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I have just begun trying to self study introductory analysis and I am just having some questions about being specific on rigour. In the book I am using, titled Introduction to Real Analysis, 4th edition by Robert G. Bartle and Donald R. Sherbert.

The second chapter begins with listing the properties of $\mathbb{R}$ and then does a bunch of the basic proofs of the properties of reals using just the given axioms.

My question is, it mentions that

It is worth noting that no smallest positive real number can exist, in fact if $$a \gt 0$$ then we have $$0 \lt \frac{1}{2}a \lt a$$ (why?)

Now, of course I understand that is true but I am just wanting to only use justifications. And I am wondering how we can say that $\frac{1}{2}a \lt a$, I mean, I am not sure that it defines any sort of rule that ordered the numbers like that or some sort of division or order of rationals.

How can that result come from Trichototmy?

I know that since $a \gt 0$ and since it can be shown that if $a \gt 0$ then $a^{2} \gt 0$ (because if $a \in P$ then $a(a) \in P$) so that settles that, and also that $\frac{1}{2} \gt 0$ since $\pm \frac{\sqrt2}{2} \in \mathbb{R}$ hence $\frac{1}{2} \gt 0$ and so $(1/2)a$ is greater then zero, but how to conclude that $(1/2)a$ is ordered less than $a$? Is there any source where I can see how the rationals are ordered anyways (but I would still be interested in how from these principals it can be concluded).

I hope it makes sense, maybe I am missing something or being very naive, but hopefully someone gets what I am trying to say.

Thanks a lot

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    $\begingroup$ Presumably (I don't have the book) one says that $x\gt y$ if $x-y\gt 0$. Let $x=a$ and $y=\frac{1}{2}\cdot a$. Then $x-y=\frac{1}{2}\cdot a$. But you have shown that $\frac{1}{2}\cdot a\gt 0$. $\endgroup$ – André Nicolas Jul 5 '15 at 5:28
  • $\begingroup$ $a>(1/2)a$ iff $a-(1/2)a=(1/2)a$>0. $\endgroup$ – Parakee Jul 5 '15 at 5:33
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By Theorem 2.1.8b, we have $1>0$.

enter image description here

By Theorem 2.1.7b, we have $1>0\implies (1)+1>(0)+1$ (in other words, $2>1$).

enter image description here

Since $\frac{1}{2}\cdot 2=1>0$ and $2>0$, by Theorem 2.1.10 we must have $\frac{1}{2}>0$.

enter image description here

Therefore, by Theorem 2.1.7c, combining the fact that $2>1$ and $\frac{1}{2}>0$ tells us that $\frac{1}{2}\cdot 2>\frac{1}{2}\cdot 1$, or in other words, that $1>\frac{1}{2}$. Now, for any $a>0$, we can again use Theorem 2.1.7c to conclude that $a\cdot 1>a\cdot \frac{1}{2}$.

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Can you use properties of inequalities such as if $c$ is positive and $a>b$ then $ca>cb$? (In my book it gives them as axioms, but I'm not familiar with yours.)

If so, note that $1> 1/2$ (I think I would assume this known), so, since $a$ is positive, $a>a/2$.

For the other one, multiply the inequality $a>0$ by $1/2$, to get $a/2>0/2=0$.

Putting these two together, you get $a>a/2>0$.

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Hmm if you already proved that $0 < \frac{a}{2}$ then you can add something to both sides...

$\frac{a}{2} + 0 < \frac{a}{2} + \frac{a}{2} = a$.

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