You are watching: Is -1 a rational number
Is it justified to say the $i =sqrt-1$ is rational?
The beginning of this concern lies in a consistent discussion I have over this t-shirt the mine:
While noticeable $pi$"s comment is completely legit, $sqrt-1$"s can be hypocritical, if the rationality the $sqrt-1$ is questionable.
It is a Gaussian rational number, however it is not rational in the conventional sense of the word since rational numbers space real.
The number $sqrt-1$ is not real. Because the rationals are simply a particular kind of real number, it can not be rational, either.
Another method to look in ~ it: were there integers $a$ and also $b$ such that $sqrt-1 = fracab$ then$$-1 = fraca^2b^2,$$and so$$a^2 = -b^2.$$Since $b^2$ is absolutely positive, that method $a^2$ is absolutely negative, i m sorry is impossible.
$sqrt-1 = i$ is an imaginary number, not lying everywhere on the genuine number line. As such as others have said, that is neither rational no one irrational in the usual senses that those words.
To discuss the specific grammar of the shirt: $pi$"s command is "Get real" i beg your pardon I would say has the connotation that "join
Echoing what other human being have already said: no $i= sqrt-1$ is not a rational number.
You have$$eginalign&invernessgangshow.netbbC ; ext the facility numbers\&cup \&invernessgangshow.netbbR ; ext the real numbers\ &cup \&invernessgangshow.netbbQ ; ext the reasonable numbers\&cup \&invernessgangshow.netbbZ ; ext the integers\&cup \&invernessgangshow.netbbN ; ext the natural numbersendalign$$Here the $cup$ denotes that the lower is had in the upper. So for instance all genuine numbers are complex numbers. And: an creature is a actual number. Note that for example not all complex numbers are real numbers. Not all complicated numbers are rational numbers. No all integers are organic numbers.
So the question now is whether $i = sqrt-1$ (which is a facility number) is a rational number. And also here we very first note the the reasonable numbers are those numbers that have the right to be expressed together a fraction $fracab$ wherein $a$ and also $b$ room integers (so belongs come $invernessgangshow.netbbZ$) ($b
eq 0$). For this reason is $i = fracab$ for any type of integers $a$, and also $b$? As listed in Austin"s fine answer, one can display that without doubt the price is no.
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To have something come think about, possibly you can answer the question: Is $sqrt2$ a rational number? (You can discover the answer below on M.SE).