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For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)

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Well, a possible but perhaps in the long run not exhaustive method, since you must find for what $k \in \mathbb{N}$ this is true; consider $$(\sqrt{n}+\sqrt{n+k})(-\sqrt{n}+\sqrt{n+k})=-n+(n+k)$$ Which of course gives $k$.

now, since $k$ is rational, it means that $\sqrt{n}$ and $\sqrt{n+k}$ is rational thus \begin{align} n&=&p^{2} \\ n+k&=& q^{2} \end{align} Perhaps you can see where I am going with this and finish it off?

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    $\begingroup$ @Maryam Here we are supposing that $\sqrt{n}+\sqrt{n+k}$ is rational, so $-\sqrt{n}+\sqrt{n+k}$ is also rational. Their sum and difference are also rational. $\endgroup$ – Element118 Jan 6 '16 at 13:58
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Hint:

If $k=2m+1$, $m\ge1$, then $\sqrt{n}+\sqrt{n+k}$ fails to be an irrational number for $n=m^2$: $$\sqrt{n}+\sqrt{n+k}=\sqrt{m^2}+\sqrt{m^2+2m+1}=m+(m+1)\in\mathbb{Q}$$

If $k=4m$, $m>1$, then $\sqrt{n}+\sqrt{n+k}$ fails to be an irrational number for $n=(m-1)^2$. $$\sqrt{n}+\sqrt{n+k}=\sqrt{(m-1)^2}+\sqrt{(m-1)^2+4m}=(m-1)+(m+1)\in\mathbb{Q}$$

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  • $\begingroup$ Just two typos, thanks. $\endgroup$ – Ángel Mario Gallegos Jan 6 '16 at 13:14
  • $\begingroup$ So to answer the question about being irrational you probably want to add that $k$ must be of the form $4i+2, i\in\mathbb{N}$. $\endgroup$ – Ian Miller Jan 6 '16 at 13:24
  • $\begingroup$ @IanMiller $k=1, 4$ also work. $\endgroup$ – Element118 Jan 6 '16 at 13:55
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This result is not true in general for example, suppose that $n$ is a square $n=l^2$ and $k=0$ $\sqrt{l^2+0}+\sqrt{l^2}$ is rational, suppose that $n=9, k=16$, $\sqrt{16+9}+\sqrt{9}$ is rational.

The question is given $n$ for what values of $k$, $\sqrt{n+k}+\sqrt{n}$ is irrrational?

Proposition

Suppose that $k,n\in N$ and $n(n+k)$ is not a square then $\sqrt{n+k}+\sqrt{n}$ is irrational.

Proof:

$(\sqrt{n}+\sqrt{n+k})^2=n+n+k+2\sqrt{n}\sqrt{n+k}$

$((\sqrt{n}+\sqrt{n+k})^2-2n-k)^2=4n(n+k)$

Consider $P(X)= (X^2-2n-k)^2-4n(n+k)=X^4-2(2n+k)X^2+(2n+k)^2-4n(n+k)=X^4-2(2n+k)X^2+k^2$ $\sqrt{n+k}+\sqrt{n}$ is a root of $P(X)$.

Consider $Q(X)=U^2-2(2n+k)U+k^2$. The discriminant of $Q(X)$ is $4(2n+k)^2-4k^2=4(4n^2+4nk+k^2)-4k^2 =16n(n+k)$, thus if $n(n+k)$ is not a square, $Q(X)$ and hence $P(X)$ is irreducible.

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Solution is $k = 2^{2v+1}*j$ where j is odd.

Obviously $k$ cannot equal $0$.

$k$ must/can be an odd number multiplied by an odd power of 2. ($k$ can not be odd. If $k$ is even, $k$ is not an odd number times an even power of 2.

If $0$ is not considered a natural number than 1 is a possible value for $k$ is the one exception. (As $\sqrt{n} + \sqrt{n+1}$ is rational $\iff n = 0$).

$\sqrt{n} + \sqrt{n + k} = r = a/b; \gcd(a,b) = 1 \implies n + k = r^2 + n - 2r\sqrt{n} \implies k = r(r - 2\sqrt{n}) \implies k = (a/b)(a/b - 2m); n = m^2$ for some integer $m$. Which implies $bk = a^2/b - 2am \in \mathbb Z \implies b = 1$.

So for any $k = a(a - 2m)$ we have a possibility of the sum being rational. Otherwise we don't.

So we can't have $k = 2j + 1$ odd or we'd have $k = k(k - 2j)$ (which would yield $\sqrt{j^2} + \sqrt{j^2 + k = j^2 + 2j + 1}$ being rational.)

We can't have $k = 2^{2v}j$ where $j$ is odd or we'd have $k = 2^vj(2^vj - 2^v(j - 1))$

On the other hand if $k = 2^{2v + 1}j$ where $j$ is odd we can't have $k = a(a - 2m)$. The powers of 2 just don't add up.

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Suppose that $$\sqrt n +\sqrt {n+k}=r, \quad r\in\mathbb{Q}$$

Now we have: $$\sqrt {n+k} =r-\sqrt n $$ $$n+k=r^2-2r\sqrt n +n$$ $$2r\sqrt n=r^2-k$$ $$\sqrt n=\frac{r^2-k}{2p}$$

This is a contradiction because the left is $\sqrt n\in\mathbb{I=\mathbb{Q^c}}$, and the right is $\frac{r^2-k}{2p}\in\mathbb{Q}$. The contradiction is due to a wrong assumption that $\sqrt n+\sqrt {n+k}$ was a rational number.

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    $\begingroup$ A few questions: What is $p$ exactly, and where did we state that $\sqrt{n}\notin \mathbb{Q}$? Furthermore, what exactly have you tried to prove? $\endgroup$ – Scounged Jan 6 '16 at 13:28
  • $\begingroup$ I'm seeing this solution. Can you tell me why it is wrong. I have a book before and was resolved the same brand $\endgroup$ – user145717 Jan 6 '16 at 13:29
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    $\begingroup$ @user145717 Look at the question again. This solution seems to have been posted based on a misunderstanding of the problem at hand, as it doesn't even try to adress which values of $k$ are allowed. $\endgroup$ – Scounged Jan 6 '16 at 13:31
  • $\begingroup$ @Scounged I want to generalized this question that for all $n\in\mathbb N$, $\sqrt{n-1}+\sqrt{n+1}$ is irrational. We should chosse $k$ in integer range. $\endgroup$ – Nosrati Jan 6 '16 at 13:34
  • $\begingroup$ @Maryam So the original assignment was to prove that $\sqrt{n-1}+\sqrt{n+1}\notin \mathbb{Q}$ for all $n\in \mathbb{N}$, and you wanted to generalize it, if I'm not mistaken? $\endgroup$ – Scounged Jan 6 '16 at 13:39

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