Why Isn't It Reducible?

Let fractions in A be denoted as $a_1, a_2, a_3, ... , a_{1007}$

Let fractions in B be denoted as $b_1, b_2, b_3, ... , b_{1006}$

But $a_1 < b_1, a_2 < b_2, a_3 < b_3 .... a_{1006} < b_{1006}, a_{1007} < 1$

Multiplying all these, $A < B$

Observe that $A \times B = \frac{1}{2014} = C^2$

So $A, C, B$ are positive numbers forming increasing G.P. (Geometric Progression)

Hence option $\boxed {2: A < C < B}$

Multiply $A$ and $B$ to get:

$\displaystyle AB=\frac{1}{2014}\,\,\, (*)$ .

Now divide $A$ and $B$ to get:

$\displaystyle \frac{A}{B}=\frac{3}{4}\times \frac{15}{16} \times \frac{35}{36} \cdots \frac{2013^2}{2012\times 2014}$

It is easy to see that the right hand side is less than 1, hence $A<B\,\,\, (**)$ .

Lets get back to $(*)$ , it is equal to $C^2$ . This shows that $A,C$ and $B$ are in geometric progression so the possible orders are: $A<C<B$ and $B<C<A$ but from $(**)$ , the only possible order is $\fbox{A<C<B}$ .

Looks like I'm the only idiot who used Stirling's approximation to solve this one (sigh).

A= $\frac{2n!!}{(2n-1)!!)}=\frac 1 {\sqrt{\pi n}}$ (simplified using Stirling's approximation) and B= $\frac{1}{2n\times A}$ , where n = $\frac {2014} 2=1007.$

Substitute n=1007 and you get:

A = 0.017779...

B = 0.27927...

C = 0.2228...

Clearly, B>C>A.

I guess as soon as I saw double factorials and a square root in the question, I jumped to Stirling's approximation. Doh!

$\frac{1}{2}$ < $\frac{2}{3}$ , $\frac{3}{4}$ < $\frac{4}{5}$ ......
A has 1007 such terms while B has 1006 such terms.
thus B>A.
$\sqrt{AB}$ =C.
by AM GM inequality A+B>2C.
Thus B>C.
if A>C then then AB= $C^{2}$ is not possible.
thus A<C<B

So, all of us have solved this amazing problem with our precious Number Theory knowledge.

But a great alternative came across my neurons. Its Programming , which would take around 5 milli seconds to calculate these big calculations!

So, here's my different C programmed Solution .

Source Code: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

$\#include<stdio.h>$

$\#include<math.h>$

int main()

{

double n=2,A=1;

printf("I am not a brilliant Mathematician.\nBut I have my programming knowledge and my great computer!\nSo it makes everything easy!");

while(n<2015){

A=A*((n-1)/n);

n=n+2;

}

printf("\n\nA=%.100lf",A);

double N=3,B=1;

while(N<2014){

B=B*((n-1)/n);

N=N+2;

}

printf("\n\nB=%.100lf",B);

printf("\n\nC=%.100lf\n\nSo, obviously\n1. B ia greater than C,\n2. C is greater than A.\n\n To express mathematically, A<C<B!!!",(1/sqrt(2014)));

return 0;

}

*Output: *

There's always thousands of ways to solve any mathematical Problem, right!!!

First note that: $A \times B = \frac{1}{2014}$

Since A and B are rational numbers, neither A nor B can equal C. Also, since A times B equals C times C, we know that C is between A and B.

We can write A and B as: $A = (\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2011}{2012}) \times \frac{2013}{2014}$ $B = (\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times ... \times \frac{2012}{2013})$

Note that for each fraction that is a factor of A, the corresponding fraction that is a factor of B is larger. Finally, the extra term that is a factor of A, (2013/2014), is less than 1. Therefore, A is less than B, so B > C > A .

notist that A=2013!!/2014!! multiply it by 2014!!/2014!! A=2014!/(2014!!)²=2014!/(1007! 1007! 2^2014)=(2014C1007)/2^2014 Same B=2012!!/2013!! 2012!!/2012!!= 2^2012 1006! 1006!/(2013 2012!) B=2^2012/(2013 (2012C1006)) so to compare A r B ,r is relation < or > (2014C1007)/2^2014 r 2^2012/(2013 (2012C1006)) i.e. 2014C1007 * 2013C1006 2013 r 2^2014 2^2012 as we know 2^2014=sum(2014Ck,k goes from 0 to 2014) so 2^2014 is at least 50 times bigger than 2014C1007,also 2^2012 is at least 50 times bigger than 2012C1006,so because 50 50/2013>1 that means that r is relation <,so A<B Similiar 1/√2014 < B because √2014 *2^2012 > 2013 2012C1006 because 50*√2014>2013 so C<B, also C>A because 2^2014> √2014 *2014C1007 beause 50>√2014 so A<C<B is correct

First we try $A\times B$ we get $\frac{1}{2} \times \frac{2}{3} \times ... \times \frac{2012}{2013} \times \frac{2013}{2014}$ .

$A\times B = \frac{1}{2014}$

If $A = B$ , then $A = B = \frac{1}{\sqrt{2014}} = C$

But $A \neq B$ , then $A > C > B$ or $A < C <B$ (2 choices left, yay!)

Consider $A$ and $B$ ;

There're 1007 terms in $A$ but there're only 1006 terms in $B$ . (Don't think that $A > B$ immediately.)

The problem of $A$ is... $A$ has $\frac{1}{2}$ , which reduces a lot. (compare $\frac{3}{4}$ of $A$ to $\frac{2}{3}$ of $B$ , and so on... that doesn't change that much.)

Therefore, $B > C > A$ (or $A < C < B$ )... $\boxed{2}$ ~~~

Note: $A \neq B$ because the numerator and denominator of $A$ and $B$ are not the same.

Simple use mathematical induction for n=1 1 We get the answer in one step

first , if a and b are positive and a<b then $\frac{a}{b}$ < $\frac{a+1}{b+1}$ use the property to show than $\frac{1}{2}$ < $\frac{2}{3}$ , $\frac{3}{4}$ < $\frac{4}{5}$ .... $\frac{2011}{2013}$ < $\frac{2012}{2014}$ then by multiplying and considering that $\frac{2013}{2014}$ < 1 we can find that A< B from AB=C we conclude that A<B<C.

note A*B=C^2

so A,B,C are in GM

so try 2,4,8

0.0177769159448591 0.0279308475796525 0.0222828258910793

A < C < B

1. A x B = 1/2014 = C²
2. so, A/C = C/B
3. Hence either A<C<B or A>C>B
4. Now, A = 0.5 x 0.75 x ......... 2007 numbers fraction and
5. B = 0.66..7 x 0.80 x ...... 2006 numbers fraction.
6. now, 0.5<0.6666...; 0.75<0.80 ...... up to 2006 numbers of terms.
7. hence if we omit 2013/2014 i.e. the last fraction of A and say it is K, till then the product of fractions in K<B.
8. also (2013/2014) <1, hence (2013/2014) x K is still less than B.
9. Hence A<B
10. so, A<C<B.

I wrote a mathlab script :-)

format long;

clear all;

A = 1;

B = 1;

C = 1/sqrt(2014);

for i = 1 : 1006

A = A*(2*i-1)/(2*i);

B = B*(2*i)/(2*i+1);

end

A = A*2013/2014;

A

B

C

Multiplying A and B gives us 1/2014, which is twice that of C. Therefore, if A and B were to be equal, then A, B, and C would all be the same.

However, if A or B was to be bigger than the other, we now have a situation where A<C<B or B<C<A. At this point, the answer is now narrowed down to 2, 4 or 8.

When looking at A and B, one may realize that every nth term in B is greater than the nth term of A, such as 1/2 < 2/3 for the first term, 3/4 < 4/5 and so on. Therefore, we can conclude B>A.

Using this information, the final answer seems to be B>C>A, or number 2. Which is right, by the way.

A clever problem, indeed.

Right away, you can eliminate choice 8, because A and B are rational, while C is not.

Generalize it. Let n = number of terms in A. Then there are n-1 terms in B. And by looking at the cases, n=1, 2, 3, you can come to some quick conclusions (keep in mind that a vacuous product is =1). In the example shown, n=1007. Now if you "cycle" each of the numerators in A one place to the left, A becomes identical to 1/B, but with an extra factor of 1/2n = 1/2014 at the end. So:

2n·A·B = 2014·A·B = 1

C, however, is just 1/√(2n) = 1/√2014. Now you can also see that each term in A is < each term in B, with that extra 2013/2014 at the end of A. And since that extra term is < 1, it's clear that A < B (scratch 3, 4, 6, & 7).

But C² = 1/(2n) = 1/2014 = A·B. So C is the mean proportional of A & B, and that means:

A < C < B (choice #2)

1st step multiply A and B which gives AB=1/2014 In the 2nd step Square C which gives C^2= 1/2014 So, C^2= AB, that implies C is geometric mean of A and B....And also we know from 1st step A<B and as c is G.M, it follows the order A < B <C

just try it with first 2 terms of each A,B,C and u will get the answe.r

simple check for A= (1/2)*(3/4) b=2/3 c=1/square root of 4

this sequence is analogous to given compare values A=0.735 b=0.666 c=0.5 A<c<b

First thing I observed is

A*B = C^2

or, C is the geometric mean of A & B, it should lie between A and B

(NOTE: A,C,B being equal doesn't make sense as A and B are rational, C is irrational)

We have to figure out whether A<B or B<A

Now if we compare each terms of A & B:

1/2 is less than 2/3 .... 3/4 is less than 4/5 .... ... and so on ... 2011/2012 is less than 2012/2013

so: (1/2) (3/4) ...(2011/2012) < (2/3) (4/5) ...(2012/2013)

Now, 'A' is the LHS of the above inequality multiplied to (2013/2014) ... as (2013/2014) is a fraction, it reduces 'A' further.

So, A<B

and as, 'C' is lying between A &B

A<C<B

if u solve A & B, we arrive at B^2 = A*C therefore, A , B and C are in G.P. so, A<C<B

B > A is easy to derive by regular way of comparing fractions. By minute inspection of numerators and denominators, B > C and C > A can be derived. Hence the result.

A= product of (2n-1)/(2n) from 1 to 1006 * 2013/2014

B= product of (2n)/(2n+1) from 1 to 1006 * 1

comparing each term, we get A<B

A*B=1/2014

C is the geometric mean of A and B and must be in between

therefore A<C<B

Because every term in $A$ is less than a corresponding term in $B$ , we have that $A<B$ . Also, we have that $AB=\frac{1}{2014}=C^2$ , so $C$ is the geometric mean of $A$ and $B$ , so $A<C<B$ . Hence, the answer is $2$ .

Clearly, A<B because 1/2 < 2/3 , 3/4 < 4/5 , and so on. Also, C is the geometric mean of A and B. So, A<C<B

We easily can see that AB = $C^{2}$

Also, A < B. So, A, C, B form an increasing G.P. Hence A < C < B.

We observe that A/B is almost 1/2014 < 1 so A<B.

Again, AB = C^2 so this implies C<B.

The hard part is to prove that A<C. For this I took the ratio A/C and as I calculated 1/2 x 3/4 x 5/6 x ... x 2013 / sqrt(2014) progressively I observed the product is decreasing (as it should be) and ultimately it falls off by 1/100 which on being multiplied with a two digit integer will definitely be < 1

For this question, there are some pointers that we can work with :

As each numerator of the fractions in $A$ is present in the denominator of the fractions in $B$ , except 2014. Hence, after cancellations, $AB = \frac{1}{2014}$ .

In addition, $C^2 = \frac{1}{2014}$ too.

Hence, $AB = C^2$ ---- Equation (1)

Now, consider each terms in $A$ and $B$ : $\frac{2}{3}$ in $B$ $> \frac{1}{2}$ in $A$ , $\frac{4}{5}$ in $B$ $> \frac{3}{4}$ in $A$ , ... until $\frac{2012}{2013}$ in $B$ $> \frac{2011}{2012}$ in $A$ . Note that the term $\frac{2013}{2014}$ in $A$ is not compared. With the following information :

As every term in $B$ > its corresponding term in $A$ except $\frac{2013}{2014}$ .

Hence, $B$ > $A'$ , where $A'$ denotes all the terms except $\frac{2013}{2014}$ .

Now as $\frac{2013}{2014}A' = A$ => $A' > A$ . Hence , as $B$ > $A'$ > $A$ , $\boxed{B > A}$

In addition, referring back to equation (1) , $AB = C^2$ , as the product of $A$ and $B$ yields the product of $C$ and itself, hence, logically, $C$ must lie between $A$ and $B$ for that to occur. Thus, as $B$ > $A$ , hence, $\boxed{B > C > A}$

A=1/2 3/4 ...2013/2014=3/2 5/4 ...2013/2012 1/2014=1/B 1/2014 Then AB=1/2014=C^2 It implies A<C<B or B<C<A B>1 then 1/B<1 also A=1/B*2014<1/2014<1 Hence A<C<B

It is a gp