proof and rules of inferences

**braddy** · September 3rd 2006, 02:55 PM

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

**ThePerfectHacker** · September 3rd 2006, 03:28 PM

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,
$ax+b=c$
There exists preciesly one such number $-b$ such as $b+(-b)=0$ because the real numbers form a group under addition.
Thus,
$(ax+b)+(-b)=c+(-b)$
Since addition is associative we have,
$ax+(b-b)=c-b$
Thus,
$ax+0=c-b$
Since,
$0$ is an identity element we have,
$ax=b$
Since, $a\not = 0$ it is an element of the real numbers under multiplication and form a group. Thus, there exist a unique, $1/a$ such as, $a(1/a)=1$ .
Thus,
$(1/a)(ax)=(1/a)(c-b)$
Since, multiplication is associative we have,
$((1/a)a)x=1/a(c-b)$
Thus,
$1x=1/a(c-b)$
Since 1 is the multiplicative identity element we have,
$x=1/a(c-b)$

**Plato** · September 3rd 2006, 04:38 PM

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 $ax + b = c,\quad a \not= 0$ .
Then
$\begin{array}{rcl} ap + b & = & c \\ aq + b & = & c \\ ap + b & = & aq + b \\ ap & = & aq \\ p & = & q \\ \end{array}$

**ThePerfectHacker** · September 3rd 2006, 04:45 PM

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 $ax + b = c,\quad a \not= 0$ .
Then
$\begin{array}{rcl} ap + b & = & c \\ aq + b & = & c \\ ap + b & = & aq + b \\ ap & = & aq \\ p & = & q \\ \end{array}$

Yes but you do not show that a solution exists

**Plato** · September 3rd 2006, 05:00 PM

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

**braddy** · September 4th 2006, 10:24 PM

Originally Posted by Plato

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

Thank you forr your Help!

Thread: proof and rules of inferences

LinkBack

Thread Tools

Display

proof and rules of inferences

Similar Math Help Forum Discussions

Inferences About Means

Proof Using Rules of Inference

Rules of Inference Proof Help

Rules of Inferences

Proof for some exponent rules

Search Tags