Which Radical Difference Is Greatest?

$\large{\sqrt{8}-\sqrt{7}= \dfrac{(\sqrt{8}-\sqrt{7})(\sqrt{8}+\sqrt{7})}{\sqrt{8}+\sqrt{7}}= \dfrac{\left(\sqrt{8}\right)^2-\left(\sqrt{7}\right)^2}{\sqrt{8}+\sqrt{7}}= \dfrac{8-7}{\sqrt{8}+\sqrt{7}}= \dfrac{1}{\sqrt{8}+\sqrt{7}}}$

$\large{\sqrt{6}-\sqrt{5}= \dfrac{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}{\sqrt{6}+\sqrt{5}}= \dfrac{\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2}{\sqrt{6}+\sqrt{5}}= \dfrac{6-5}{\sqrt{6}+\sqrt{5}}= \dfrac{1}{\sqrt{6}+\sqrt{5}}}$

We certainly have:

$\large\sqrt{8}+\sqrt{7}> \sqrt{6}+\sqrt{5} \iff \dfrac{1}{\sqrt{8}+\sqrt{7}} < \dfrac{1}{\sqrt{6}+\sqrt{5}} \iff \boxed{\sqrt{8}-\sqrt{7} < \sqrt{6}-\sqrt{5}}$

Thus $\sqrt{6}-\sqrt{5}$ is greater.

I'm sure there are plenty of beautiful algebraic solutions to this. But I am a creature of habit, so I will be using calculus.

Both numbers take on the form $f(x) = \sqrt{x + 1} - \sqrt{x}$ . This function's derivative is $f'(x) = \frac{1}{2\sqrt{x + 1}} - \frac{1}{2\sqrt{x}},$

which is negative for all $x > 0$ , implying $f(x)$ is decreasing on its entire domain and that $f(5) > f(7)$ . Therefore, $\sqrt{6} - \sqrt{5} > \sqrt{8} - \sqrt{7}$

I pictured $f(x) = \sqrt{x}$ . The larger value of $x$ , the smaller the change between $f(x) \text{ and } f(x-1)$ . Therefore, $\boxed{\sqrt{6}-\sqrt{5} > \sqrt{8}-\sqrt{7}}$ .

$\sqrt{8} - \sqrt{7} \text{ }\boxed{?} \sqrt{6} - \sqrt{5}$ $\sqrt{8} + \sqrt{5} \text{ } \boxed{?} \sqrt{7} + \sqrt{6}$ Square both sides subtract 13 and divide by two to get $\sqrt{40} \text{ } \boxed{?} \sqrt{42}$ $\sqrt{40} \text{ }\boxed{<} \sqrt{42}$

and we are done.

You can get some intuition from noting that d/dx of sqrt(x) is decreasing, but the proof is a lot of squares.

assume the right answer

$\sqrt{8} - \sqrt{7} < \sqrt{6} - \sqrt{5}$

$(\sqrt{8} - \sqrt{7})^{2} < (\sqrt{6} - \sqrt{5})^{2}$

$8 - 2\sqrt{56} + 7 < 6 - 2\sqrt{30} +5$

$4 < 2\sqrt{56} - 2\sqrt{30}$

$2 < \sqrt{56} - \sqrt{30}$

$(2)^{2} < (\sqrt{56} - \sqrt{30})^{2}$

$4 < 56 - 2\sqrt{56*30} +30$

$4 < 86 - 2\sqrt{1680}$

$-82 < - 2\sqrt{1680}$

$41 > \sqrt{1680}$

$(41)^{2} > (\sqrt{1680})^{2}$

$1681 > 1680$

This is clearly true, so the original statement is true

Sqrt (2) - Sqrt(1) = 0.4+

Sqrt(4) - Sqrt (3) = 0.3-

Sqrt(10) - Sqrt(9) = 0.16+

Therefore, the smaller figures' must be of greater difference.

Simple. Just consider the value of "√(n+1) - √n" when increase "n", it tends to 0.

we can say that both √8-√7 and √6-√5 are positive then, let √8-√7=x ⇒8+7-2√56=x²⇒15-√224=x².........[1] again let √6-√5=y ⇒6+5-2√30=y²⇒11-√120=y²..........[2] then from [1]and[2] we can say √224>√120⇔[2]>[1] therefore √6-√5 >√8-√7

d/dx(sqrt{X}) = 1/2sqrt(x). As x increases, slope of curve decreases. Draw 2 lines at x=k and x=k+1 for any k>=0. Draw two lines parallel to x axis which pass thru the intersections of curve with X=k and x=k+1. Let these lines intersect y axis at y1,y2. Let z=y2-y1 (y2>y1). If k increases, you'll find that z decreases. This is due to the nature of the curve that was deduced in the first step. Therefore, (c) is correct.

I did this graphically. The reasoning is that the function is concave down and increasing. When a function is concave down and increasing OR concave up and decreasing, the difference between successive terms decreases as x increases.

The square root function has a decreasing slope. This is easily seen by eye and also by looking at the derivative function.

Therefore, the steps between the value of the function at adjacent X are becoming smaller as X grows with 0 as a limit as X approaches infinity.

The gaps between the two X's are same between the two different answer choices.

Thus the second choice in the list is the right answer, the one with 6 and 5.

well i pictured it this way.. see the more the valve in the root the lesser its numerical value is... take examples of root 4 root 9 and root 16 see the pattern

This solution is similar to Nihar Mahajan's solution.

Another interesting thing about these equations, ie. $\sqrt{x}-\sqrt{x-1}$ is the complex conjugates are also the multiplicative inverse so $\sqrt{x}-\sqrt{x-1}=\frac{1}{\sqrt{x}+\sqrt{x-1}}$ , so we can simply observe $\sqrt{8} + \sqrt{7} > \sqrt{6}+\sqrt{5}$ , thus $\frac{1} { \sqrt{8} + \sqrt{7} }= \boxed{ \sqrt{8} - \sqrt{7} < \sqrt{6} -\sqrt{5} } = \frac{1}{ \sqrt{6} + \sqrt{5} }$

Observe that for consecutive integers $1, 2, 3, 4, \ldots , n$ their squares are increasing at an increasing rate $1, 4, 9, 16, \ldots , n$ . Since square roots must follow the opposite (i.e. inverse) pattern to squares, observe then that square roots of consecutive numbers increase at a decreasing rate. As an instant of this consequence, $\sqrt{6} - \sqrt{5} > \sqrt{8} - \sqrt {7}$ .

Rationalize both surds and then it can be clearly seen that sq.root 6 - sq. Root 5 is greater since its denominator comes to be smaller.

Just look at x^1/2 graph ... You can see as the slope goes on decreasing f(6)-f(5) will be greater than f(8)-f(7)

I found using logarithms better and easy.. ie.,

Since all the given numbers, ie \sqrt{8}, \sqrt{7}, \sqrt{6}, \sqrt{5}, are irrational numbers, it is obvious that \sqrt{8}-\sqrt{7} is not equal to \sqrt{6}-\sqrt{5}.

Then, assuming \sqrt{8}-\sqrt{8} to be greater than \sqrt{6}-\sqrt{5}, we have /[ \dfrac{{\sqrt{8}-\sqrt{7}}{ \sqrt{6}-\sqrt{5}} > 1/]

Implying, 0.5 x( ln8 -ln7 -ln6 + ln5 ) < 0...

It is obvious fro the above statement that it is true, since 8>7>6>5 and hence ln8>ln7>ln6>ln5 follows.

Hence ln5 cannot be greater than (ln8 +ln7 +ln6 )..

Hence our assumption is true.

If it would have been false, our assumption would have been false... :)

By taking log it can be easily solved. log(6^1/2-7^1/2)>log(8^1/2-7^1/2)

This was displayed incorrectly. It looked like root 6 minus 1. The whole root 5 wasn't shown.

This is my first written feed back on this site. I may make mistakes and would like some input.
I set up:
8^(1/2) - 7^(1/2) -->
ln(8^(1/2)) - ln(7^(1/2)) -->
(1/2)(ln(8)) -(1/2)(ln(7))-->
(1/2)(ln(8)-ln(7))-->
(1/2)(ln(8/7)-->
I know that if I were to input y=ln(1), y=0. The graph increases ad infinitum afterwards. I would look at the quotients within the ln(x) and see which ratio is greater. 6/5 > 8/7-->
1.2 >1.14... -->
The ratio between 6:5 is greater and therefore would have a greater difference.

Here is a theoretical approach. We know that 8-7=6-5 because integral numbers grow at a fix rate of 1. But how do these numbers' square root behave? Do they also grow at a fix rate? Considering that: Sqrt (1) = 1 and Sqrt (4) = 2: it takes 3 numbers (from 1 to 4) to its Sqrt grow 1 Sqrt (4) = 2 and Sqrt (9) = 3: it takes 5 numbers (from 4 to 9) to its Sqrt grow 1 So square roots from integer numbers grow at a decreasing rate, not a fix one. Therefore the difference between two consecutive bigger numbers’ square roots ( Sqrt(8) – Sqrt (7) ) will be smaller than the difference between two consecutive smaller numbers’( Sqrt(6) – Sqrt (5) )

Considering the question, We imagine (^4 - ^3) and ( ^2 - ^ 1),we find that ( ^ 2 -^ 1) is bigger. on same lines we find (^6 - ^5) is bigger tha ( ^ 8 - ^ 7)

Simple, just becouse of the rational fuction tendency. When greater the number smaller is his growth

Just picture a graphical representation of the function and the answer will be obvious.