Linked Questions

2
votes
7answers
2k 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 ...
3
votes
2answers
905 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?
3
votes
3answers
2k 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 ...
2
votes
3answers
1k 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. ...
0
votes
2answers
958 views

Are the results 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 ...
0
votes
4answers
120 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?
0
votes
1answer
5k 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 ...
0
votes
2answers
357 views

Prove there is no rational number p/q whose square is 2. [duplicate]

I need to prove there is no rational number p/q whose square is 2.I think there may be an easy way to prove this by contradiction,But i cannot solve it.Please help
1
vote
1answer
130 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 ...
0
votes
1answer
110 views

non-perfect square of number [duplicate]

Possible Duplicate: $a^{1/2}$ is either an integer or an irrational number I would like to know the better proof for the following one. question: non perfect square of any integer is an ...
1
vote
1answer
53 views

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. $2^1 > 1$ ...
1
vote
0answers
41 views

Are all the square roots of non-square numbers surds? [duplicate]

Quite self explanatory really, basically, are $\sqrt5 ,\sqrt3$ and $\sqrt7 $ and surds? (So basically, every square root of any non-square number)
0
votes
1answer
37 views

Let $n$ be contained in $\mathbb N$. Show that $ n^{1/2}$ is either an integer or it is irrational. [duplicate]

My textbook shows the proof of $\sqrt{2}$ but I'm having trouble doing the same exact thing for $\sqrt{n}$. So, I have $A:= \{x$ contained in $\mathbb R: x^2 < n\}$. First if $x^2 < n$ then $x ...
23
votes
9answers
4k 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. ...
26
votes
7answers
3k views

How can you prove that the square root of two is irrational?

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?

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