PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoIn grade \(9\), we have learnt about the concept of irrational numbers and their properties. We have also learnt how to plot irrational numbers in the number line. In this topic, we are going to learn how to prove irrational numbers.
A number \(x\) is said to be an irrational number if it cannot be expressed in the form of \(\frac{p}{q}\) where \(q \neq 0\).
Example:
\(\sqrt{2}\), \(\sqrt{5}\), \(\sqrt{7}\) are some examples of irrational numbers.
Important!
In this topic, we are going to prove that \(\sqrt{p}\) is irrational where \(p\) is a prime. Now, we shall prove the theorem using the fundamental theorem of arithmetic.
Theorem:
Proof: The prime factorization of \(a\) can be written as:
\(a = p_1p_2 ….p_n\)
Where \(p_1, p_2, …., p_n\) are unique prime factors of \(a\).
Then, the prime factorization of \(a^2\) can be written as:
\(a^2 = a \times a\)
\(a^2 = (p_1p_2 ….p_n)(p_1p_2 ….p_n)\)
\(a^2 = p_1^2 p_2^2 ….p_n^2\)
It is given that \(p\) divides \(a^2\). We know that from the fundamental theorem of arithmetic that \(p\) is one of the prime factors of \(a^2\). That is, \(p\) is one of (\(p_1,p_2, ….,p_n\)).
Since \(a = p_1p_2 ….p_n\) and \(p\) is one of (\(p_1,p_2, ….,p_n\)), then \(p\) divides \(a\).
Hence, we proved.
Register for free to see more content