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Hey Akshay try this for a change The positive rational numbers may be arranged in the form of a simple series as follows ( 1 / 1 ), ( 2 / 1 ), ( 1 / 2 ) , ( 3 / 1 ), ( 2 / 2 ), ( 1 / 3 ) , ( 4 / 1 ) , ( 3 / 2 ), ( 2 / 3), ( 1 / 4 ), ... Show that ( p / q ) is the [ ( 1 / 2 ) ( p + q - 1 ) ( p + q - 2 ) + q ] th term of the series. I see you like solving problems on mathematics so give it an honest try on your own Also to those boasting about years of experience in mathematics, please stop boasting about yourself before the rest of the teachers. If you are good, you are good and I do not doubt that. We all are here to help and not boast and this space is for questions only and not boasting. Hope you all keep this in mind.

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Dear G, I agree with the partitioning concept and what you have said is correct. Also, there is yet another way, as is often the case with mathematics, which is rather elementary, based on counting the number of terms. Anyways, Mr. Akshay was very interested in sequences and series and therefore I...
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Dear G, I agree with the partitioning concept and what you have said is correct. Also, there is yet another way, as is often the case with mathematics, which is rather elementary, based on counting the number of terms. Anyways, Mr. Akshay was very interested in sequences and series and therefore I put a real problem for him to work over for a while. @Sunil, I picked up this problem from "A course of pure mathematics" by G H Hardy, the very first page of chapter 1. Thanks G, I do agree with your solution. Now we are talking real mathematics :-) Cheers, read less
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As is visible, this relates to the partitioning of numbers without repetition. Let p+q = t Let us see the partitioning of 't' for 1st term 't' = 2 = 1+1 2nd term 3= 2+1 = 1+2 And so on. As one can see each 't' can be partitioned in 't-1' ways. So, till the end of group where sum...
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As is visible, this relates to the partitioning of numbers without repetition. Let p+q = t Let us see the partitioning of 't' for 1st term 't' = 2 = 1+1 2nd term 3= 2+1 = 1+2 And so on. As one can see each 't' can be partitioned in 't-1' ways. So, till the end of group where sum of numerator and denominator = 't-1' there would be total of 1/2 * (t-2)(t-1) terms. Further the 1st term of every t = p+q starts with q = 1 Thus, the (p/q) = would be at [1/2 * (t-2)(t-1) + q] position. where t = p+q Note:-This is the way in which Cantor proved that rational numbers are countably infinite. read less
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The question does not seem to be well framed. Rational numbers include (for example) 1.35,2.398,7.5 etc. Since it appears you(Santosh) don't understand what it means by irrational number, I tell you a few examples: square root of the number 2, that of 3, that of 5, that of 6 etc. They are irrational...
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The question does not seem to be well framed. Rational numbers include (for example) 1.35,2.398,7.5 etc. Since it appears you(Santosh) don't understand what it means by irrational number, I tell you a few examples: square root of the number 2, that of 3, that of 5, that of 6 etc. They are irrational because, unlike square roots of 4 (which is 2.0) and 1.21 (which is 1.10), square roots of 2,3,5 etc.(theoretically too) don't have any end to the digits coming after decimal points. Cube root of 2,3 and 4 etc also are irrational (whereas, again cube root of 8 is 2.0 which is not irrational) numbers. Also, value of pi [The ratio (Circumference)/(Diameter)] of a circle (one way of expressing it is (4 times as arctan(1)) is an irrational number (hope you know 22/7 is only an approximation discovered by Bhaskaracharya, which is valid up to 2 digits after decimal point only); besides, Ramanujam had another way of expressing it (much more complicated expression, but understandable)]. Since you said rational number, I defeat you by setting p=1.4 and q=3.79 and saying p/q corresponding to the mentioned values of p and q is not there in the series provided by you. Evidently, the series presented has to be elements of set of whole numbers and q starting from natural numbers. Is that what you wanted to say? Even if that is the case, I also can provide any number of series of numbers, the rank of some p/q can have more complicated expression than what you provided. Any way, I am ready to attend your challenge if you provide me a qualifying (to answer) question. Now, since (as of now) we are dealing with a subject as precise as mathematics, I will like to ask you where is the proof that teachers are boasting. Is mentioning the experience a boasting? Quite a few among us are professionals valuing the time we spent (if time is money, then if I stop spending time with silly guys like you, I would have earned 3 lakh rupees per month by teaching in an IIT entrance coaching center who invited me). So don't waste our time with the silly stuffs you blabber. Besides, all great scientists and mathematicians (who launched Mangalyan and lock the space craft with an orbit around Mars (which requires a precision of sending a golf ball from Mumbai and putting it in a pit in Srinagar) and do such works as laudable throughout the world) rather also surprise me with their humility. Also, I have never seen someone with your trivialities and provoking attitude ever coming up to become anything. Before coming to us (who are Pofessional teachers) with questions, please go to a Professional psychologist for a counselling. If you don't do that in future, I am afraid I may not waste my time with you any more. All the best for you. read less
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Chemistry Wizard/English Expert

Yes very true
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Physics & Mathematics Expert for JEE Mains,Advanced & Olympiads with 13 yrs experience

True, idea from numberphile
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Spoken English, BTech Tuition, C Language, C++ Language, CAD etc., with 4 years of experience

So true
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