Maths - Proof Roots of Primes are Irrational
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Flashcards
When proving irrationality of a surd $x$, what is the setup?
Assume $x = \frac{a}{b}$ where $a$ and $b$ are integers that do not share any factors.
Why can’t $a$ and $b$ share any factors when proving the irrationality of a surd?
Because it would mean you could simplify $\frac{a}{b}$.
How do you get rid the square root over $x$ when proving it’s irrationality?
Square both sides.
What’s the next step after assuming $\sqrt{3} = \frac{a}{b}$.
\[3 = \frac{a^2}{b^2}\]How can you rearrange $3 = \frac{a^2}{b^2}$?
Given something like $xa^2 = b^2$, what does this tell us about $b$?
$x$ is a factor of $b$.
Why is $3$ a factor of $b$ in $3a^2 = b^2$?
Since $b^2$ is equal to some integer ($a^2$) multiplied by $3$.
If $x$ is a prime factor of $b^2$, then why must it also be a prime factor of $b$?
Consider that $b^2$’s prime factorisation must come in pairs, since it’s the same as the product of $b$’s prime factors squared.
If $b$ is a multiple of $3$, then how could you rewrite $b$?
What’s another way of expressing $3a^2 = b^2$?
- $3a^2 = (3c)^2$
- $3a^2 = 9c^2$
What does the statement $3a^2 = 9c^2$ tell you about $a$?
- Simplifying: $a^2 = 3c^2$
- $3$ must be a factor of $a$.
What is the core contradiction at the heart of the proof the roots of primes are irrational?
If $a$ and $b$ share factors, then the root can’t be written as $\frac{a}{b}$.
In one sentence, what is the core reason that the square root of a prime is irrational?
If $\sqrt{x} = \frac{a}{b}$, you can show $x$ is a factor of $a$ and $b$, which means the fraction could be simplified.
The proof the square roots of primes are irrational is a proof by…?
Contradiction.