Can we prove these bounds for this fraction involving surds?

Can we prove these bounds for this fraction involving surds?

Question

1. The variables $x$ and $y$ are connected by the relation $y = ax^n$, where $a$ and $n$ are constants; $y = 3$ when $x = 4$ and $y = 2$ when $x = 9$. Find the exact values of $n$ and $a$.

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1. Express $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ as a fraction having a rational denominator. Show that, if $n$ is a positive integer, then $n(n+2)$ lies between $n^2$ and $(n+1)^2$. If $n$ is a positive integer, use the results in (i) and (ii) to find in terms of $n$ two consecutive integers between which $\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ must lie.

Suggestion

The question asks us to get rid of the square roots in the denominator; what method might we use?

Can we multiply $\sqrt{a}-\sqrt{b}$ by something so that we lose the square roots?

What is $n(n+2) - (n+1)^2$?

The question tells us to use results we’ve already proved… like many exam questions!

Solution

By substituting the information we’re given in the question, we obtain the equations

$$\begin{equation} 3 = 4^n a \qquad \end{equation}$$

and

$$\begin{equation} 2 = 9^n a. \qquad \end{equation}$$

Dividing the second by the first to eliminate $a$ gives $\frac{2}{3} = \left(\frac{9}{4}\right)^n.$

To work with this, it will be more convenient to write everything as powers of $2$ and $3$.

That is, $2^{2n+1}=3^{2n+1}$, which can only be the case when $2n+1=0$ (as $2$ and $3$ have highest common factor $1$).

Therefore, $n=-\frac{1}{2}$.

Now we can go back to equation $(1)$ or equation $(2)$ to find that $a=6$.

Alternatively, if you know about logarithms…

Taking logs to base $10$ with our starting equations (any base will do, in fact) we get $\log3 = \log a + 2n\log2, \log2 = \log a + 2n \log3.$

Subtracting these gives $\log3 - \log2 = 2n(\log2-\log3)$, and so $n = -\dfrac{1}{2}$.

Substituting back into one of our starting equations gives $a = 6$.

If we multiply both the numerator and the denominator by $\sqrt{a} + \sqrt{b}$, then the fraction becomes $\dfrac{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{(\sqrt{a}+\sqrt{b})^2}{a-b}=\dfrac{a+b+2\sqrt{ab}}{a-b}.$ This has a rational denominator.

Note $n(n+2)-n^2 = 2n > 0$, so $n(n+2)>n^2$.

We also have $n(n+2)-(n+1)^2=n^2 + 2n - n^2 - 2n -1=-1 < 0,$ so $n(n+2)<(n+1)^2$.

So $n(n+2)$ lies between $n^2$ and $(n+1)^2$.

1. If $n$ is a positive integer, use the results in (i) and (ii) to find in terms of $n$ two consecutive integers between which $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ must lie.

By using (i) we get $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}=\dfrac{2n+2+2\sqrt{n(n+2)}}{2}=n+1+\sqrt{(n+2)n}.$

Now we know $(n+1)^2>n(n+2)>n^2$, which implies $n+1>\sqrt{n(n+2)}>n$.

So adding $n+1$ to each part of the inequality, we get $2n+2>n+1+\sqrt{(n+2)n}>2n+1,$

and so $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ lies between $2n+1$ and $2n+2$.