Square Root of Primes

Prove that the square root of a prime number is irrational.

Solution

Condition 1\text{Condition 1}

Let pp be any prime number. For p\sqrt{p} to be rational, it must be expressible as the quotient of two coprime integers.

p=mn\sqrt{p}=\frac{m}{n}

p=m2n2p=\frac{{m}^{2}}{{n}^{2}}

pn2=m2p{n}^{2}={m}^{2}

Since n,m,n2,m2n,m,{n}^{2},{m}^{2} are integers, this implies that mm has a factor of pp. Therefore, if the expression m=pmm=pm' is substituted into the third equation, then n2=p(m)2{n}^{2}=p{(m' )}^{2}. By a similar argument, the integer nn must possess a factor of pp as well. This demonstrates the fact both mm and nn are not coprime, which contradicts Condition 1\text{Condition 1}. Hence, the square root of a prime number is irrational.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
5 years, 1 month ago

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FYI - To display text in Latex, use \text{XXX}. Otherwise, it will appear italicized

Calvin Lin Staff - 5 years, 1 month ago

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Do you mean the equations in paragraphs? I actually like my equations italicized. They stand out.

Steven Zheng - 5 years, 1 month ago

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Oh, I was refering to "(Condition1) (Condition 1) ". I decided to remove the Latex brackets, so it displays as "Condition 1" instead. If you wanted to use latex, you could do " \alpha \text{ Condition } \beta" to get α Condition β \alpha \text{ Condition } \beta.

Calvin Lin Staff - 5 years, 1 month ago

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@Calvin Lin Okay. I will fix it.

Steven Zheng - 5 years, 1 month ago

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