# Proving One Statement Cannot Equal Another

• Mar 6th 2013, 04:38 AM
Proving One Statement Cannot Equal Another
I've just currently started MATH187 at university and was flipping through the exercises we were given in our subject notebook when I came upon a particular question that I am unsure how to approach.

The question is:

"Let a, b be non-zero real numbers for which a + b cannot = 0. Show that 1/a + 1/b is never equal to 1/a+b"

I was never particularly good at this style of question at school and need some direction as to how to set it out and approach it.

Any help would be particularly appreciated :)
• Mar 6th 2013, 05:21 AM
HallsofIvy
Re: Proving One Statement Cannot Equal Another
Use a common denominator to add 1/a+ 1/b. Then set it equal to 1/(a+ b). What does that tell you?
• Mar 6th 2013, 02:25 PM
Re: Proving One Statement Cannot Equal Another
So once you do use the common denominator it becomes b/ab + a/ab which simplifies to (a + b)/ab. When I equate it to 1(a + b) I get (a + b)/ab = 1/(a + b). How do I progress from here to complete the final proof? Is this where the "a + b cannot = 0" comes into play?
• Mar 6th 2013, 02:37 PM
Plato
Re: Proving One Statement Cannot Equal Another
Quote:

So once you do use the common denominator it becomes b/ab + a/ab which simplifies to (a + b)/ab. When I equate it to 1(a + b) I get (a + b)/ab = 1/(a + b). How do I progress from here to complete the final proof? Is this where the "a + b cannot = 0" comes into play?

You know that $\displaystyle ab\ne 0~.$ So what if:
\displaystyle \begin{align*}\frac{1}{a}+\frac{1}{b}&=\frac{1}{a+ b} \\(a+b)^2&=ab \\a^2+b^2&=-ab\end{align*}

What is wrong with that?
• Mar 6th 2013, 02:50 PM
Re: Proving One Statement Cannot Equal Another
Quote:

Originally Posted by Plato
You know that $\displaystyle ab\ne 0~.$ So what if:
\displaystyle \begin{align*}\frac{1}{a}+\frac{1}{b}&=\frac{1}{a+ b} \\(a+b)^2&=ab \\a^2+b^2&=-ab\end{align*}

What is wrong with that?

Is it because $\displaystyle ab\ne 0~.$ and therefore $\displaystyle -ab\ne 0~.$ in:

$\displaystyle a^2+b^2&=-ab$

or I am on the completely wrong track?
• Mar 6th 2013, 03:06 PM
Plato
Re: Proving One Statement Cannot Equal Another
Quote:

Is it because $\displaystyle ab\ne 0~.$ and therefore $\displaystyle -ab\ne 0~.$ in:

$\displaystyle a^2+b^2&=-ab$

or I am on the completely wrong track?

You are correct.
That means $\displaystyle a^2+b^2\ge 0$ BUT $\displaystyle -ab\le 0$.

So what?
• Mar 6th 2013, 03:20 PM
Re: Proving One Statement Cannot Equal Another
Quote:

Originally Posted by Plato
You are correct.
That means $\displaystyle a^2+b^2\ge 0$ BUT $\displaystyle -ab\le 0$.

So what?

So theoretically the only value in the inequalities in which they overlap is 0, however $\displaystyle -ab\ne 0~.$. So would that be the final line of the proof as LHS cannot equal RHS?
• Mar 6th 2013, 03:48 PM
Plato
Re: Proving One Statement Cannot Equal Another
Quote:

So theoretically the only value in the inequalities in which they overlap is 0, however $\displaystyle -ab\ne 0~.$. So would that be the final line of the proof as LHS cannot equal RHS?