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I have been fiddling around with this question and a definite proof would highly be appreciated.

My reasoning in a nutshell is

Due to the nature of irrationals; cannot be simplified, this means that they are not divisible by any other number with exception to 1 and self which means they must be primes.

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closed as unclear what you're asking by user223391, BruceET, user91500, Henrik, Daniel W. Farlow Dec 13 '16 at 17:54

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  • $\begingroup$ Prime numbers are integers, and these are rational numbers. $\endgroup$ – Xam Dec 11 '16 at 2:24
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    $\begingroup$ Yes, it is for real; complex numbers cannot be categorized as prime. ;) $\endgroup$ – Jed Dec 11 '16 at 2:34
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    $\begingroup$ @ZelosMalum Yes this is a serious question and judging from it you can tell that I am definitely a newbie. Everyone has to begin somewhere and this is where I chose to begin. If it's a joke to you then simply scroll over to another question. No need for such comments. $\endgroup$ – eshirima Dec 11 '16 at 2:44
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    $\begingroup$ @Jed I know you're kidding, but you can classify complex numbers as primes. The Gaussian integers contain primes distinct from those in the rational integers. In a generic ring, we can call an element $p$ prime if $(p)$ is a prime ideal. $\endgroup$ – Mark Dec 11 '16 at 3:48
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    $\begingroup$ Further consider. $\ \alpha = \sqrt\alpha \cdot \sqrt \alpha $ $\endgroup$ – Bill Dubuque Dec 11 '16 at 5:36
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Prime numbers are by definition integers with certain special properties. An irrational number is not an integer, and so cannot be a prime number.

To give some explanation for why only integers are defined to be prime numbers, note that the concept of divisibility is highly sensitive to what kind of numbers you allow yourself to use. For instance, $3$ is not divisible by $2$ as an integer: there is no integer $x$ such that $2x=3$. However, $3$ is divisible by $2$ as a real number, since there does exist a real number $x$ such that $2x=3$. So our usual notions of divisibility only behave the way we would expect when we restrict our attention to integers.

(More generally, you can define "prime" or "irreducible" elements in any integral domain which generalize the notion of "prime numbers" to other settings. But "prime number" always refers to these notions in the setting of the integers, and in any case there is not any natural setting in which all irrational numbers would be prime or irreducible. In the real numbers, for instance, every nonzero element is a "unit" and is not considered prime, similar to how $1$ is not considered a prime number.)

Finally, let me remark that I don't know what you mean when you say irrational numbers "cannot be simplified". The definition of an irrational number is a real number which cannot be written as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers. This doesn't particularly have anything to do with "simplification", whatever you might mean by that, nor does it have anything to do with divisibility. For instance, the number $2\pi$ is irrational: would you say it is divisible by $\pi$? Would you say it is divisible by $2$? In fact, it is reasonable to say it is divisible by every nonzero real number, if you are working in the context of real numbers, since if $d$ is a nonzero real number then $2\pi/d$ is also a real number.

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  • $\begingroup$ I sincerely thank you for actually taking me seriously in spite of other members taking the question for a joke. Your explanation was clear and truly helped me understand. With regards to my "simplification" remark, you read my mind and thank you for the clarification. $\endgroup$ – eshirima Dec 11 '16 at 2:40
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From the Wikipedia page "Prime number":

"A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself."

Irrational numbers are not natural numbers, so they are not prime.

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