Azhaghu always compares

If we decide to just stick to $A$ and $B$ and find out which of the two is bigger, we can do it as follows:-

We know that $\sqrt{65}+8 > \sqrt{50} +7$ From the properties of inequalities:- $\dfrac{1}{\sqrt{65}+8}< \dfrac{1}{\sqrt{50}+7}$ Now we can rationalize the radical expression on both the sides. $\sqrt{65}-8 < \sqrt{50} - 7$ $A<B$ Hence $B$ is Bigger.

First, we will assume that $A > B$ . Then, if that is true, then $A - B >0$ .

Thus,

$\sqrt {65} - \sqrt {50} -8 + 7 > 0$

$\sqrt {65} - \sqrt {50} > 1$

However, the inequality does not hold true on the left side. We know that $\sqrt {64} - \sqrt {49} = 1$ , and an increment in the first radical will not yield a greater (or even an equal) increase with the same increment in the second radical. Thus, $A> B$ is false and we arrive at the fact that $B > A$ .

Solution to Bonus: Generalize this $A = \sqrt{65} - 8 = \sqrt{64 + 1} - 8 = \sqrt{8^2 + 1} - 8$ $B = \sqrt{50} - 7 = \sqrt{49 + 1} - 7 = \sqrt{7^2 + 1} - 7$ Let $n = 7$ and $n + 1 = 8$ We have two equations shown: $A = \sqrt{(n + 1)^2 + 1} - (n + 1)$ $B = \sqrt{n^2 + 1} - n$ Rationalize $B$ , and you'll get: $B = \dfrac{(\sqrt{n^2 + 1} - n) (\sqrt{n^2 + 1} + n)}{(\sqrt{n^2 + 1} + n)}$ $B = \dfrac{n^2 + 1 - n^2}{(\sqrt{n^2 + 1} + n)}$ $B = \dfrac{1}{(\sqrt{n^2 + 1} + n)}$ Do the same for $A$ (try it yourself, I'm too lazy to type out the steps) and you'll get: $A = \dfrac{1}{\sqrt{(n + 1)^2 + 1} + (n + 1)}$ Now, for all positive $n$ values, you should be able to see that $\sqrt{n^2 + 1} + n < \sqrt{(n + 1)^2 + 1} + (n + 1)$ Therefore $\dfrac{1}{\sqrt{n^2 + 1} + n} > \dfrac{1}{\sqrt{(n + 1)^2 + 1} + (n + 1)}$ We can conclude the general rule $\boxed{\sqrt{n^2 + 1} - n > \sqrt{(n + 1)^2 + 1} - (n + 1)}$ for all positive $n$ values

For this question, we can use the general rule found to say that $\sqrt{50} - 7 > \sqrt{65} - 8$ which gives $\boxed{B > A}$ Other examples that would follow this general rule: $\sqrt{82} - 9 > \sqrt{101} - 10$ $\sqrt{37} - 6 > \sqrt{50} - 7$ $\sqrt{5} - 2 > \sqrt{10} - 3$

Here's a different solution

I'll use the result

$\displaystyle (1 + x)^n \approx 1 + nx, x \ll 1$

Here

$\displaystyle A = (64 + 1)^{\frac{1}{2}} - 8 = 8( 1 + \frac{1}{64})^{\frac{1}{2}} - 8$

Taking power down to get

$A = 8( 1 + \frac{1}{128}) - 8 = \frac{1}{16}$

Similar work for B

$B = 7(1 + \frac{1}{49})^{\frac{1}{2}} - 7 = \frac{1}{14}$

Hence $B > A$

Note that these value are close to original values but not equal to it.

Mine is a graphical solution.

$y = \sqrt{x}$ is half a sideways parabola in the first quadrant. Notice that for positive real numbers $a > b$ the secant between $x = a$ and $x = a+1$ is slightly gentler (less steep) than the secant between $x = b$ and $x = b+1$ .

Since their width is equal ( $=1$ ) their heights must then be unequal. It's easy, then, to see that the difference $\sqrt{a+1}-\sqrt{a} < \sqrt{b+1}-\sqrt{b}$ .

Consider $a = 64$ and $b = 49$ . We can then say $\sqrt{64+1}-\sqrt{64} = \sqrt{65}-8 < \sqrt{50}-7 = \sqrt{49+1}-\sqrt{49}$ . This is equivalent to saying

$\boxed{B > A}$

A = $\sqrt{65} - 8$ ; B = $\sqrt{50} - 7$

Let's define a function : f(x) = $\sqrt{x^{2} + 1} - x$

f'(x) $= \dfrac{x}{\sqrt{x^{2} + 1}} - 1$ and f'(x) $< 0 \quad \forall x \in Domain$

$\therefore$ f(x) is a decreasing function .

Hence we can conclude that f(7) > f(8) $\implies$ B>A

I'm working on a generalization, but for the moment my intuitive solution for this problem. First, let's analize the difference between the square of the numbers involved: $8^2-7^2=9=(8+7)*(8-7)$ . It becomes clear from that the difference between the square of two consecutive numbers grows as the numbers get bigger. Then, as the numbers get bigger, $sqrt(x^2+1)-x$ will become smaller (since it's "harder" for the square root to "reach" the next number, because the gap is much bigger). From that we can clearly see that $sqrt(65)-8$ will be smaller than $sqrt(50)-7$ . (I don't know how to write in LaTex, sorry about that).

General Case:

√(x^2+1) > √((x-1)^2+1)

x > x-1

√(x^2+1) + x > √((x-1)^2+1) + (x-1)

1/(√(x^2+1) + x) < 1/(√((x-1)^2+1) + (x-1))

(√(x^2+1)-x)/(x^2+1-x^2) < (√((x-1)^2-1)-x)/((x-1)^2+1-(x-1)^2)

√(x^2+1) - x < √((x-1)^2+1) - (x-1)

Specific Example:

Let x = 8:

√(8^2+1) - 8 < √((8-1)^2+1) - (8-1)

√(65) - 8 < √(50) - 7

The square root grows slower when we have greater numbers. Since we know sqrt(64)=8 and sqrt(49)=7, we are looking for the distance between sqrt(49+1) and sqrt(49) and for the distance between sqrt(64+1) and sqrt(64). Because of my first sentence, the first distance must be greater than the second distance.

Here is yet another (albeit less elegant) solution that invokes calculus.

Since $A = \sqrt{8^2 + 1} - 8$ and $B = \sqrt{7^2 + 1} - 7$ , both $A$ and $B$ are of the form $f(n) = \sqrt{n^2 + 1} - n$ . Differentiating $f$ with respect to $n$ , we find $\frac{{\rm d}f}{{\rm d}n} = \frac{n}{\sqrt{n^2 + 1}} - 1$ $\frac{{\rm d}f}{{\rm d}n} = \frac{1}{\sqrt{1 + \frac{1}{n^2}}} - 1.$ Given positive $n$ , $n^2 > 0$ , so the denominator in the first term is always greater than one and $\frac{{\rm d}f}{{\rm d}n} < 0$ . In other words, $f(n)$ takes on smaller and smaller values as $n$ increases. It follows that since $8 > 7$ , $\sqrt{8^2 + 1} - 8 < \sqrt{7^2 + 1} - 7$ .

As we can tell ( 65=64+1, 50=49+1), we can tell easily that 50 is more close to being a perfect square( add 14) than 65( add 16). Hence the fractional part is greater in 50's case. i.e 1/14 > 1/16. hence B is greater than A.

Let's forget about root sign for some time. Then, it is 50 and 65. The numbers are 1 more than the multiples( i.e. 49 and 65) . 7 is smaller so it will have larger deviation for multiple+1(50) than 8, multiple +1(65). So it is like big-small and small- big. Even if root sign is applied it will not make much difference.

I somehow find my solution easier than some solutions mentioned below (I'm sorry if somebody has posted a similar solution. I read top 3 and this was not there) So, as we know

• $2^{2} = 4$
• $3^{2} = 9$
• $4^{2} = 16$

So the difference between squares of 3 and 4 is more than the difference between squares of 2 and 3.

This trend is common for all numbers. We call this exponential rise. So, when comparing $\sqrt{65}$ and $\sqrt{50}$ , we can see that, from the above exponential growth knowledge, $\sqrt{65}$ will be closer to 8 than $\sqrt{50}$ will be from 7 as both are just one ahead of their perfect squares, but as we go higher, lesser is needed for a greater increase. So we will need a number nearer to 8 to get to 65 when compared to getting to 50 from 7. Hence A<B

My idea is to know $1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25$ From those result of square, we get a pattern: $1, 4, 9, 16, 25$ From that pattern, we see the space between numbers is getting bigger:
$2, 4, 6, 8$

What's the space pattern for?

I'll give you an illustration: $\sqrt{1} • • \sqrt{4} • • • • \sqrt{9}••••••\sqrt{16}$ And if I convert itu numbers: $1 (1.3) (1.6) 2 (2.2) (2.4) (2.6) (2.8) 3 (3.1) (3.3) (3.4) (3.6) (3.7) (3.9) 4$

We can see that the dot's gap between $\sqrt{1} to \sqrt{2}$ is wider than $\sqrt{4} to \sqrt{5}$ Ex: $\sqrt{5}-\sqrt{4} ? \sqrt{2}-\sqrt{1}$ $2.2-2 ? 1.3-1$ $0.2 < 0.3$

Based on the pattern in illustration, we can conclude that the delta between square root is decreasing as the number of square root is rising.

So, B is bigger.

The greater the number ,the greater the rate at which it increases when squared. Same case with roots root of lower number can have more decimal value than higher value when the difference is same.

Squaring and adding both the equations we get, A(square) = 14B-16A since A(Square) is positive, 14B>16A or 7B>8A. It is obvious that B should be greater since 8 is grater than 7.

▲y = dy/dx * ▲x Checking 65^1/2 Y = X^1/2

X = 64 ▲X = 1

Dy/dx = 1/ 2x^1/2 = 1/(2 * 64^1/2 ) = 1/16 = 0.06

THEREFORE Y + ▲Y =( X + ▲X)^1/2 65^1/2 = 8 + 1/16 = 8.06

Similarly for 50^1/2 THE VALUE COMES OUT TO BE = 7.07

A = 8.06-8 = .06 B = 7.07-7 = .07

THEREFORE B>A

In case of 65^2, the remainder 1 requires to be divided by 16 whereas incase of 50^2,the remainder to be divided by 14. So the figure beyond decimal in B is greater than A.

The difference between 7 & 8 & 9 is 1 each

However differce between 49(7 squared) and 64(8 squared) is 15 And the differnce between 64 & 81 is 17

That is the difference between 7 squared and 8 squared is less than that between 8 and 9

So √65= √(64+1) can be thought of having 1 unit as 1/17 nd √50=√(49+1) has 1 as 1/15

1/17<1/15 So √65 is less farther from 8 then √50 is from 7

Therefore √(65-1) < √(50-1)

Given to compare: √65 -8, √50-7

Lets multiply, (√65 -8) with √10 and (√50-7) with √13

So we get (√650-8√10) and (√650-7√13)

Above solution can also be written as (√650-√640) and (√650-√637).

Now it’s easier to compare. Hence √65 -8 ˂√50-7.

My solution is more "empirical".

We know that:

• $64 < 65 < 81$
• $49 < 50 < 64$

The differences between the central part and the left part of each inequality are equals to 1: let call that absolute difference $\sigma$ .

But our intervals are not the same:

• $I_A = (64,81)$
• $I_B = (49,64)$

So there are two relative differences:

• $\epsilon_A = \frac { \sigma }{ I_A }$
• $\epsilon_B = \frac { \sigma }{ I_B }$

Obviously $\epsilon_A < \epsilon_B$ , thus $\sqrt{65} - 8 < \sqrt{50} - 7$ .