Articles of rationality testing

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 […]

Understanding the proof of “$\sqrt{2}$ is irrational” by contradiction.

I have some difficulties in understanding the proof of “$\sqrt{2}$is irrational” by contradiction. I am reading it in 10th class(in India) Mathematics book( available online, here ) This is the snapshot of it: The proof starts with assuming that $\sqrt{2}$ can be written as a ratio of two integers and then that this fraction can […]

Question on a constructive proof of irrationality of $\sqrt 2$

Here is the constructive proof of $\sqrt 2 \not \in \mathbb Q$ found on this page : Given positive integers $a$ and $b$, because the valuation (i.e., highest power of 2 dividing a number) of $2b^2$ is odd, while the valuation of $a^2$ is even, they must be distinct integers; thus $|2 b^2 – a^2| […]

Find conditions on positive integers so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational

Find conditions on positive integers $a, b, c$ so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational. My solution: if $ab$ is not the square of an integer, then the expression is irrational. I find it interesting that $c$ does not come into this at all. My solution is modeled (i.e., copied with modifications) from dexter04’s solution to Prove […]

Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational

How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational? I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?