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 :)

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?

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?

Re: Proving One Statement Cannot Equal Another

Quote:

Originally Posted by

**JimmyyLAD** 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?

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?

Re: Proving One Statement Cannot Equal Another

Quote:

Originally Posted by

**JimmyyLAD** 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?

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?

Re: Proving One Statement Cannot Equal Another

Quote:

Originally Posted by

**JimmyyLAD** 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?

**By George, you got it.** Now is that a proof?

Re: Proving One Statement Cannot Equal Another

Thanks for the help with that :) Really appreciated :)