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.

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- Jan 25th 2009, 04:45 PMnoles2188Proof of Theorems
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. - Jan 25th 2009, 04:54 PMo_O
(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 $\displaystyle a \cdot \frac{1}{a} = 1$. - Jan 26th 2009, 04:08 AMarchidi

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