \frac{q}{2} = \frac{{p + q}}{2} \Rightarrow \frac{{p + q}}{2}>q\)

 

[Steven G]

No 2 is NOT the lower bound. There are many lower bounds! 1 is a lower bound, -17.5 is a lower bound, ... Maybe 2 is the least lower bound but 2 is not the lower bound. Your proof must be flawless since that is what a mathematical proof is all about.