# Five Properties: basic proofs

• Apr 22nd 2009, 08:39 PM
glover_m
Five Properties: basic proofs
I'm not expecting anyone to answer all of these, but hopefully someone can do one for me. It seems pretty straightforward and I wouldn't be posting if I had time to talk to my prof. before my exam tomorrow. Again, only using the 5 rules .

(1) For any real numbers a and b and any positive real number c.

a < b => ac < bc

(2) For any real number a,

a is positive <=> -a is negative

(3) For any real numbers a, b, c

if a < b and c is negative, then ac > bc

(4) For any real numbers a and b

ab is positive <=> a and b are both positive or both negative

(5) -1 < 0 < 1

USING ONLY THESE RULES

(a) a^2>=0 for any a

(b) For any real number a, if a is positive, then 1/a is positive.

(c) For any positive real numbers a and b, if a < b, then 1/a > 1/b

Thanks guys.
• Apr 23rd 2009, 05:24 AM
Inequalities
Hello glover_m
Quote:

Originally Posted by glover_m
I'm not expecting anyone to answer all of these, but hopefully someone can do one for me. It seems pretty straightforward and I wouldn't be posting if I had time to talk to my prof. before my exam tomorrow. Again, only using the 5 rules .

(1) For any real numbers a and b and any positive real number c.

a < b => ac < bc

(2) For any real number a,

a is positive <=> -a is negative

(3) For any real numbers a, b, c

if a < b and c is negative, then ac > bc

(4) For any real numbers a and b

ab is positive <=> a and b are both positive or both negative

(5) -1 < 0 < 1

USING ONLY THESE RULES

(a) a^2>=0 for any a

(b) For any real number a, if a is positive, then 1/a is positive.

(c) For any positive real numbers a and b, if a < b, then 1/a > 1/b

Thanks guys.

(a) If $\displaystyle a = 0, a^2 = 0$. Otherwise, using Rule (4), $\displaystyle \forall \,a , b \in \mathbb{R},\, (b = a)\Rightarrow (a$ and $\displaystyle b$ are both positive or both negative) $\displaystyle \Rightarrow aa = a^2$ is positive.

(b) $\displaystyle (a \cdot \frac1a = 1 > 0)$ from (5) $\displaystyle \Rightarrow a$ and $\displaystyle \frac1a$ are both positive or both negative, using Rule 4, with $\displaystyle b = \frac1a$. So if $\displaystyle a>0, \frac1a>0$.

(c) $\displaystyle a>0 \Rightarrow \frac1a > 0$, from (b). So $\displaystyle 0<a<b \Rightarrow a\cdot \frac1a < b\cdot \frac1a$, using Rule (1) with $\displaystyle c = \frac1a$

$\displaystyle \Rightarrow 1 < b\cdot\frac1a$

Also $\displaystyle b>0 \Rightarrow \frac1b>0$ from (b), so $\displaystyle 1\cdot\frac1b < \frac1b\cdot b\cdot \frac1a$, using (1)

$\displaystyle \Rightarrow \frac1b < \frac1a$