Consider the process of long division. Does this algorithm give any in- sight as to why rational

Consider the process of long division. Does this algorithm...

Key Concepts

Remainder Cycles

During long division, once a remainder repeats, it indicates that the sequence of digits in the quotient will also repeat. Since there are only a limited number of possible remainders (fewer than the divisor), the repetition of a remainder is inevitable unless the remainder becomes zero, culminating in a repeating decimal expansion.

Long Division Algorithm

The long division algorithm is a step-by-step method for dividing two numbers, where at each stage a new digit of the quotient is determined by dividing a partial dividend by the divisor. This process involves repeatedly multiplying and subtracting, which leads to a sequence of remainders that helps reveal the structure of the decimal expansion for fractions.

Rational Numbers

Rational numbers, defined as the quotient of two integers, inherently possess decimal expansions that either terminate or repeat. This outcome is directly linked to the finite number of possible remainders produced during the division process, reflecting a fundamental property of numbers expressed in this form.

Terminating and Repeating Decimals

A decimal expansion terminates if the remainder eventually becomes zero during the long division process, indicating that no further digits will be produced. Conversely, if a remainder recurs before reaching zero, the division enters a cycle, thereby generating a repeating pattern in the decimal expansion.

Transcript

00:01 Problem we have been given that without actually performing the long division state whether the following rational numbers will have a terminating decimal expansion or a non -terminating repeating decimal expansion.

00:08 All right now we need to know that is terminating so just a minute so it is terminating if p and q are co -prime also q is off from 2 raise to the power of n into 5 to the power of m so where m and n are integers so these are the two conditions now let us begin with the first question in the first part we have been given to find that for 13 by 3 125 now checking the condition so i just write this as condition number 1 and the second condition where q is of palm 2 x x to m as condition 2.

01:29 So now for condition number 1 and for condition number 2.

01:35 So condition number 1 is satisfied because 13 and 312 are co -prime.

01:40 And condition number 2 is also true because 8 is equal to a big a pardon, not 8, just a minute.

01:54 So, 3125 is equal to 0 x0 into 5 x x x to 5.

02:06 So therefore it is also the second condition is also satisfied.

02:10 So there is terminating.

02:22 Similarly, we can see that for the other questions as well.

02:29 In the second part, for 17 divided by 8, so we need to check it for the first condition and for the second condition the first condition is satisfied because 17 and 8 are co -prime and in the second condition 8 is equal to 2 cube and 2 2 to 5 to the power of 0 so this is also satisfied so therefore it is terminating now for the third question i will just pick up the speed so that because all of them have to be just we just need to check both the conditions so therefore in the third question condition in the third question 64 divided by 455 checking condition number one and condition number two condition number one yes 64 and 455 are co -prime but 455 is equal to 5 into 7 into 13 it is not of the just a minute and write it clearly into 13 and it is not of the form 2 rest 2 n into 5 yes 2m so therefore it is not and repeating and similarly for other questions for the fourth question 15 divided by 1600 which can be written as equal to 3 divided by 320 checking condition number 1 and condition number 2 3 and 320 are co -prime and 320 is equal to in the second condition therefore it is terminating therefore it is also terminating now again for the fifth question where we have been asked that for 29 divided by 343.

04:36 So checking condition number 1 and condition number 2.

04:40 Condition number 1 is satisfied because 29 and 343 are co -prime but 343 in the second condition 343 is equal to 7 to the power of 3.

04:49 So it is not of the form 2 race to and into 5 race 2 m.

04:51 So therefore it is not satisfied.

04:54 So therefore it is not terminating and this is not terminating and repeating.

04:59 So non -terminating.

05:02 And for the next question, sixth part, we have been asked that for 23 divided by 2 cube into 5 square.

05:14 So checking condition number 1 and condition number 2, we can clearly see that condition 2 is correct because it is of the form 2 race 2 n into 5 raise to m...

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