MHB Properties of "less Than" & "Less Than or Equals" - Bloch Theorem 1.2.9 - Peter

[Math Amateur]

I am reading Ethan D. Bloch's book: The Real Numbers and Real Analysis ...

I am currently focused on Chapter 1: Construction of the Real Numbers ...

I need help/clarification with an aspect of Theorem 1.2.9 (6) ...

Theorem 1.2.9 reads as follows:
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View attachment 6979
In the above proof of (6) we read the following:

" ... ... Suppose that $$a \lt b$$ and $$a = b$$. It then follows from Part (3) of this theorem that $$a \lt a$$ ... ... "Can someone please explain how Part (3) of Theorem 1.2.9 leads to the statement that $$a \lt b$$ and $$a = b \Longrightarrow a \lt a$$ ... ...

... ... ... ...Further ... why can't we argue this way ...

... because $$a = b$$ we can replace $$b$$ by $$a$$ in $$a \lt b$$ giving $$a \lt a $$ ... which contradicts Part (1) of the theorem ...
Hope someone can help ...

Peter

 

[I like Serena]

Re: Properties of 'less than" and "less than or equals" - Bloch Theorem 1.2.9 ... ...

In the above proof of (6) we read the following:
" ... ... Suppose that $$a \lt b$$ and $$a = b$$. It then follows from Part (3) of this theorem that $$a \lt a$$ ... ... "Can someone please explain how Part (3) of Theorem 1.2.9 leads to the statement that $$a \lt b$$ and $$a = b \Longrightarrow a \lt a$$ ... ...

Further ... why can't we argue this way ...
... because $$a = b$$ we can replace $$b$$ by $$a$$ in $$a \lt b$$ giving $$a \lt a $$ ... which contradicts Part (1) of the theorem ...

Hi Peter,

You are quite correct.
Part (3) about transitivity is not involved whatsoever on $a<b$ and $a=b$.
Instead we can simply substitute as you surmised.
In other words, we can leave out the reference to part (3) and it will be correct.

It looks as if Bloch accidentally gave a reference to part (3) when that only applies to the part that comes after: about $a<b$ and $b < a$.

 

[Evgeny.Makarov]

Re: Properties of 'less than" and "less than or equals" - Bloch Theorem 1.2.9 ... ...

Part (3) can be used because $a=b$ (rather, $b=a$) implies $b\le a$ by definition, and then $a<b$ and $b\le a$ imply $a<a$ by part (3). But yes, it is easier to replace $b$ with $a$ in $a<b$ to get $a<a$.

 

[Math Amateur]

Re: Properties of 'less than" and "less than or equals" - Bloch Theorem 1.2.9 ... ...

Part (3) can be used because $a=b$ (rather, $b=a$) implies $b\le a$ by definition, and then $a<b$ and $b\le a$ imply $a<a$ by part (3). But yes, it is easier to replace $b$ with $a$ in $a<b$ to get $a<a$.

Thanks I Like Serena and Evgeny ...

Appreciate your help ...

Peter