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Consider the vector space $\mathbb R$ over the field $\mathbb Q$.Consider the irrational number $e$ and let $c_1 , c_2 , ... , c_n$ be distinct rational numbers.Then show that $e^{c_{1}} ,e^{c_{2}} ,... , e^{c_{n}}$ are linearly independent.

I have tried but I fail.Please help me.

Thank you in advance.

Sungjin Kim
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Arnab Chattopadhyay.
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3 Answers3

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This is false. Indeed, every positive real number has the form $e^c$ for some $c\in\mathbb{R}$, so $e^{c_1},e^{c_2},\dots,e^{c_n}$ could be any sequence of distinct positive real numbers at all, including sequences which are linearly dependent over $\mathbb{Q}$.

Eric Wofsey
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Suppose that $e^{c_1}, \ldots, e^{c_n}$ are linearly dependent over $\mathbb{Q}$. Let $b_1,\ldots, b_n$ be rational numbers, not all zero, such that $$ b_1 e^{c_1} + \ldots + b_n e^{c_n} = 0. \ \ (1) $$ Let $N$ be a common denominator of $c_i$'s, so that $N$ is a positive integer. Then $(1)$ implies that $e^{1/N}$ is an algebraic number. This is impossible since $e$ is a transcendental number.

Sungjin Kim
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    The numbers $c_i$s are $d_i/N$ for some integer $d_i$. So, with (1), we can find a nonzero polynomial with rational coefficients that $e^{1/N}$ is a root. – Sungjin Kim May 25 '17 at 01:01
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Consider the real funtion $e^x$, so for 6 and 3 there is a $c_1,c_2\in \mathbb {R}$, with $3=e^{c_1}$ and $6=e^{c_2}$, but 3 and 6 are dependents.

user407940
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