How to show that $1>0$

Question

Hello to everyone i just started to proofs in math, and i need a little help in this question

Show that $1>0$ by using ( Algebraic Prooerties of the real numbers) Such as ( Commutativity of Addition & Associativity of Addition & Existence of Additive Identity etc)

@RSerrao Such as ( Commutativity of Addition & Associativity of Addition & Existence of Additive Identity etc) — Scott, Nov 28 '16 at 10:53
You need to tell us exactly what the properties you have are, with no "etc" to hid the interesting parts. Do you have properties for $>$ too, or is it a defined concept for you? — hmakholm left over Monica, Nov 28 '16 at 11:00
No sir, i dont have properties of >. I only have (Commutativity of Addition & Associativity of Addition & Existence of Additive Identity & Existence of ADditive Inverses & Commutativity of Multiplication & Associativity of Multiplication & Existence of Multilipcative Identity & Existence Of Multiplicative Inverses And Distributivity) — Scott, Nov 28 '16 at 11:03
Possible duplicate of how do I prove that $1 > 0$ in an ordered field? — InsideOut, Nov 29 '16 at 09:35

Mauro ALLEGRANZA · Accepted Answer · 2016-11-28T13:15:47.830

Sketch of the proof.

We need some preliminary results :

Lemma 0 : $0 \cdot a = 0 = a \cdot 0$.

By the identity law for addition ($x+0 = x$) we have : $0+0 = 0$, and so $a(0+ 0) = a \cdot 0$. By the distributive law : $a \cdot 0+a \cdot 0 = a \cdot 0$, and by identity law for addition we have : $a \cdot 0+a \cdot 0 = 0+a \cdot 0$.

Then, by the cancellation law for addition : $a \cdot 0=0$. The commutative law for multiplication finally implies that : $0 \cdot a = 0$.

Lemma 1 : $(-a)b=-ab=a(-b)$.

We prove it using : $(-1) \cdot a = -a$.

In turn, this is proved as follows : $a+a(-1)=a \cdot 1+ a(-1)= a[1+(-1)] = a \cdot 0 = 0$.

Lemma 2 : $-(-a)=a$.

Lemma 3 : If $a \ne 0$, then $a^2 > 0$.

Suppose $a \ne 0$; by trichotomy : $a > 0$ or $a < 0$.

(i) Suppose that $a > 0$. Then by the multiplication law for order (if $x < y$ and $z > 0$, then $xz < yz$), we have that : $a \cdot a > 0 \cdot a$. Then, by L.0 : $a^2 > 0$.

(ii) Suppose that $a < 0$. We have $-a > 0$, and thus (by (i)) : $(-a)^2 > 0$. Applying L.1 twice we have : $(-a)^2 = (-a)(-a) = -[a(-a)] = -[-a^2]$, and thus by L.2 : $(-a)^2 = a^2$. It follows that $a^2 > 0$.

Now for te main result :

$1 \ne 0$ (non-triviality axiom) and thus by L.3 : $1 \cdot 1 > 0$.

Applying the identity law for multiplication : $x \cdot 1 = x$, we finally have :

$1 > 0$.

score 0 · Answer 2 · answered Nov 28 '16 at 11:23

We assume that $\mathbb{R}$ is a totally ordered field $(\mathbb{R},+,\cdot,\leq)$. We define $x < y$ to be "$x \leq y$ and $x \neq y$". I will assume the following properties:

$\quad$(TO1)$\quad$ For any $x$ and $y$, either $x \leq y$ or $y \leq x$ (or both).

$\quad$(TO2)$\quad$ ($x \leq y$ and $y \leq z$) $\implies$ $x+z \leq y + z$.

$\quad$(TO3)$\quad$ ($x \leq y$ and $0 \leq z$) $\implies$ $xz \leq yz$.

$\quad$(F1)$\quad$ For any $x$ we have that $1 \cdot x = x$.

$\quad$(F2)$\quad$ For any $x$ we have that $0 \cdot x = 0$.

$\quad$(F3)$\quad$ For any $x$ we have that $0 + x = x$.

$\quad$(F4)$\quad$ $0 \neq 1$.

$\quad$(F5)$\quad$ $1$ has an additive inverse, $-1$, i.e. $1 + (-1) = 0$.

Assume that $0 \not< 1$. From (F4) we know that $0 \neq 1$, hence $0 \not\leq 1$. Then from (TO1) we have that $1 \leq 0$. Again, since $1 \neq 0$, we now know that $1 < 0$.

Because our field is totally ordered, we have $$1 < 0 \implies 1 + (-1) < 0 + (-1) \implies 0 < -1,\tag{1}$$ where I used (TO2) in the first step, and (F5) and (F3) in the second step.

But since $0 < -1$, we also have that $$ 1 < 0 \implies 1 \cdot (-1) < 0 \cdot (-1) \implies -1 < 0,\tag{2}$$ where I used (TO3) in the first step, and (F1) and (F2) in the second step.

Equations $(1)$ and $(2)$ together pose a contradiction.

Note: Properties (F2) and (F4) you may still have to prove yourself.

How to show that $1>0$

2 Answers2