A Farey sequence of order n is the sequence of fractions

Thanks Bunnel, I will message the same to Jeff Sackmann also so that he can deleat this question from their 100 challenging questions on Algebra.
Regards.

GMAT doesn't expect you to know what a Farey sequence is but it could describe the characteristics of the sequence for you and then ask you a question on it as is done in this question.
Say, a +-1 sequence is a sequence where each term is one of +1 or -1. What is the sum of all the terms of this sequence?... and then two statements or something similar.
So I would say that make sure you understand the logic tested in this question.

Very true. Though I very much doubt that the GMAT will throw you a question on such a topic as a Farey sequence, which is a bit out of the GMAT scope.

Thank you Bunnel and Karishma...Great insight by you both...indeed valuable.

Hi

Can anyone tell me how the individual terms of a farey sequence are arrived at? Do we have to memorize the terms or is there an actual way to get the terms.

I know this may be out of scope for GMAT, but there is no harm in understanding this.

Thank you

They have given you the rules:
Terms must be between 0 and 1
For order n, the denominator must be less than or equal to n.
The terms should be arranged in increasing order.

So say we need the Farey sequence of order 2.
Terms will be: {0, 1/2, 1}

For order 4, terms will be: 0, 1/2, 1, 1/3, 2/3, 1/4, 3/4
Arrange them in increasing order: {0, 1/4, 1/3, 1/2, 2/3, 3/4, 1}

Basically, 0 and 1 will always be there (the example shows us that).
Then find the terms you will have with each denominator (2, 3, 4 in this case).
The terms where the numerator will be less than the denominator will be included. The fractions need to be in the lowest terms. If a fraction is repeated, we put it in only once (we obtained 2/4 above but omitted it since we already have 1/2)

With denominator 2, we have only 1/2
With denominator 3, we have 1/3 and 2/3
With denominator 4, we have 1/4, 2/4 (already there), 3/4

Similarly, you can find the sequence of order 5. Note that there will be a unique sequence for every order.
Also, you don't need to memorize it since if GMAT does have a question related to any specific sequence, the question will explicitly define the sequence and explain how to arrive at the terms (as done here).

Thank you, here is proof that I have understood it

Are the terms of the Farey sequence of order 6 as below,( not arranged in increasing order ).

{ 0,1/6,1/5,1/4,1/3,1/2,2/3,3/4,2/5,3/5,4/5,5/6,1}

Well if at least the terms are correct then I think I have got it, thanks to you

but is there an easier way to arrange them in increasing order? All I know of is cross multiplying the fractions and then comparing both , e.g between 2/3 and 3/4 , I cross multiply( denominator * numerator) and get 8 and 9 so I know 3/4 >2/3 , but to use this method with 13 terms seems tedious.
So:
a) Have I at least got the terms of the farey sequence of order 6 correct?
b) Please share any miraculous way you know to arrange them in increasing order ?

Thank you +1

Yes, the terms are correct.
I don't really have a miraculous method to arrange them though some logic goes a long way in reducing our work.
Two things help us here:
1. Many fractions already have the same numerator or denominator. We can compare them by just looking at them.
Same numerator - Larger the denominator, smaller the fraction.
Same denominator - Larger the numerator, larger the fraction.
2. The fractions are simple so we already know their percent equivalents. 2/3 is 66.7%; 3/5 = 60%, 3/4 is 75%, 5/6 is 83%; 2/5 is 40%; 1/3 is 33% etc

First split them into two groups - less than 1/2 and greater than 1/2
Now just arrange them according to the discussion above.
{ 0,1/6,1/5,1/4,1/3, 2/5,1/2,3/5,2/3,3/4,4/5,5/6,1}

Thank you, I think you may have unknowingly lit a bulb in me.
the idea of converting fractions to percents in order to compare them never dawned on me.
I always used to cross multiply or convert to common denominator to compare, now I can convert to percent too!

I think you have given the miraculous way I was searching for.

also the same denominator and same numerator hint has really helped.

below is proof of my understanding,it is only because of percent conversion as shown by you, I could attempt something as herculean as this , earlier I couldn't even have imagined attempting something as tedious as this.Thank you.

Farey sequence of order 7 in increasing order.
{0, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1}

Hope these are correct, full credit to you, thank you a ton, +1

(2) The second element of sequence S is 1/5 --> there is a Farey sequence with 1/5 as the second element in it as well as there are infinitely many other type of sequences also with 1/5 as the second element in it. Not sufficient.. I think there can only be one such sequence in which 1/5 is the second number coz the first digit will be 0 and if 1/5 is the second number then it is sure that S is not greater than 5. Please clarify, I didn't understand this.

Important:

1. "Fractions between 0 and 1" must (and will) be considered as "fractions between 0 and 1, both included" so that the example given satisfies the definition presented.
2. The { } notation will always denote here a finite sequence. (In other words, the order of the elements presented is relevant.)

\(S\,\,\mathop = \limits^? \,\,{\rm{Farey}}\)

\(\left( 1 \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,S = \left\{ {0;{1 \over 3};{1 \over 2};{2 \over 3};1} \right\}\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\left( {{\rm{given}}} \right)\,\, \hfill \cr \\
\,{\rm{Take}}\,\,S = \left\{ {0;1;{1 \over 3};{1 \over 2};{2 \over 3}} \right\}\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\left( {{\rm{wrong}}\,\,{\rm{order}}} \right)\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,S = \left\{ {0;{1 \over 5};{2 \over 5};{3 \over 5};{4 \over 5};1;{1 \over 4};{2 \over 4} = {1 \over 2};{3 \over 4};{1 \over 3};{2 \over 3}} \right\}\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\left( {{\rm{wrong}}\,\,{\rm{order, }}\,{\rm{although}}\,\,{1 \over 5}\,\,{\rm{IS}}\,\,{\rm{the}}\,{\rm{ second}}\,\,{\rm{here!}}} \right)\,\, \hfill \cr \\
\,{\rm{Take}}\,\,S = \left\{ {0;{1 \over 5};{2 \over 5};{3 \over 5};{4 \over 5};1;{1 \over 4};{2 \over 4} = {1 \over 2};{3 \over 4};{1 \over 3};{2 \over 3}} \right\}\,\,{\rm{but}}\,\,{\rm{rewritten}}\,\,\,{\rm{in}}\,\,{\rm{increasing}}\,\,\,{\rm{order}}\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\left( {{1 \over 5}\,\,{\rm{WILL}}\,\,{\rm{BE}}\,\,{\rm{the}}\,\,{\rm{second!}}} \right)\,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\left( {{1 \over 5} \in \,\,S\,\,{\rm{Farey}}\,\,\,\, \Rightarrow \,\,\,{\rm{S}}\,\,{\rm{has}}\,\,{\rm{more}}\,\,{\rm{than}}\,\,{\rm{10}}\,\,{\rm{elements}}!} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hello

In the second statement, we are given that 1/5 is the second term of sequence S. But we cannot already assume that S is a Farey sequence. There can be many sequences in this world where 1/5 is the second term of that sequence without it being a Farey sequence.

Eg, there can be an AP.. (1/10, 1/5, 3/10, 2/5... this has a common difference of 1/10)
or there can be a GP.. (1, 1/5, 1/25, 1/125,... this has a common ratio of 1/5)

Thats why there can be infinite sequences with second term as 1/5, we cannot say on the basis of that whether S is a Farey sequence or not.

Are we really supposed to list out all the elements in our heads? I mean wouldn't that be much of an overhead and time-consuming?

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