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16answers
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How can you prove that the square root of two is irrational? [closed]

I have read a few proofs that $\sqrt{2}$ is irrational. I have never, however, been able to really grasp what they were talking about. Is there a simplified proof that $\sqrt{2}$ is irrational?
37
votes
9answers
25k views

Prove $2^{1/3}$ is irrational.

Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer. Assume $2^{1/2}$ is irrational. $2^{...
15
votes
5answers
4k views

Proof of irrationality of square roots without the fundamental theorem of arithmetic

Here is an elementary proof (adapted from Hardy's A Course of Pure Mathematics) that for any integer $k$, $\sqrt{k}$ is either irrational or integral. Suppose $\sqrt{k}$ is rational, $\sqrt{k} = \...
6
votes
3answers
2k views

The sum of square roots of non-perfect squares is never integer [duplicate]

My question looks quite obvious, but I'm looking for a strict proof for this: Why can't the sum of two square roots of non-perfect squares be an integer? For example: $\sqrt8+\sqrt{15}$ isn't an ...
42
votes
9answers
99k views

Prove that $\sqrt 2 + \sqrt 3$ is irrational [duplicate]

I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. ...
30
votes
10answers
8k views

Proving $\sqrt 3$ is irrational. [duplicate]

There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum: Suppose $$\...
4
votes
2answers
7k views

Is $n^{th}$ root of $2$ an irrational number? [duplicate]

Possible Duplicate: $a^{1/2}$ is either an integer or an irrational number. Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?
24
votes
8answers
27k views

What rational numbers have rational square roots?

All rational numbers have the fraction form $$\frac a b,$$ where a and b are integers($b\neq0$). My question is: for what $a$ and $b$ does the fraction have rational square root? The simple answer ...
20
votes
1answer
34k views

The square root of a prime number is irrational [duplicate]

If $p$ is a prime number, then $\sqrt{p}$ is irrational. I know that this question has been asked but I just want to make sure that my method is clear. My method is as follows: Let us assume that ...
1
vote
4answers
600 views

Proving $\sqrt2$ is irrational

I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, ...
20
votes
5answers
53k views

Prove that $\sqrt 5$ is irrational

I have to prove that $\sqrt 5$ is irrational. Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common ...
5
votes
2answers
26k views

Can we ever get an irrational number by dividing two rational numbers?

If we try to divide any two random arbitrarily long rational numbers like 103850.2387209375029375092730958297836958623986868349693868398659825528365... and <...
3
votes
3answers
2k views

The contradiction method used to prove that the square root of a prime is irrational

The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is ...
4
votes
4answers
4k views

Why do irrationality proofs of $\sqrt x$ not apply when $x$ is a perfect square?

When trying to prove that a particular root (say $\sqrt{2}$ or $\sqrt{10}$) cannot be rational, I always see a particular indirect proof that goes something like this: Suppose $\sqrt{x}$ were ...
4
votes
3answers
4k views

If $P$ and $Q$ are distinct primes, how to prove that $\sqrt{PQ}$ is irrational?

$P$ and $Q$ are two distinct prime numbers. How can I prove that $\sqrt{PQ}$ is an irrational number?
3
votes
3answers
6k views

Proving that for each prime number $p$, the number $\sqrt{p}$ is irrational [duplicate]

Possible Duplicate: $\sqrt a$ is either an integer or an irrational number. I'm a total beginner and any help with this proof would be much appreciated. Not even sure where to begin. Prove ...
5
votes
3answers
643 views

Irrationality of $\sqrt[n]2$ [duplicate]

I know how to prove the result for $n=2$ by contradiction, but does anyone know a proof for general integers $n$ ? Thank you for your answers. Marcus
1
vote
1answer
3k views

$\sqrt{2}$ is irrational proof using well-Ordering Property

I am having a problem understanding this proof. Suppose that $\sqrt{2}$ was rational. Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\...
36
votes
1answer
959 views

Is $\ln(\ln(n))$ irrational for any integer $n>1$?

Is there $n\in \mathbb{N}$ such that $\ln(\ln(n)) \in \mathbb{Q}$? If such $n$ exists, we will get $$\ln(\ln(n)) = \frac{p}{q}, \quad p, q \in \mathbb{Z}.$$ Hence we will get $n = e^{e^{p/q}},$ ...
8
votes
5answers
947 views

Shorter proof of irrationality of $\sqrt{2}$?

Euclid's proof of the irrationality of $\sqrt{2}$ via contradiction involves arguments about evenness or odness of $a^2 = 2 b^2$ which is then lead to contradiction in using few more steps. I wonder ...
6
votes
3answers
1k views

Prove $2^{1/3} + 2^{2/3}$ is irrational

What's the nice 'trick' to showing that the following expression is irrational? $2^{1/3} + 2^{2/3}$
11
votes
0answers
348 views

What is known about the sum $\sum\frac1{p^p}$ of reciprocals of primes raised to themselves? [duplicate]

Consider the following series: $$\sum_{p\in\mathcal{P}}\frac{1}{p^p}$$ where $\mathcal{P}$ is the set of all prime numbers: $\mathcal{P}=\{2,3,5,7,11,13,\ldots\}$. My question is: Is this a known ...
5
votes
3answers
10k views

Is the cube root of a prime number rational?

The question is: if $P$ is prime, is $P^{1/3}$ rational? I have been able to prove that if $P$ is prime then the square root of $P$ isn't rational (by contradiction) how would I go about the cube ...
5
votes
2answers
697 views

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 ...
4
votes
5answers
577 views

Please explain this step in proving the square root of 3 is irrational

Assume that $$3 = \frac{p^2}{q^2}$$ So, $$ 3 q^2 = p^2$$ So $p^2$ is divisible by $3$. How we can conclude this?
11
votes
0answers
1k views

Novel (?) proof of the irrationality of $\sqrt3$ [duplicate]

A student of mine offered the following proof of the irrationality of $\sqrt{3}$: Suppose $(a/b)^2 = 3$ with $a,b$ having no common factor. Since $a^2=3b^2$, an easy parity argument (using the fact ...
3
votes
3answers
210 views

Find number of integral solutions of $\sqrt{n}+\sqrt{n+7259}$

How many integers $n$ are there such that $\sqrt{n}+\sqrt{n+7259}$ is an integer? No idea on this one.
6
votes
3answers
3k views

Prove $\log_7 n$ is either an integer or irrational

I have been trying to prove a certain claim and have hit a wall. Here is the claim... Claim: If $n$ is a positive integer then $\log_{7}n$ is an integer or it is irrational Proof (so ...
4
votes
1answer
127 views

Irrationality of the concatenation of the rightmost nonzero digits in $n!$

Surfing the internet I bumped into a very interesting problem, which I tried to solve, but got no results. The problem is following: let $h_n$ be the most right non-zero digit of $n!$, for example, $...
3
votes
3answers
358 views

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

How to show that $(2^n-1)^{1/n}$ is irrational for all integer $n\ge 2$? If $(2^n-1)^{1/n}=q\in\Bbb Q$ then $q^n=2^n-1$ which doesn't seem right, but I don't get how to prove it.
10
votes
2answers
5k views

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 ...
8
votes
2answers
616 views

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

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 ...
6
votes
1answer
252 views

A proof that $\sqrt{2}$ is not a rational number.

Is this proof correct? Suppose that $\sqrt{2}=\frac{a}{b}$, where $a,b \in \mathbb{N}$ and $a$ is as small as possible. Then $\sqrt{2}b=a$ which means $2b=\sqrt{2} a$. So we rewrite $\sqrt{2}=\frac{a}...
4
votes
4answers
331 views

irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$

irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$ My attempt: $\sqrt{n}+\sqrt{n+1}=\frac{p}{q}$ $2n+1+2\sqrt{n(n+1)}=\frac{p^2}{q^2}$ Now we have to show $2n+1+2\sqrt{n(n+1)}$ cannot ...
4
votes
6answers
288 views

Show that there are no rationals $r$ such that $r^3 = 6$

Show that there are no rationals $r$ such that $r^3 = 6$ We were asked this for a real analysis assignment. I just assumed that it would be the same as if it were $r^2$, but now I'm second guessing ...
3
votes
6answers
1k views

Is there a way, in general, to tell whether the nth root of a integer is rational?

Is there a way, in general, to tell whether the $n^{th}$ root of a integer is rational? More explicitly, is it possible to elegantly determine whether the result of $k^{1/n}$ is rational for $k,n \in ...
2
votes
3answers
1k views

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{...
2
votes
6answers
2k views

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 ...
2
votes
4answers
402 views

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 ...
3
votes
2answers
137 views

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(...
3
votes
2answers
121 views

$\sqrt{13a^2+b^2}$ 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 ...
0
votes
3answers
2k views

Prove that $6 - \sqrt{2}$ is Irrational by contradiction

What is a Proof by Contradiction, and how to prove by contradiction that $6 - \sqrt{2}$ is an irrational number?