Integer Parts and Fractional Parts

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Root two plus one is about two point four one. Raise it to the power six. What is its integer part, the whole number before the decimal point? No calculator! One hundred and ninety seven? Or one hundred and ninety eight? Too close to guess! Today, one clever partner number settles it exactly. This comes up often in JEE, so let's begin.

0:28Integer part plus fractional part

So, take a plus root b, to the power n; the tick-shaped sign is the square root sign. We split it into capital I, the integer part, plus small f, the fractional part. The less than sign with a line below means less than or equal to, so f is at least zero, and less than one. Okay? The conditions: n, a and b are positive integers, and root b is less than one away from a. So a minus root b, or root b minus a, lies between zero and one. Right? Now the trick. We give our number a partner, its conjugate. f prime, f with a small dash, is the modulus of a minus root b, to the power n. The vertical bars mean modulus, the size without the sign. With me? Why must f prime lie between zero and one? Yes, exactly! The base is between zero and one, and its powers only get smaller, but never reach zero. So f prime is a small positive number. Clear?

1:30Case 1: the sum is an integer, so f+f'=1

Case one: a is bigger than root b, so f prime is just a minus root b, to the power n. Remember Chapter one? In the conjugate sum, the odd powers of root b cancel. So the sum, I plus f plus f prime, is an integer. Got it? So f plus f prime is an integer too. But f is less than one, and f prime is between zero and one. So f plus f prime lies strictly between zero and two. Which integer can it be? One, exactly! So f plus f prime equals one, and the integer part is simply the sum, minus one. Okay? Now, our puzzle! Pair root two plus one with root two minus one. In Chapter one, the sum of their sixth powers came to one hundred and ninety eight. And root two minus one is less than one, so its sixth power is tiny, and positive. Right? So the integer part is one ninety eight minus one: one hundred and ninety seven. Not one ninety eight! The tiny partner was hiding inside the sum. Very good! One more result. Since one minus f is f prime, I plus f, times one minus f, is the number times its partner. Multiply the pair first: a squared minus b, to the power n. With me?

2:53Case 2: the difference is an integer, so f=f'

Now case two. Sometimes the number minus its partner is the integer. Then I plus f, minus f prime, is an integer. We will meet such a number in Example two. Okay? So f minus f prime is an integer. Both are at least zero and less than one, so the difference lies strictly between minus one and one. Which integer is it? Zero! So f equals f prime: the fractional part is the partner itself. Do not mix up the two cases! Clear?

3:26Fractional part of a^N/m: find the remainder

Now, a new kind of question: a to the power capital N, over m. Write a to the power N as m q plus s: quotient q, and remainder s, at least zero and less than m. Since s over m is less than one, the fractional part, shown by these curly brackets, is s over m. The double arrow means, so it follows. Got it? From the course: the fractional part of two to the power four hundred and three, over fifteen. Two to the power four is fifteen plus one. So two to the power four hundred is fifteen plus one, to the power one hundred: fifteen k plus one, since every term except the last has a factor fifteen. Okay? Multiply by eight; the dot means times. Two to the power four hundred and three is fifteen times eight k, plus eight. Remainder eight, so the fractional part is eight over fifteen. Perfect!

4:24Example 1: integer part of (2+√3)^5

Alright, Example one. Find the integer part of two plus root three, to the power five. Pause the video and try it. Done? Okay, let's check. Root three lies between one and two, so two minus root three is between zero and one, and its fifth power is our small f prime. With me? Add the pair: odd powers of root three cancel, even ones double. Two times the bracket: two to the power five, plus five C two, times two cubed, times root three squared, plus five C four, times two, times root three to the power four. The C with a small five at its top left is the binomial coefficient. Clear? Five C two is ten, five C four is five. Ten times eight times three is two hundred and forty, and five times two times nine is ninety. With thirty two, the bracket is three hundred and sixty two. The cross means times: two times that is seven hundred and twenty four. Okay? So the integer part is seven hundred and twenty four, right? No, no, careful! Seven twenty four is the number plus its partner, I plus f plus f prime. And f plus f prime is one. So the integer part is seven hundred and twenty three. The same trap as our puzzle. Very good!

5:48Example 2: prove Rf = 20^(2n+1) for R = (6√6+14)^(2n+1)

Now Example two, a harder classic. Capital R is six root six plus fourteen, to the power two n plus one, and f is R minus the integer part of R, shown by the square brackets. Prove that R times f is twenty to the power two n plus one. Pause the video and try it. Hint: sum or difference? Done? Okay, let's check. The partner f prime is six root six minus fourteen, to the power two n plus one. Six root six is root two hundred and sixteen; the wavy equals sign means approximately: about fourteen point seven. So f prime is between zero and one. Right? Now, which one is an integer here: R plus f prime, or R minus f prime? The minus sits on the fourteen, so when we subtract, even powers of fourteen cancel, and odd powers double. R minus f prime is two times the sum, over odd r, of two n plus one C r, times six root six to the power two n plus one minus r, times fourteen to the power r. The big Greek letter sigma means add up. Got it? Two n plus one is odd, and r is odd, so the power of six root six is even. And six root six, squared, is two hundred and sixteen. So no root survives! Okay? So R minus f prime, the integer part plus f minus f prime, is an integer. Case two! So f equals f prime. With me? Now finish. R times f is R times f prime: the pair multiplied, all to the power two n plus one. Two hundred and sixteen minus one hundred and ninety six is twenty. So R times f is twenty to the power two n plus one. That's it, proved!

7:40Quick revision: f + f' = 1, f = f', fractional part of a^N/m

So, quick revision. One: write the number as I plus f, and pair it with its small partner, f prime. Two: number plus partner an integer, f plus f prime is one; number minus partner an integer, f equals f prime. Three: the fractional part of a to the power N over m is the remainder over m. Now try the practice problems on this page. All the best!