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My homework question:

From the order axioms for $\mathbb{R}$, show that $0 < 1$. [Hint: From the field axioms, $0 \not=1$. By the trichotomy property, either $0<1$ or $4<0$. Assuming $1 < 0$, get $0 < -1$. Now use Exercise 4.]

Exercise 4 from my textbook problems states:

"From the order axioms for $\mathbb{R}$, show that the set of positive real numbers, {$x \in \mathbb{R} : x > 0$}, is closed under addition and multiplication."

How am I expected to use Exercise 4 as directed?

user174707
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    The hint says to use Exercise 4 right after it says $0<-1$, so perhaps the idea is to see what happens when $-1$ is added or multiplied with itself? – Jonas Meyer Sep 09 '14 at 05:37
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    What Jonas said. If the set is closed under multiplication, then the result of multiplying any member by any member should be in the set. – Nagase Sep 09 '14 at 05:39
  • @Nagase But $0 < -1$ does not seem to fit under the criteria of being closed under addition and multiplication. Is this an indication of setting a proof by contradiction? – user174707 Sep 09 '14 at 05:45
  • @user174707 Yep, that's probably the way to go. Incidentally, there seems to be a typo in the post. I think you meant $1 < 0$, not $4 <0$. – Nagase Sep 09 '14 at 05:50
  • Yes that was indeed a typo, $1 < 0$. – user174707 Sep 09 '14 at 06:03
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    Can you guys help me prove that $0<1$ ? - Nope. Sorry. Way above our pay grade. :-) – Lucian Sep 09 '14 at 07:22
  • I'm not trying to be pessimistic, but the "hint" is pretty much the full solution. You should really be able to work out the missing step yourself at this point. Did you try solving the problem yourself? I don't see any personal input in your question. – Najib Idrissi Sep 09 '14 at 07:24

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How about proving that $x^2 >0$ whenever $x \ne 0$ and then noticing that $1 = 1^2$.

So, let $x \ne 0$. Then $x > 0$ or $x < 0$. If $x>0$ then $x^2 = x\cdot x > 0$. If $ x < 0$ then adding $-x$ to both terms we get $0 < -x$ and therefore $0 < (-x)(-x) = x^2$.

One should also check the sign rule $(-a)(-b) = a b$ in an introductory course.

orangeskid
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By contradiction assume $1<0$ then $0<-1 $ ,so $ -1 $ is a positive number so(by the fact that positive reals is closed under multiplication) $(-1)(-1) $must also be positive , but by assumption 1 is not positive . contradiction

Avrham Aton
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