What is an irrational sort 63 rewritten as the product of integer and another irrational sort in simplest form

Answer 1

To rewrite the irrational number \( \sqrt{63} \) as the product of an integer and another irrational number, we can first factor 63. The prime factorization of \( 63 \) is:

\[ 63 = 9 \times 7 \]

We know that \( 9 \) is a perfect square. Therefore, we can write \( \sqrt{63} \) as follows:

\[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} \]

Since \( \sqrt{9} = 3 \), we have:

\[ \sqrt{63} = 3 \times \sqrt{7} \]

Thus, \( \sqrt{63} \) can be rewritten as the product of an integer and another irrational number in simplest form as:

\[ \sqrt{63} = 3\sqrt{7} \]

Here, \( 3 \) is the integer and \( \sqrt{7} \) is the irrational number.