# Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).

Mathematics Asked by Mardia on November 6, 2020

The question is "Show that the solution of the equation $$x^5-2x^3-3=0$$ are all less than 2."

I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.

My attempt:

Assume to the contrary that solution of this equation is greater then or equal to 2.

Let $$x=frac 2p$$.

Then we have

$$(frac 2p)^5-2(frac 2p)^3-3=0$$

$$frac {2^5}{p^5} – frac {2^4}{p^3} – 3 = 0$$

We now consider two case: when $$p=1$$ and $$p<1$$.

when $$p=1$$:

$$2^5 – 2^4 – 3 = 32 – 16 – 3 = 13$$.

Since $$x = 2$$ is not a solution this is a contradiction.

When $$p<1$$:

If $$p<1$$, then we know $$1/p>1.$$ This implies $$frac {1}{p^{n+1}} > frac {1}{p^n}.$$

Since $$frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$$, it is clear that the inequality

$$frac 1{p^5}(2^5)- frac 1{p^3} 2^4-3 > frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$$ holds

Since for all $$x > 2$$ is not a solution this is a contradiction.

Thus, solution $$x$$ is less then 2.

Let $$x$$ be a root and $$x>2$$.

Thus, $$x^5-2x^3-3=x^5-2x^4+2x^4-4x^3+2x^3-3>0,$$ which is a contradiction.

Correct answer by Michael Rozenberg on November 6, 2020