Articles of rationality testing

Proving that the square root of 5 is irrational

Prove that $\sqrt{5}$ is irrational. I begin with the identity $(\sqrt{5} + 2 )(\sqrt{5} – 2 ) = 1$. Then I am told to extract $\sqrt{5}$ from the first or second factor and consider it to be $\frac{m}{n}$ so I should replace it in both sides. I have $$\frac{m}{n} = (\frac{1}{\frac{m}{n}} + 2) + 2.$$ […]

Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational

Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$? $\sqrt{n^2-4} \in \mathbb{Q} \\ \sqrt{n^2-4} = \frac{p}{q} \\ (\sqrt{n^2-4})^2 = \left(\frac{p}{q}\right)^2 \\ q^2\left( n^2-4\right)=p^2 \\ \text{p is divisible by} \left (n^2-4 \right) \Rightarrow p=k\left (n^2-4 \right) \\ q^2 \left (n^2-4 \right)=k^2 \left(n^2-4 \right)^2 \\ q^2=k^2 \left (n^2-4 \right )\Rightarrow \text{it follows that q is […]

Why $\sqrt{{2 + \sqrt 5 }} + \sqrt{{2 – \sqrt 5 }}$ is a rational number?

This question already has an answer here: How to show this equals 1 without “calculations” 3 answers

Adding or Multiplying Transcendentals

Is it possible to add or multiply (no subtraction) only positive transcendental numbers and yield a solution that is algebraic? Exponential manipulation is excluded from this question, as $e^{\ln2} = 2$ EDIT: Also, excluding reciprocals of other transcendental numbers and subtraction?

Proving the irrationality of $\sqrt{5}$: if $5$ divides $x^2$, then $5$ divides $x$

I am working on proving that $\sqrt{5}$ is irrational. I think I have the proof down, there is just one part I am stuck on. How do I prove that $x^2$ is divisible by 5 then $x$ is also divisible by $5$? Right now I have $5y^2 = x^2$ I am doing a proof by […]

Show that $(\sqrt{2}-1)^n$ is irrational

Show that for all $n\in \mathbb{N}$ the number $(\sqrt{2}-1)^n$ is irrational. I do not get the idea of the proof at all, any help appreaciated. edit: I am also thinking whether it will be possible to show $(\sqrt{2}-1)^n=\sqrt{m+1}-\sqrt{m}$

How would you prove that $\sqrt{2}$ is irrational?

This question already has an answer here: Is $n^{th}$ root of $2$ an irrational number? [duplicate] 2 answers

Using Modularity Theorem and Ribet's Theorem to disprove existence of rational solutions

This is likely overly optimistic, but can one take the results from the Modularity theorem and Ribet’s theorem, and distill these down to an undergrad math level of a way to check if certain rational polynomials have no non-trivial solutions? For instance, I do not understand the details of either theorem, but if I can […]

$\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational

Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational. If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions.

Simple proof that $\pi$ is irrational – using prime factors of denominator

Simple proof that $\pi$ is irrational Consider the Gregory – Leibniz series for $\pi/4$: $$\frac \pi 4 = 1 – \frac 1 3 + \frac 1 5 + \cdots $$ Let $A_n/B_n$ be the irreducible fraction given by partial sum $S_n$ up to the $n$th term $\pm 1/(2n-1)$. It can be shown that largest prime […]