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Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$? Squaring both sides will give me something but I could not go any further.

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12  
Use calculator to this end. –  user64494 Nov 11 '13 at 10:16
3  
Here is a generalisation of this question. –  TonyK Nov 11 '13 at 14:02
    
I have answered a more generalized question. –  Shay Ben Moshe Nov 11 '13 at 18:44
    
Just need a bit of common sense: square roots rise slower the larger the number being square rooted –  EpicGuy Nov 14 '13 at 8:07

15 Answers 15

up vote 37 down vote accepted

As $(\sqrt7+\sqrt5)^2=12+2\sqrt{35}$ and $(\sqrt6+\sqrt6)^2=12+2\sqrt{36}$

$$(\sqrt7+\sqrt5)^2<(\sqrt6+\sqrt6)^2$$

$$\implies \sqrt7+\sqrt5<\sqrt6+\sqrt6\text{ as } a^2>b^2\iff a>b\text{ for }a,b>0$$

$$\implies \sqrt7-\sqrt6<\sqrt6-\sqrt5$$

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1  
To the downvoter, can you please pinpoint the mistake which is more important to me than mere $"-2"$ –  lab bhattacharjee Nov 12 '13 at 18:18
    
I beg the down-voter to point out the mistake, which I wish to rectify –  lab bhattacharjee Nov 24 '13 at 16:40

Hint. $$\sqrt{a+1}-\sqrt{a}=\frac{1}{\sqrt{a+1}+\sqrt{a}}$$

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Consider the function $f(x) = \sqrt{x}$. Then the second derivative is $$f''(x) = -\frac{1}{4}x^{-3/2} < 0$$

Hence this function is concave down, so we see that $\sqrt{7} - \sqrt{6}$ is the smaller of the two numbers.

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3  
If we wanted to go the calculus route (which is probably too advanced for OP), we could just note that the derivative of $\sqrt{x+1} - \sqrt{x}$ is $> 0$ for $x > 0$ –  BlueRaja - Danny Pflughoeft Nov 11 '13 at 19:52

To take your idea of squaring both sides, you can do one thing to make that more successful. We are comparing $\sqrt 7-\sqrt 6$ with $\sqrt 6-\sqrt 5$ so $$ \sqrt 7-\sqrt 6\stackrel{?}{\lt\gt}\sqrt 6-\sqrt 5\\\sqrt 7 + \sqrt 5\stackrel{?}{\lt\gt}2\sqrt 6\\ 12 + 2\sqrt{35} \stackrel{?}{\lt\gt}24 $$ but since $\sqrt {35} \lt 6$ the left is less.

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4  
My post is in the spirit of exploration, trying to find the answer. lab bhattacharjee's is a good example of how to write up what you discovered in this approach. –  Ross Millikan Nov 11 '13 at 4:19
2  
+1 for $\stackrel{?}{\lt\gt}$ –  miracle173 Nov 14 '13 at 7:07

$1-$ I will leave this part as it is; please read the comments below:

$$\lim_{n\rightarrow\infty} \sqrt{n+1}-\sqrt{n}=0$$

$2-$ let

$$f(n)=\sqrt{n+1}-\sqrt{n}$$

$$f^{'}(n)=\frac{1}{2}\left[\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n(n+1)}}\right]$$ We see that $f^{'}(n)<0$ for all $n$. This shows that $f$ is monotonically decreasing.

As a result for any pair $n_1$ and $n_2$ s.t. $n_1>n_2$ we have $f(n_2)>f(n_1)$

Now take $n_1=6$ and $n_2=5$. this leads to $f(5)>f(6)\Rightarrow \sqrt{6}-\sqrt{5}>\sqrt{7}-\sqrt{6}$

I hope this help now.

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6  
Point 1 is irrelevant to the demonstration, point 2 just generalises the statement without indication of how the more general statement is easier to prove that the particular one. –  Marc van Leeuwen Nov 11 '13 at 9:03
    
@Marc van Leeuwen you are completely right. It is only the idea why one at first look could think thar $2$ should hold in a very simple way. Please don't get me wrong. I don't say that $1$ implies $2$. It gives the idea of $2$ together with the fact that $\sqrt{n+1}-\sqrt{n}$ is a simple function. –  Seyhmus Güngören Nov 11 '13 at 9:11
    
Sharing ideas leading to the solution is in my opinion more important than everything. For example: Hint.... How did you find that hint? What lead you to give such a hint? Only these information will teach the owner of the question how to think for the next time. Otherwise every person becomes a cipher to be decoded. –  Seyhmus Güngören Nov 11 '13 at 12:44
    
@MarcvanLeeuwen see the edit. –  Seyhmus Güngören Nov 12 '13 at 23:01

Hint: $$ \begin{align} \frac1{\sqrt7+\sqrt6}\le\frac1{\sqrt6+\sqrt5} \end{align} $$

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Short is fine. UpVote $0$ k. –  Felix Marin Nov 14 '13 at 6:32

Consider the shape of the graph of the root function. It is monotonously rising but getting flatter all the time. Hence the differences between two values ($1$ apart on the x-axis) near the y-axis is greater than between two values ($1$ apart on the x-axis) further away from the y-axis.

enter image description here

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Nice to have a plot. You do realize that this is exactly what T.Bongers's is saying :-) –  Jyrki Lahtonen Nov 11 '13 at 17:12
    
Yeah, his answer proves the second part of my second sentence. But I'm pretty sure my answer was here before his. Just the plot took me a little longer and was added later ;-) –  Alfe Nov 12 '13 at 11:16

Brute force solution, Only Squarings:

Let's assume: $$\sqrt{7}-\sqrt{6} \geq \sqrt{6}-\sqrt{5}$$ after squaring $$13 -2 \sqrt{42} \geq 11 - 2 \sqrt{30}$$ after simplification: $$1 -\sqrt{42} \geq -\sqrt{30}$$ reversing: $$\sqrt{30} \geq \sqrt{42}-1$$ Squaring 2nd time: $$30 \geq 43 - 2 \sqrt{42}$$ Simplification: $$2 \sqrt{42} \geq 13$$ Last squaring: $$ 168 \geq 169$$ Contradiction, thefore $$\sqrt{7}-\sqrt{6} < \sqrt{6}-\sqrt{5}$$

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You can also use the inequality between the arithmetic mean and the quadratic mean: $$ \frac{x+y}{2} \leq \sqrt{\frac{x^2 + y^2}{2}}. $$ Setting $x = \sqrt{5}$ and $y = \sqrt{7}$ gives you $$\frac{\sqrt{5} + \sqrt{7}}{2} \leq \sqrt{6},$$ which is equivalent to $$ \sqrt{7} - \sqrt{6} \leq \sqrt{6} - \sqrt{5}.$$

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Here's how to figure it out without any arithmetic:

  1. As you know, the graph of the square function grows steeper and steeper: The difference between $(n+1)^2$ and $n^2$ grows ever larger as $n$ increases.

  2. The square root function is the inverse of the square function, so the difference grows ever smaller as $n$ increases. Therefore, the difference between $\sqrt{7}$ and $\sqrt{6}$ is smaller than the difference between $\sqrt{6}$ and $\sqrt{5}$.

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Let $A = (\sqrt{7}-\sqrt{6}) - (\sqrt{6}-\sqrt{5} )$

This comes to comes to $A = (\sqrt{7}+\sqrt{5})-2\sqrt{6}$

The square of the first is $12+2\sqrt{35}$, the square of the second is $12+2\sqrt{36}$

Since $35 \lt 36$, then $A \lt 0$, hence $\sqrt{6}-\sqrt{5}$ is bigger.

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$\sqrt{7} - \sqrt{6} = (\sqrt7 - \sqrt6 )\frac{\sqrt7 + \sqrt 6}{\sqrt7 + \sqrt 6} = \frac1{\sqrt7 + \sqrt 6} $

Similarily,
$\sqrt6 - \sqrt 5 = \frac1{\sqrt6 + \sqrt 5}$

Since $ \sqrt 7 > \sqrt 5$,
$$\sqrt7 + \sqrt6 > \sqrt6 + \sqrt5$$ $$\implies \frac1{\sqrt7 + \sqrt6} < \frac1{ \sqrt6 + \sqrt5}$$ $$\implies \sqrt{7} - \sqrt{6} < \sqrt{6} - \sqrt{5}$$

So $ \sqrt6 - \sqrt5$ is the greater one.

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$\sqrt 6−\sqrt 5$ is greater than $\sqrt 7−\sqrt 6$ because the value of $\sqrt 6$ is 2.449, $\sqrt 5$ is 2.236 $\sqrt 7$ is 2.645 on subtracting $\sqrt 6−\sqrt 5$ you get 0.213 whereas on subtracting $\sqrt 7−\sqrt 6$ you get 0.196.

So, $\sqrt 6−\sqrt 5$ is greater.

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$7-6=6-5$

or, equivalently $(\sqrt7-\sqrt6)(\sqrt7+\sqrt6)=(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)$

hence, $$\frac{\sqrt7-\sqrt6}{\sqrt6-\sqrt5}=\frac{\sqrt6+\sqrt5}{\sqrt7+\sqrt6}<1 \text{ }\text{ }\text{ } (\text{since } \sqrt7>\sqrt5)$$

$\therefore \sqrt6-\sqrt5>\sqrt7-\sqrt6$

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$$\frac{f(5)+f(7)}{2} \lt f(\frac{5+7}{2}) $$ for a concave function $f$, and so for $f(x)=\sqrt x$.

But even if you proceed on the way you have started you will get the result.

$$\begin{eqnarray} \sqrt{7}-\sqrt{6} &\lt& \sqrt{6}-\sqrt{5} &\mid& ^2 \\ 13-2\,\sqrt{6}\,\sqrt{7} &\lt& 11-2\,\sqrt{5}\,\sqrt{6} &\mid& -11+2\sqrt{6}\sqrt{7} \mid \div 2 \\ 1 &\lt& \sqrt{6}\,\sqrt{7}-\sqrt{5}\,\sqrt{6} & \mid& ^2 \\ 1&\lt& 72-12\,\sqrt{5}\,\sqrt{7} & \mid& +12\sqrt{5}\sqrt{7} \\ 12\,\sqrt{5}\,\sqrt{7} &\lt& 71 & \mid& ^2 \\ 12^2 \cdot 35 &\lt& 71^2 \end{eqnarray}$$ All squaring operations are reversible and don't change the inequality relation because the LHS and the RHS of the inequalities are positive. But $$12^2 \cdot 35 =12 \cdot 6 \cdot 2 \cdot 35=72 \cdot 70 = (71+1)(71-1)=71^2-1$$

So the last and therefore all inequalities are true.

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