The square root of 21, written as \( \sqrt{21} \), is an irrational number.
To determine the rationality of \( \sqrt{21} \), we can use the following reasoning:
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A rational number is defined as a number that can be expressed in the form \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
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If \( \sqrt{21} \) were rational, then it could be expressed as \( \frac{p}{q} \) in its simplest form, where \( p \) and \( q \) are integers with no common factors other than 1.
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Squaring both sides gives: \[ 21 = \frac{p^2}{q^2} \] which rearranges to \( p^2 = 21q^2 \).
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This implies that \( p^2 \) is divisible by 21. Since 21 can be factored as \( 3 \times 7 \), it follows that \( p^2 \) must also be divisible by both 3 and 7. Thus, \( p \) itself must also be divisible by 3 and 7 (because a prime factor appearing in the square of a number must appear at least once in the number itself).
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Let \( p = 3k \) for some integer \( k \). Substituting into the equation gives: \[ (3k)^2 = 21q^2 \ 9k^2 = 21q^2 \ k^2 = \frac{21q^2}{9} = \frac{7q^2}{3} \] This means \( k^2 \) must also be divisible by 3, leading us back to the conclusion that \( p \) is divisible by 3.
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However, this creates a contradiction, as we began with the assumption that \( p \) and \( q \) have no common factors.
Since our initial assumption leads to a contradiction, \( \sqrt{21} \) cannot be expressed as a fraction of integers, and therefore, \( \sqrt{21} \) is irrational.
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