# Help with a few questions about spivak calculus

1. Sep 19, 2011

### mendem03

Question
if a < b, then -b < -a

proof
if a < b then a-b<0 and b-a>0
so a-b<0<b-a
so -b<-a

Question
if a<b & c<0, then ab>bc

proof
if ac<ab then ab>bc
then ac<ab>bc
then ac>bc

Question
if a>1 then a^2>a

proof
a*a > 1*a
lemma 1: a*a = a^2
(a*a)*a^-1 = a^2 * a^-1
a*(a*a^-1) = a
a * 1 = a
a = a
so a*a>1*a
a^2>a

2. Sep 19, 2011

### issacnewton

in the first proof,

when you go from $a-b<0$ to $b-a>0$ , you have to do this first

$$a-b<0 \Rightarrow \; -(b-a)<0 \;$$

but to go from here to $b-a>0$, you have to assume the thing which you set out to prove.

in the last proof, when you go from $\; a>1\;$ to $a*a > 1*a$, you are
assuming that $1>0$. have you proved it before ?

3. Sep 20, 2011

### mendem03

Thank you for replying and helping me out
I have not proven 1 greater than 0 before
but can you give me a pointer on how to prove that or another way of doing this proof

4. Sep 25, 2011

### issacnewton

order axioms say that between two real numbers a and b , three relations are possible.
a>b or a=b or a<b. this is called property of trichotomy. to prove 1>0, you assume negation, assume $1\ngtr 0$. since one of the three possibilities between 1 and 0 is true and we have assumed $1\ngtr 0$ , it must be true that
$1=0\;\mbox{or}\;1<0$ . since the field axioms explicitly say that
$1\neq 0$ , it must be true that $1<0$ now work with axioms to find a contradiction.........

are you studying analysis on your own or you are taking class ?