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?
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?
26k 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 ...
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 ...
8k 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 ...
411 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. ...
456 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 $\sqrt{t} \in \mathbb{N}$
127 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". ...
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 $\sqrt{n}$ ...
340 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 $ra^2$.
165 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. $2^1 > 1$ ...