4k views

How to prove that $\sqrt 3$ is an irrational number? [duplicate]

Possible Duplicate: $a^{1/2}$ is either an integer or an irrational number I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?
27k views

Prove that the square root of 3 is irrational [duplicate]

I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How ...
2k 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?
6k views

Prove that if $n$ is not the square of a natural number, then $\sqrt{n}$ is irrational. [duplicate]

Possible Duplicate: $\sqrt a$ is either an integer or an irrational number. I have this homework problem that I can't seem to be able to figure out:Prove: If $n\in\mathbb{N}$ is not the square ...
9k views

Are the square roots of all non-perfect squares irrational? [duplicate]

I was asked to define a non-perfect square. Now obviously, the first definition that comes to mind is a square that has a root that is not an integer. However, in the examples, 0.25 was considered a ...
425 views

all prime numbers have irrational square roots [duplicate]

How can I prove that all prime numbers have irrational square roots? My work so far: suppose that a prime p = a*a then p is divisible by a. Contradiction. Did I begin correctly? How to continue?
16k views

Is square root of an integer, either an integer or an irrational number, but never (non-integer) rational? [duplicate]

Possible Duplicate: $\sqrt a$ is either an integer or an irrational number. $\sqrt{2}$ is irrational number, but $\sqrt{9} = 3$ is an integer. Are there such integers whose square root is a (...
4k 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. ...
482 views

question about the proof about the square root of natural numbers [duplicate]

Could someone please help me to prove that for $t \in \mathbb{N}$ , $\sqrt{t} \in \mathbb{Q}$ if only if $\sqrt{t} \in \mathbb{N}$
131 views

Prove that $\sqrt{3}$ is not a rational number [duplicate]

There is a similar question however that question asks why $3 |p^2$. Here the question is about $3 | p^2 \rightarrow 3 | p$. It is a simple exercise (1.2.1) from Abbot's "Understanding Analysis". ...
355 views

Prove that if n is a natural number and if n has a rational square root then in fact the square root of n is an integer [duplicate]

$n = (\frac{a}{b})^2$, where $a$ and $b$ have no common divisors. This yields $nb^2 = a^2$ $ra^2b^2 = a^2$ (because $n = ra^2$) I don't understand why $n$ is equal to $ra^2$.
100 views

If $\sqrt{n}$ is not an integer, are there any rational numbers $x$ such that $x^2 = n$? [duplicate]

I have a feeling this has something to do with prime numbers, but I'm totally lost. I'm not sure how to write the proof that proves (or disproves) the following: "Let $n$ be an integer. If $\sqrt{n}$ ...
166 views

Prove $\sqrt{k}$ is not a rational number. [duplicate]

Suppose $k>1$ is an integer, and k is not a square number, then $\sqrt{k}$ is not a rational number. Proof: Let $\sqrt{k}=\frac{p}{q}$, and $(p,q)=1$,So $q^2|p^2$, $p\neq 1$, $k$ is not an ...
How to check this number $\sqrt{47}$ is irrational [duplicate]
Prove that $\sqrt{47}$ is irrational number. I know that a rational number is written as $\frac{p}{q}$ where $p$ & $q$ are co-prime numbers. But I do not have any idea to prove it irrational ...
How do you prove $\sqrt{n}$ is an integer or it is irrational? [duplicate]
I have tried this problem five times but I keep getting stuck. I keep following the proof for $\sqrt{2}$. I know that I have to prove that the set is nonempty. Which I do by induction. $2^1 > 1$ ...