Teresa Manning

2023-02-28

In Euclid's Division Lemma when $a=bq+r$ where $a,b$ are positive integers then what values $r$ can take?

gelo1368m6

Beginner2023-03-01Added 5 answers

Find the integer's value.

According to Euclid's Division Lemma if we have two integers $a$ and $b$ then there exist unique integer $q$ and $r$ which satisfy the condition $a=bq+r$ where $0\le r<b$

$r$ is the remainder and $b$ is the divisor and remainder is always less than divisor and greater than equal; to $0$

Example: $a=59$ and $b=5$

After according to Euclid's division lemma $a$ can be stated as

$a=bq+r\Rightarrow 59=11\xb75+4$

then take $a=60$ and $b=5$

$a=bq+r\Rightarrow 60=12\xb75+0$

We can see value of $r$ ranges from $0$ to less than $b$

Thus the value of $r$ is $0\le r<b$

According to Euclid's Division Lemma if we have two integers $a$ and $b$ then there exist unique integer $q$ and $r$ which satisfy the condition $a=bq+r$ where $0\le r<b$

$r$ is the remainder and $b$ is the divisor and remainder is always less than divisor and greater than equal; to $0$

Example: $a=59$ and $b=5$

After according to Euclid's division lemma $a$ can be stated as

$a=bq+r\Rightarrow 59=11\xb75+4$

then take $a=60$ and $b=5$

$a=bq+r\Rightarrow 60=12\xb75+0$

We can see value of $r$ ranges from $0$ to less than $b$

Thus the value of $r$ is $0\le r<b$