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Let us prove the theorem by assuming its contrary.
Example:
1. Prove that \(\sqrt{7}\) is an irrational number.
Proof: Let us assume that \(\sqrt{7}\) is a rational number.
Therefore, it can be written as \(\sqrt{7} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
Thus, \(\sqrt{7} = \frac{p}{q}\)
\(q \sqrt{7} = p\)
Squaring on both sides, we have:
\(7q^2 = p^2\) ---- (\(1\))
Therefore, \(p^2\) is divisible by \(7\).
And, \(p\) is also divisible by \(7\) [If \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.]
So, we can write \(p = 7a\) for some integer \(a\).
Substituting the value of \(p\) in equation (\(1\)), we have:
\(7q^2 = 49a^2\)
\(q^2 = 7a^2\)
This implies that \(q^2\) is divisible by \(7\) and \(q\) is also divisible by \(7\). [If \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.]
Therefore, \(p\) and \(q\) have atleast \(7\) as a common factor.
This contradicts the fact that \(p\) and \(q\) are coprime.
Therefore, our assumption is wrong.
Thus, \(\sqrt{7}\) is an irrational number.
Hence, we proved.
2. Prove that \(8 - \sqrt{3}\) is an irrational number.
Proof: Let us assume that \(8 - \sqrt{3}\) is a rational number.
Therefore, it can be written as \(8 - \sqrt{3} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
Thus, \(8 - \sqrt{3} = \frac{p}{q}\)
\(8 - \frac{p}{q} = \sqrt{3}\)
\(\frac{8q - p}{q} = \sqrt{3}\)
Since \(8\), \(p\) and \(q\) are integers, then \(\frac{8q - p}{q}\) is rational. This implies that \(\sqrt{3}\) is a rational number.
But, we know that \(\sqrt{3}\) is an irrational number.
Therefore, our assumption is wrong.
Thus, \(8 - \sqrt{3}\) is an irrational number.
Hence, we proved.
Important!
In grade \(9\), we have learnt that
1. The sum or difference of a rational and an irrational number is irrational.
2. The product and quotient of a non-zero rational and irrational number is irrational.
Example:
Show that \(\frac{3}{\sqrt{5}}\) is an irrational number.
Proof:
Let us assume that \(\frac{3}{\sqrt{5}}\) is a rational number.
Therefore, it can be written as \(\frac{3}{\sqrt{5}} = \frac{p}{q}\) where \(p, q\) are integers and \(q \neq 0\).
Here, \(p\) and \(q\) are coprime(no common factors between \(p\) and \(q\)).
Thus, we have:
\(\frac{3}{\sqrt{5}} = \frac{p}{q}\)
\(3q = \sqrt{5}p\)
\(\frac{3q}{p} = \sqrt{5}\)
Since we know that \(3\), \(q\) and \(p\) are integers and \(\frac{3q}{p}\) is a rational number. This implies that \(\sqrt{5}\) is a rational number.
But, we know that \(\sqrt{5}\) is an irrational number.
Therefore, our assumption is wrong.
Thus, \(\frac{3}{\sqrt{5}}\) is an irrational number.
Hence, we proved.
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