Archimedean property application with logarithms

Asked Dec 13 '19 at 03:00

Active Dec 13 '19 at 03:00

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I am attempting this proof but do not know what to do about assuming the sign of $a$. Is this proof correct?I don't think it is because the result is supposed to hold for all $a$

The problem is prove if $a \in \mathbb{R}$,$\exists n \in \mathbb{N}$ such that $a<10^n$.

My attempt assume $a>0$ then $log(a) \in \mathbb{R}$

Then by the Archimedean property there exists a natural number such that

$n>log(a) \implies 10^n>a$

asked Dec 13 '19 at 03:00

user707991

Looks good to me. – Rushabh Mehta Dec 13 '19 at 03:02
@DonThousand is it okay here to assume $a$ is positive, what about when it is negative or zero? – user707991 Dec 13 '19 at 03:03
1

Don't need to assume. $n=1$ works when $a\leq0$. – Rushabh Mehta Dec 13 '19 at 03:05
sounds good thanks – user707991 Dec 13 '19 at 03:06

Archimedean property application with logarithms

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