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

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- Sep 3rd 2006, 02:55 PMbraddyproof 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 PMThePerfectHackerQuote:

Originally Posted by**braddy**

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 PMPlato
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 PMThePerfectHackerQuote:

Originally Posted by**Plato**

- Sep 3rd 2006, 05:00 PMPlato
No, you are correct!

But you showed a solution existed.

I simplified its uniqueness. - Sep 4th 2006, 10:24 PMbraddyQuote:

Originally Posted by**Plato**

:)