Given: $a$ and $b$ are positive integers such that $a = bq + r$, where $0 \leq r < b$ and $q$ is a whole number.

To Prove: HCF $(a, b) = \text{HCF}(b, r)$.

Proof:

Let $h = \text{HCF}(b, r)$. This means $h$ is the highest common factor of $b$ and $r$.

Since $h$ divides both $b$ and $r$, we can write:

$$b = k_1 h \quad \text{and} \quad r = k_2 h$$

where $k_1$ and $k_2$ are coprime integers.

Now, from the given relation $a = bq + r$, we substitute the values of $b$ and $r$:

$$a = (k_1 h)q + k_2 h$$ $$a = h(k_1 q + k_2)$$

This shows that $h$ divides $a$. Since $h$ already divides $b$, we conclude that $h$ is a common factor of $a$ and $b$.

Now, we need to prove that $h$ is the highest common factor of $a$ and $b$.

Assume there exists a common factor $d$ of $a$ and $b$ such that $d > h$.

Since $d$ divides both $a$ and $b$, and $r = a - bq$, it follows that $d$ divides $r$ as well.

This means $d$ is a common factor of $b$ and $r$ greater than $h$.

But this contradicts our assumption that $h$ is the HCF of $b$ and $r$.

Therefore, no such $d$ exists, and $h$ is indeed the highest common factor of $a$ and $b$.

∴ Hence, HCF $(a, b) = h = \text{HCF}(b, r)$.