# Thread: Basic proof where x is a real number

1. ## Basic proof where x is a real number

We just started proofs in my mathematical structures class so this is the first time i am dealing with proofs.

The question i am having trouble proving is:

Suppose x is a real number and x $\displaystyle \neq$ 0. Prove that if $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ then $\displaystyle x \neq 8.$

This is what i have started:

Givens:
x is a real number
x $\displaystyle \neq$ 0
$\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$

Goals:
$\displaystyle x \neq 8.$

i feel like to complete this proof you would solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ for x and then this would show that whatever this value of x is, it will not be equal to 8.

The problem i am running into is that i cannot solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ for x equaling a value, so i feel like i may be going about this wrong.

Any help / tips for proofs would be very great

Thank you

2. Originally Posted by mybrohshi5
Suppose x is a real number and x $\displaystyle \neq$ 0. Prove that if $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ then $\displaystyle x \neq 8.$

Prove:

$\displaystyle \dfrac{8^{1/3} + 5}{8^2+6} \neq \dfrac{1}{8}$

Fernando Revilla

3. Originally Posted by mybrohshi5
We just started proofs in my mathematical structures class so this is the first time i am dealing with proofs.

The question i am having trouble proving is:

Suppose x is a real number and x $\displaystyle \neq$ 0. Prove that if $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ then $\displaystyle x \neq 8.$

This is what i have started:

Givens:
x is a real number
x $\displaystyle \neq$ 0
$\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$

Goals:
$\displaystyle x \neq 8.$

i feel like to complete this proof you would solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ for x and then this would show that whatever this value of x is, it will not be equal to 8.

The problem i am running into is that i cannot solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}$ for x equaling a value, so i feel like i may be going about this wrong.

Any help / tips for proofs would be very great

Thank you
Have you thought about the contra positive of the statement.
Remember that If a, then b is logically equivilent to if not b, then not a

If $\displaystyle x=8$, then $\displaystyle \displaystyle \frac{x^{1/3} + 5}{x^2+6} \ne \frac{1}{x}$