# Proof of Theorems

• January 25th 2009, 04:45 PM
noles2188
Proof 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.
• January 25th 2009, 04:54 PM
o_O
(1) $\exists a,b \in \mathbb{R}$ such that $a + b = 0$, then $b = -a$ .

Let $b = -a$ and the conclusion follows.

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(2) Assume $a \neq 0$. If we assume $\frac{1}{a} = 0$, try to arrive at a contradiction. So we know $\frac{1}{a} \neq 0$ .

For the second part, use the fact that $a \cdot b = 1 \Rightarrow \ b = \frac{1}{a}$ where $a \neq 0, b \in \mathbb{R}$ and consider $a \cdot \frac{1}{a} = 1$.
• January 26th 2009, 04:08 AM
archidi
Quote:

Originally Posted by noles2188
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.

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 ( $A\neq 0\rightarrow$(1/A).A=1)........................................... ...................................11

Now to prove that 1/a $\neq 0$.

Assume 1/a =0 and a $\neq 0$ ,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 $\neq 0$

Since 1/a $\neq 0$ ,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 $A\neq 0$ 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