 # AP SSC Class 10 Maths Chapter 1 Real Numbers

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 science 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 factorization of q is of the form $2^{n}5^{m}$, where n, m are non-negative 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 non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

Below, we have provided a few chapter questions with solutions,

1. Express each number as a product of its prime factors.
1. 140 – 140 can be expressed as a product of $2\times 2\times 5\times 7=140$
2. 3825 – 3825 can be expressed as a product of $5\times 5\times 3\times 3\times 17=3825$
3. 5005 – 5005 can be expressed as a product of $5\times 7\times 11\times 13=5005$
1. Using the stated theorems, without actual division, state whether the following rational numbers are terminating or non-terminating decimals.
1. $\frac{4}{45}$

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 non-terminating decimal.

2. $\frac{25}{32}$

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.

1. Write the following in logarithmic form.
1. $3^{5}=243$

Solution: $3^{5}=243$ in logarithmic form is written as $\log_{3}(243)=5$

2. $10^{-3}=0.001$

Solution: $10^{-3}=0.001$ in logarithmic form is written as $\log_{10}(0.001)=-3$

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