Real numbers are a set of rational and irrational numbers. In the AP SSC Class 10 Maths Chapter 1 Real Numbers, we learn about various theorems that are used to explore the properties of rational and irrational numbers. We also, study about a type of function known as logarithms and see their applications in Maths and everyday life.
Theorems On Real Numbers
 The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
 Let x = \(\frac{p}{q}\) be a rational number, such that the prime factorisation of q is of the form \(2^{n}5^{m}\), where n, m are nonnegative integers. Then x has a decimal expansion which terminates.
 Let x = \(\frac{p}{q}\) be a rational number, such that the prime factorization of q is not of the form \(2^{n}5^{m}\), where n, m are nonnegative integers. Then, x has a decimal expansion which is nonterminating repeating (recurring).
Below, we have provided a few chapter questions with solutions,
 Express each number as a product of its prime factors.
 140 – 140 can be expressed as a product of \(2\times 2\times 5\times 7=140\)
 3825 – 3825 can be expressed as a product of \(5\times 5\times 3\times 3\times 17=3825\)
 5005 – 5005 can be expressed as a product of \(5\times 7\times 11\times 13=5005\)
 Using the stated theorems, without actual division, state whether the following rational numbers are terminating or nonterminating decimals.
 \(\frac{4}{45}\)
Solution: \(\frac{4}{45}\) can also be written as \(\frac{4}{3^{2}\times 5^{1}}\), since it is not in the form of \(2^{n}5^{m}\), it is a nonterminating decimal.

 \(\frac{25}{32}\)
Solution: \(\frac{25}{32}\) can be expressed as \(\frac{25}{2^{5}\times 5^{0}}\), since it is the form of \(2^{n}5^{m}\), it is a terminating decimal.
 Write the following in logarithmic form.
 \(3^{5}=243\)
Solution: \(3^{5}=243\) in logarithmic form is written as \(\log_{3}(243)=5\)

 \(10^{3}=0.001\)
Solution: \(10^{3}=0.001\) in logarithmic form is written as \(\log_{10}(0.001)=3\)
4. Find the HCF and LCM of 12 and 18 by the prime factorization method.
Solution:Â Given that 12 = 2 x 2 x 3 = 2^{2} x 3^{1} andÂ 18 = 2 x 3 x 3 = 2^{1} x 3^{2}
HCF (12, 18) = 2^{1} x 3^{1Â }= 6 = Product of the smallest power of each common prime factors in the numbers.
LCM (12, 18) = 2^{2} x 3^{2} = 36 = Product of the greatest power of each prime factors, in the numbers.
So, here HCF (12, 18) x LCM (12, 18) = 12 x 18.
Thus,Â we can verify that for any two positive integers a and b, HCF (a,b) x LCM (a, b) = a x b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers
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