# Can a rational number be represented by a combination of irrational numbers?

Any real number is some power of $e$ (because $\ln(x)$ has values in the range $(-\infty , + \infty)$.

Say, $5$ is a rational number. So there is some $x$ which makes $\exp(x)= 5$.

1. What is $x$? rational , irrational?

2. A power of $e$ leads to $5$, Will it be an infinite series converging to the value $5$?

3. Can a combination of irrational number result into a rational number?

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I don't really know what the question is, but $\ln(5)$ is irrational, a consequence of the fact that $e$ is transcendental. If $e^{p/q}=5$, then $e^p=5^q$. Slightly more generally, if $e^x$ is rational, then $x$ is irrational. Related question where you'll find a much more general statement: math.stackexchange.com/questions/15285/… –  Jonas Meyer Jan 11 '13 at 6:22
What @Jonas said. Regarding Can a combination of irrational number(s) ... result into a rational number?, consider $\pi+(42-\pi)$. –  Did Jan 11 '13 at 6:30
$(1/2 + \pi) + (1/2 - \pi) = 1$. –  mjqxxxx Jan 11 '13 at 6:30
@Jonas, also $x$ must be nonzero. –  sdcvvc Jan 11 '13 at 6:58
Also, if x is rational, then e^x is irrational. This statement is different to Jonas'. –  Adam Rubinson Jan 11 '13 at 8:53

(When he was 12) George Bergman proved that any integer can be represented by a finite sum of integral powers of the golden ratio (irrational number), http://en.wikipedia.org/wiki/Golden_ratio_base. And any rational can be written as the ratio of two integers.

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which implies that you

1. can always find a sequence of irrational numbers that converges to a given rational

2. can always find a sequence of rational numbers that converges to a given irrational

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Obviously $\sqrt{2}$ and $-\sqrt{2}$ are both irrational, and if "combining" them allows for the operation of addition, then just observe the sum is the rational number $0$.