5
$\begingroup$

Is it possible to prove that this equation is false:

$$ \sum_{i=0}^n a_i p^i = 0 $$

with following conditions:

$a_i \in [-1;1]$; [Might $a\in\{-1,1\}$ have been intended here?]
$p$ is a prime number;
$n > 0$.

$\endgroup$
  • $\begingroup$ Might you have meant $a\in[-1,1]$ or "$a$ is in $[-1,1]$" where you wrote "$a$ is $[-1,1]$"? ${}\qquad{}$ $\endgroup$ – Michael Hardy May 7 '15 at 15:06
  • $\begingroup$ The first option $\endgroup$ – Sofia May 7 '15 at 15:08
  • $\begingroup$ @Sofia You didn't have to choose: $a\in[-1,1]$ and $a$ is in $[-1,1]$ means the same. $\endgroup$ – ajotatxe May 7 '15 at 15:10
  • 1
    $\begingroup$ @MichaelHardy: $a\in[-1,1]$ means the same as "$a$ is in $[0,1]$". I suspect the OP meant to say "$a_i \in \{-1,1\}$". At least that makes the statement true! $\endgroup$ – TonyK May 7 '15 at 15:10
  • $\begingroup$ @TonyK : I now suspect you're right. Six answers have appeared so far, and not all of them construe the question that way. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 7 '15 at 15:16
4
$\begingroup$

What is true is that if $a_n = 1$, $p \ge 2$ and all $a_i \in [-1,1]$, then $\sum_{i=0}^n a_i p^i > 0$.

$\endgroup$
7
$\begingroup$

I assume you meant $a_i \in \{-1,1\}$. The same argument works if $ a_n = \pm 1$ and the rest of the $a_i$'s are in the interval $[-1,1]$ or even more generally if $\vert a_ n\vert \geq \vert a_i \vert$ for all $i \in \{0,1,2\ldots,n-1\}$.

If so, assume what you have is true and we will obtain a contradiction.

We have $$\sum_{i=0}^{n-1} a_i p^i = - a_n p^n$$ This gives us $$\left \vert \sum_{i=0}^{n-1} a_i p^i \right \vert = \left \vert - a_n p^n \right \vert = p^n$$ since $\vert -a_n \vert = 1$. We have $$p^n = \left \vert \sum_{i=0}^{n-1} a_i p^i \right \vert \leq \sum_{i=0}^{n-1} \left \vert a_i p^i\right \vert = \sum_{i=0}^{n-1} p^i = \dfrac{p^n-1}{p-1} \leq p^n-1$$ which gives us a contradiction.

$\endgroup$
6
$\begingroup$

How about this?

$$0.01p-0.03=0$$

$\endgroup$
1
$\begingroup$

No. $p=2$, $n = 1$, $a_0 = 1$ and $a_1 = -\frac 12$ gives $$ a_0 + a_1p^1 = 1 -\frac 12 \cdot 2 = 0. $$

$\endgroup$
1
$\begingroup$

Let $n=1$, $a_n=-\frac1p$, $a_0=1$.

$\endgroup$
1
$\begingroup$

If you mean $a_i \in \{-1, 1\}$ then you can just use the rational root's test to show that the only possible rational roots are $\pm 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.