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 $$\frac{m}{n} = (\frac{1}{\frac{m}{n}} + 2) + 2.$$ […]

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 […]

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

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?

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 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 $(\sqrt{2}-1)^n=\sqrt{m+1}-\sqrt{m}$

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

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 […]

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 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 […]

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