Question

In Euclid’s division Lemma for positive integers a and b, the unique integers q and r are obtained such that a = bq + r is:
(a) 0 < r < b
(b) \[0\le r\le b\]
(c) \[0< r\le b\]
(d) \[0\le r< b\]

Answer 1

Hint: To find the appropriate answer of this question, we should know about Euclid’s division lemma which states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that a = bq + r, where a is the dividend, b is the divisor, q is the quotient and r is the remainder.

Complete step-by-step answer:
In this question, we have to find the range of r in terms of b, when it is given that a = bq + r where a, b, q, and r are whole numbers and \[a,b\ne 0\]. To find the correct answer, we should know that Euclid’s division lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that a = bq + r, where a is the dividend, b is the divisor, q is the quotient and r is the remainder. Or we can say that it states that any positive integer ‘a’ can be divided by any other positive integer ‘b’ in such a way that it leaves the remainder, r.
Now, let us consider option (a) that is 0 < r < b. In this, \[r\ne 0\], but we know that sometimes ‘a’ is completely divisible by ‘b’, that is ‘b’ is a factor of ‘a’. In that case, the remainder is equal to zero. So option (a) is the wrong answer.
Now, let us consider option (b), that is \[0\le r\le b\]. In this, r = 0 but at the same time it states that r = b. So, let us put r = b in Euclid’s division lemma, so we get,
a = bq + b
a = b (q + 1) + 0
So, r will become 0 if we put r = b or we can say \[r\ne b\] in any possible situation. So option (b) is the wrong answer.
Now, let us consider option (c), \[0< r\le b\]. In this case, we can see that \[r\ne 0\] and also this case states that r = b and in the above case, we proved that r can never be equal to b. So, option (c) is also the wrong answer.
Now, let us consider option (d), that is \[0\le r< b\]. In this case, we have been given that r is not equal to b. Also, this states that r = 0 is possible. So, this is the correct answer.
So, we can say that after observing all the options, we concluded that the option (d) is the correct answer.

Note: In this question, we can mark the correct answer by observing them like we know that sometimes ‘a’ is completely divisible by ‘b’. In that case, we will get r = 0. So, \[r\ge 0\]. Now, if we consider r = b, then we can write a = bq + r as a = bq + b, we get, a = b (q + 1) + 0 which again gives r = 0. So, r < b.