Relatively prime integers are:
(a). 6, 9
(b). 4, 9
(c). 7, 28
(d). 8, 12
Answer
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Hint: Here we may use the concept that relative prime numbers are those numbers whose greatest common divisor is 1. So, here we have to check all the options whether they have a common divisor as 1 or not by using Euclid’s division algorithm.
Complete step-by-step answer:
So, let us check all the options one by one whether the numbers are relatively prime not.
So, now we may use the Euclid’s division Lemma to find the greatest common divisor of the two numbers given in each option.
Now, we should know that in Euclid’s division algorithm we apply the Euclid’s division Lemma in succession several times until we get a remainder zero. The last non-zero remainder that is obtained in this process is termed as the great common divisor of the two numbers.
So, first we should know that the according to Euclid’s division Lemma:
If we have two integers ‘a’ and ‘b’ then for these numbers there exist ‘q’ and ‘r’ such that: -
$a=bq+r$ , where q is the quotient and r is the remainder.
So, on applying this algorithm on option (a):
Using Euclid’s division Lemma we can write:
$9=6\times 1+3$
$6=3\times 2+0$
So, the last non-zero remainder obtained here is=3. So, the greatest common divisor of 9 and 6 is 3.
Hence, 9 and 6 are not relative prime numbers.
Now applying the algorithm on option (b):
Again using Euclid’s division Lemma we can write:
$9=4\times 2+1$
$4=2\times 2+0$
Here, the last non-zero remainder is 1. So, the greatest common divisor of 4 and 9 is 1.
Hence, 4 and 9 are relatively prime.
Now applying the algorithm on option (c):
Again using Euclid’s division Lemma we can write:
Here, the greatest common divisor of 28 and 7 is 7.
So, they are not relatively prime.
Now applying the algorithm on option (d):
Again using Euclid’s division Lemma we can write:
Here, the last non-zero remainder is 4. So, the greatest common divisor of 12 and 4 is 4.
Hence, they are not relatively prime.
Hence, option (b) 4, 9 is the only correct answer.
Note: Students should note here that if we get a zero remainder in the first step itself as in option (c), then the smaller of the two numbers becomes their greatest common divisor.
Complete step-by-step answer:
So, let us check all the options one by one whether the numbers are relatively prime not.
So, now we may use the Euclid’s division Lemma to find the greatest common divisor of the two numbers given in each option.
Now, we should know that in Euclid’s division algorithm we apply the Euclid’s division Lemma in succession several times until we get a remainder zero. The last non-zero remainder that is obtained in this process is termed as the great common divisor of the two numbers.
So, first we should know that the according to Euclid’s division Lemma:
If we have two integers ‘a’ and ‘b’ then for these numbers there exist ‘q’ and ‘r’ such that: -
$a=bq+r$ , where q is the quotient and r is the remainder.
So, on applying this algorithm on option (a):
Using Euclid’s division Lemma we can write:
$9=6\times 1+3$
$6=3\times 2+0$
So, the last non-zero remainder obtained here is=3. So, the greatest common divisor of 9 and 6 is 3.
Hence, 9 and 6 are not relative prime numbers.
Now applying the algorithm on option (b):
Again using Euclid’s division Lemma we can write:
$9=4\times 2+1$
$4=2\times 2+0$
Here, the last non-zero remainder is 1. So, the greatest common divisor of 4 and 9 is 1.
Hence, 4 and 9 are relatively prime.
Now applying the algorithm on option (c):
Again using Euclid’s division Lemma we can write:
Here, the greatest common divisor of 28 and 7 is 7.
So, they are not relatively prime.
Now applying the algorithm on option (d):
Again using Euclid’s division Lemma we can write:
Here, the last non-zero remainder is 4. So, the greatest common divisor of 12 and 4 is 4.
Hence, they are not relatively prime.
Hence, option (b) 4, 9 is the only correct answer.
Note: Students should note here that if we get a zero remainder in the first step itself as in option (c), then the smaller of the two numbers becomes their greatest common divisor.
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