Prove the following two theorems.
1) If x is any number, then --x = x.
2) If x is any number distinct from 0, then 1/x is not equal to 0, and 1/(1/x) = x.
(1)$\displaystyle \exists a,b \in \mathbb{R}$ such that $\displaystyle a + b = 0$, then $\displaystyle b = -a$ .
Let $\displaystyle b = -a$ and the conclusion follows.
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(2) Assume $\displaystyle a \neq 0$. If we assume $\displaystyle \frac{1}{a} = 0$, try to arrive at a contradiction. So we know $\displaystyle \frac{1}{a} \neq 0$ .
For the second part, use the fact that $\displaystyle a \cdot b = 1 \Rightarrow \ b = \frac{1}{a}$ where $\displaystyle a \neq 0, b \in \mathbb{R}$ and consider $\displaystyle a \cdot \frac{1}{a} = 1$.
These kind of questions although look simple can become very tricky,depending what axioms we choose within the frame of real Nos.
So if we choose as an axiom the following:
....................for all A , A+(-A)=0...........................................1
Then the -(-x)=x problem can be solved as follows:
In (1) put A= x and (1) becomes : x + (-x)=0................................2
And if we put A= -x then (1) becomes: (-x )+ [-(-x)] =0....................3
Hence from (3) and (2) and using cancellation law : x=-(-x)
But however if we choose as an axiom the following:
................for all x ,there exists y such that ,x+y=0........................4
then to prove the above we must do the following:
a) prove the uniqueness of y........
b) Then introduce in our system a new operation symbol " - " ,which in our 1st approach (equation (1) ) we consider as one of the primitive symbols of our system.
Hence the following definition:
............for all x,y ( -x = y iff x+y =0 ).......................................... 5
OR ,using different variables for convenience we have:
............for all A,B ( -A = B <====> A+B =0 )....................................6
And if we put now , A= -x.........B= -xTHEN (6) becomes:
................-x =-x <====> x + (-x) = 0............................................7
While if we put A = -x and B= -(-x) ,then 6 becomes:
.................... -(-x) = -(-x) <=====> (-x) + [-(-x)] =0.......................8
But since -x = -x and -(-x) = -(-x) ,then (7) and (8) became:
..........................x + (-x) =0................................................ ........9
.................................-(-x) + [-(-x)] = 0......................................10
Which are exactly the same with equations (2) and (3).That in a way explains why some systems start with (1) as their axiom
Now for the 2nd part :
Firstly we choose as our axiom :
...............................for all,A ($\displaystyle A\neq 0\rightarrow$(1/A).A=1)........................................... ...................................11
Now to prove that 1/a$\displaystyle \neq 0$.
Assume 1/a =0 and a$\displaystyle \neq 0$ ,then (1/a).a=0 and also by using (11),where we put A=a we have (1/a).a=1 and hence 1=0,a contradiction. Thus 1/a$\displaystyle \neq 0$
Since 1/a$\displaystyle \neq 0$ ,we put in (11) A = 1/a and we get 1/(1/a)*1/a=1. And multiplying that equation by a and using the fact (1/a)*a=1,we get:
..................1/(1/a)= a................................................. .........
Now if we choose as our axiom the following:
...................for all A $\displaystyle A\neq 0$ there exists B such that AB=1 ,THE PATH of the proof changes completely .
However following more or less the path shown above we can get the desired result. You try it