Prove that $\sqrt{5}$ is an irrational number. Also, examine whether the decimal expansion of $\frac{13}{3125}$ is terminating or non-terminating repeating.

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Part 1: Proof that $\sqrt{5}$ is Irrational Introduction and Assumption:
To prove that $\sqrt{5}$ is an irrational number, we will use the method of contradiction. Let us assume, to the contrary, that $\sqrt{5}$ is a rational number. Definition of Rational Numbers:
Since $\sqrt{5}$ is assumed to be rational, it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q 
eq 0$, and $p$ and $q$ are co-prime (meaning their highest common factor, HCF, is 1). Mathematical Manipulation: $$\sqrt{5} = \frac{p}{q}$|
Squaring both sides of the equation, we get: $$5 = \frac{p^2}{q^2}$| $$p^2 = 5q^2$$
This equation implies that $5$ divides $p^2$. According to the fundamental theorem of arithmetic, if a prime number divides the square of an integer, it also divides the integer itself. Therefore, $5$ divides $p$. Substitution:
Since $5$ divides $p$, we can write $p = 5c$ for some integer $c$. Substituting this value of $p$ into the equation for $p^2$, we get: $$(5c)^2 = 5q^2$$ $$25c^2 = 5q^2$$ $$q^2 = 5c^2$$
This equation shows that $5$ divides $q^2$, which means $5$ also divides $q$. Conclusion of Contradiction:
From our deductions, both $p$ and $q$ have $5$ as a common factor. This contradicts our initial assumption that $p$ and $q$ are co-prime (having no common factors other than 1). This contradiction arises because of our incorrect assumption that $\sqrt{5}$ is rational. Hence, we conclude that $\sqrt{5}$ is an irrational number. Part 2: Decimal Expansion of $\frac{13}{3125}$ Prime Factorization of the Denominator:
Let the given rational number be $r = \frac{p}{q}$, where $q = 3125$. We find the prime factors of the denominator $3125$: $$3125 = 5 	imes 5 	imes 5 	imes 5 	imes 5 = 5^5$$ Application of Rational Number Theorem:
According to the theorem in real numbers, 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 $n$ and $m$ are non-negative integers. Then $x$ has a terminating decimal expansion. Comparison and Result:
Here, the denominator $3125 = 2^0 	imes 5^5$, which is clearly of the form $2^n 5^m$ (where $n = 0$ and $m = 5$). Therefore, the decimal expansion of $\frac{13}{3125}$ is terminating.

CBSE · Class 10 · Mathematics · Real NumbersProve that $\sqrt{5}$ is an irrational number. Also, examine whether the decimal expansion of $\frac{13}{3125}$ is terminating or non-terminating repeating.

Part 1: Proof that $\sqrt{5}$ is Irrational

Part 2: Decimal Expansion of $\frac{13}{3125}$