Rational and Integral Terms

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take root two plus cube root three, the whole thing to the power twelve. Expand it, and you get thirteen terms, full of square roots and cube roots. But some of them come out as plain whole numbers, with no roots left. How many? Will you expand all thirteen to check? No way! Today we count them just by looking at the powers. This comes up often in JEE, so let's begin.

0:31When is a term rational?

So, here is the general case: a to the power one over p, plus b to the power one over q, the whole thing to the power n, where n is a positive integer. The power one over p means the p th root. Okay? Here a and b are coprime positive integers, that is, with no common factor other than one, and neither is a perfect power. Remember this condition; a trap is hiding in it. Right? Now the general term. T r plus one, capital T with a small r plus one below, equals n C r, with the small n at the top left of capital C, times a to the power n minus r, over p, times b to the power r over q. And r runs zero, one, up to n; the three dots mean and so on. With me? Now think. n C r is always a whole number. So when is the whole term rational? Yes, exactly! When both powers are whole numbers. n minus r, over p, is whole when p divides n minus r; the vertical bar means divides. And r over q is whole when q divides r. Since a and b are coprime and not perfect powers, a fractional power on either one leaves the term irrational. Clear? And then, n C r is an integer, and a and b get whole number powers. So such a term is not just rational, it is an integer: an integral term.

1:58Counting the rational terms

Got it? So, how do we count them? Three steps. Step one: write the two conditions on r. q divides r, and p divides n minus r. Okay? Step two: list the values of r that satisfy both, from the set zero, one, up to n, written in curly brackets. Right? And here is the topper's trick. These values of r form an A P, an arithmetic progression, with common difference the LCM, the least common multiple, of p and q. So find the first good r, and keep adding the LCM until you pass n. With me? Now tell me: how many terms does the expansion have in all? n plus one, correct! Because r runs from zero to n. So step three: irrational terms equal n plus one, minus the rational terms. Oh no, careful: the total is not n; the term with r equal to zero counts too. Clear?

2:52Back to our puzzle

Now let's solve our puzzle. Root two plus cube root three, to the power twelve; the cube root has a small three on the root sign. Root two is two to the power one half, and cube root three is three to the power one third. So the general term is twelve C r, times two to the power twelve minus r, over two, times three to the power r over three. Okay? So p is two and q is three. Your turn: which values of r give rational terms? Let's see. Two divides twelve minus r, and three divides r. Twelve is even, so r must be even; even and a multiple of three means a multiple of six. The double arrow means, which gives: r equals zero, six and twelve. They go up by six, the LCM of two and three. Right? So three rational terms. There are thirteen terms in all, so thirteen minus three gives ten irrational terms. Puzzle solved, without expanding anything! Got it?

3:52Trap: simplify perfect-power roots first

Now the trap I promised. Count the integral terms in fourth root of four, plus sixth root of five, to the power one hundred twenty. The small four and six on the root signs mark the fourth and sixth roots. So, is p equal to four here? No, no, wait! Four is a perfect power: it is two squared. So the fourth root of four is just root two, and p is two, not four. Remember the condition: the bases must not be perfect powers, so simplify first. Okay? Now the conditions: one hundred twenty minus r is even, and six divides r. Since six divides r, r is even, so one hundred twenty minus r is even too. So r is zero, six, and so on, up to one hundred twenty. Right? Twenty multiples of six, plus r equal to zero: twenty one integral terms. Take p as four, and you would miss many of them. Got it?

4:50Example 1: terms of (x^(1/5) + y^(1/10))^55 free of radicals

Alright, example one. How many terms of x to the power one fifth, plus y to the power one tenth, the whole thing to the power fifty five, are free from radical signs? Free from radical signs means no root is left in the term. Pause the video and try it. Done? Okay, let's check. General term: fifty five C r, times x to the power fifty five minus r, over five, times y to the power r over ten. With me? So we need ten divides r, and five divides fifty five minus r. Now think: if ten divides r, is the second condition automatic? Yes, exactly! If ten divides r, then five divides r, and five divides fifty five, so five divides fifty five minus r. So r is any multiple of ten up to fifty five: zero, ten, twenty, up to fifty. Okay? Count them: zero, ten, twenty, thirty, forty, fifty. Six terms, with r equal to zero counted too. Clear?

5:57Example 2: irrational terms in (5^(1/6) + 2^(1/8))^100

Now a harder one. How many terms of five to the power one sixth, plus two to the power one eighth, the whole thing to the power one hundred, are irrational? Pause the video and try it. Hint: count the rational terms first, then subtract. Done? Okay, let's check. General term: one hundred C r, times five to the power one hundred minus r, over six, times two to the power r over eight. Clear? Now, quick. Eight divides r, and six divides r, so r must be a multiple of twenty four, right? No, no, careful! The six sits under one hundred minus r, not under r. So we need eight divides r, and six divides one hundred minus r. With me? One hundred is ninety six plus four, and ninety six is a multiple of six. So r must leave remainder four on division by six. The sign with three lines means congruent: r is congruent to four, mod six, in round brackets. Okay? Now, among the multiples of eight, zero, eight, sixteen, up to ninety six, keep those that leave remainder four: sixteen, forty, sixty four, eighty eight. They go up by twenty four, the LCM of six and eight. The topper's trick again! Got it? So only four rational terms, out of one hundred plus one, that is one hundred and one terms. Clear? Irrational terms: one hundred and one minus four, ninety seven. Always pair each root with its own power. Right?

7:27Quick revision: divisibility test, lcm gap, irrational count

So, quick revision. One: T r plus one is rational when p divides n minus r, and q divides r. Two: the good values of r form an A P, with common difference the LCM of p and q. Three: irrational terms equal n plus one, minus the rational terms. Four: simplify roots of perfect powers first. Now go and try the practice problems on this page. All the best!