Thread: 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

Originally Posted by braddy

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,
$\displaystyle x=1/a(c-b)$

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
$\displaystyle \begin{array}{rcl}
ap + b & = & c \\
aq + b & = & c \\
ap + b & = & aq + b \\
ap & = & aq \\
p & = & q \\
\end{array}$

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

No, you are correct!
But you showed a solution existed.
I simplified its uniqueness.

Originally Posted by Plato

No, you are correct!
But you showed a solution existed.
I simplified its uniqueness.

Thank you forr your Help!