(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(!) Assuming only arithmetic (not the quadratic formula or calculus), prove that

[tex]

\left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\}

[/tex]

2. Relevant equations

[tex]

\left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\}

[/tex]

3. The attempt at a solution

Again, new and trying to learn how exactly to do this to proper mathematical standards.

I think this will just be a simple plug'n'play:

Test case, 0 and 2 since using boundaries results in 0 < 0:

(0)^2-2(0)-3 < 0

-3 < 0

True

(2)^2-2(2)-3 < 0

4-4-3 < 0

-3 < 0

True

Thoughts? This seemed too easy but when I manipulated with adding and subtracting/multiplying I started to lose focus from the core inequality ie:

x^2-2x-3 < 0

x^2-2x < 3

-1 < x < 3

0 < x+1 < 3+1

0 < x^2+x < 4x

Et cetera...

Sincerely,

NastyAccident

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Using only arithmetic prove

**Physics Forums | Science Articles, Homework Help, Discussion**