# proof and rules of inferences

• Sep 3rd 2006, 02:55 PM
proof and rules of inferences
Hi,
Please can someone help me with this problem.

show that a,b,c are real numbers and a#0, then there is a unique solution of the equation ax+b=c.

the uniqueness of the solution is my problem.
Thank you
B
• Sep 3rd 2006, 03:28 PM
ThePerfectHacker
Quote:

Hi,
Please can someone help me with this problem.

show that a,b,c are real numbers and a#0, then there is a unique solution of the equation ax+b=c.

the uniqueness of the solution is my problem.
Thank you
B

Okay.
If,
$\displaystyle ax+b=c$
There exists preciesly one such number $\displaystyle -b$ such as $\displaystyle b+(-b)=0$ because the real numbers form a group under addition.
Thus,
$\displaystyle (ax+b)+(-b)=c+(-b)$
Since addition is associative we have,
$\displaystyle ax+(b-b)=c-b$
Thus,
$\displaystyle ax+0=c-b$
Since,
$\displaystyle 0$ is an identity element we have,
$\displaystyle ax=b$
Since, $\displaystyle a\not = 0$ it is an element of the real numbers under multiplication and form a group. Thus, there exist a unique, $\displaystyle 1/a$ such as, $\displaystyle a(1/a)=1$.
Thus,
$\displaystyle (1/a)(ax)=(1/a)(c-b)$
Since, multiplication is associative we have,
$\displaystyle ((1/a)a)x=1/a(c-b)$
Thus,
$\displaystyle 1x=1/a(c-b)$
Since 1 is the multiplicative identity element we have,
$\displaystyle x=1/a(c-b)$
• Sep 3rd 2006, 04:38 PM
Plato
Here is the way that I would expect this to be done.
Suppose that each of p and q is a solution to the equation $\displaystyle ax + b = c,\quad a \not= 0$.
Then
$\displaystyle \begin{array}{rcl} ap + b & = & c \\ aq + b & = & c \\ ap + b & = & aq + b \\ ap & = & aq \\ p & = & q \\ \end{array}$
• Sep 3rd 2006, 04:45 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
Here is the way that I would expect this to be done.
Suppose that each of p and q is a solution to the equation $\displaystyle ax + b = c,\quad a \not= 0$.
Then
$\displaystyle \begin{array}{rcl} ap + b & = & c \\ aq + b & = & c \\ ap + b & = & aq + b \\ ap & = & aq \\ p & = & q \\ \end{array}$

Yes but you do not show that a solution exists :eek:
• Sep 3rd 2006, 05:00 PM
Plato
No, you are correct!
But you showed a solution existed.
I simplified its uniqueness.
• Sep 4th 2006, 10:24 PM