
If r=0, then what is the relationship between $a,b{\text{ }}and{\text{ }}q$ in $a = bq + r$.
Answer
479.1k+ views
Hint: We will substitute the value of $r = 0$ given in the question to the equation given in the question and find the relationship between $a,b{\text{ }}and{\text{ }}q$ in $a = bq + r$ . Euclid’s Division Lemma states that if two positive integers $a$ and $b$, then there exist two unique integers $q$ and $r$ such that $a = bq + r$ where $0 \leqslant r < b$.
Complete step-by-step solution:
The given value of $r$ is $0$ in the question.
The given equation is $a = bq + r$ .
This equation is known as Euclid’s division lemma.
Now we will substitute the value of $r = 0$ in the equation, we will get,
$a = bq + 0$
Adding any number with $0$ gives us the same number unchanged.
Therefore, $a = bq$
Therefore, the relationship between $a,b{\text{ }}and{\text{ }}q$ from the equation after substituting $r = 0$ is $a = bq$.
Since, $b$ divides $a$ , we can conclude that $b$ is a factor of $a$.
Note: According to Euclid’s division lemma:
$a = bq + r$ where $a$ is dividend, $b$ is divisor, $q$ is quotient and $r$ is the remainder.
Here, remainder is always smaller than the divisor and it is equal to or more than $0$.
Therefore, $0 \leqslant r < b$ , when $r = 0$ , $a$ is completely divisible by $b$.
Complete step-by-step solution:
The given value of $r$ is $0$ in the question.
The given equation is $a = bq + r$ .
This equation is known as Euclid’s division lemma.
Now we will substitute the value of $r = 0$ in the equation, we will get,
$a = bq + 0$
Adding any number with $0$ gives us the same number unchanged.
Therefore, $a = bq$
Therefore, the relationship between $a,b{\text{ }}and{\text{ }}q$ from the equation after substituting $r = 0$ is $a = bq$.
Since, $b$ divides $a$ , we can conclude that $b$ is a factor of $a$.
Note: According to Euclid’s division lemma:
$a = bq + r$ where $a$ is dividend, $b$ is divisor, $q$ is quotient and $r$ is the remainder.
Here, remainder is always smaller than the divisor and it is equal to or more than $0$.
Therefore, $0 \leqslant r < b$ , when $r = 0$ , $a$ is completely divisible by $b$.
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