Revision: Number Systems >> Number Systems Maths English Medium Class 9 CBSE

All rational numbers and all irrational numbers together form the set of real numbers.

Numbers that can be written in the form \[\frac{p}{q}\], where p and q are integers and q ≠ 0.

Numbers that cannot be written in the form\[\frac{p}{q}\]. Their decimal expansion is non-terminating and non-repeating.

1) `a^m . a^n = a^(m + n)`

2) `(a^m)^n = a^(mn)`

3) `a^m / a^n = a^(m - n)  ,m>n `

4)  `a^mb^m = (ab)^m`

5)  `(a)^0 = 1`

6) `1/a^n = a^(-n)`

Statement:

Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof:

Step 1: Let the prime factorisation of a be
a = p₁ p₂…p_n (where p₁,p₂,…,p_n are prime numbers)

Step 2: Squaring both sides,
\[a^2=(p_1p_2\ldots p_n)^2=p_1^2p_2^2\ldots p_n^2\]​

Step 3: p divides a²
So, p must be one of the prime factors of a².

Step 4: By the uniqueness of prime factorisation, the prime factors of a² are exactly
p₁,p₂,…,p_n.

Step 5: Hence, p is one of p₁,p₂,…,p_n.
Therefore, p divides a.

\[\sqrt{2}\] is irrational.

Step 1: Assume \[\sqrt{2}\] is rational.

\[\sqrt{2}\] = \[\frac{a}{b}\]

where a and b are integers and b ≠ 0

Step 2: Square both sides.

\[2=\frac{a^2}{b^2}\Rightarrow a^2=2b^2\]

Step 3: 2 divides a².

Since 2 is prime, by the divisibility property of primes,

2 divides a.

So, let a = 2c.

Step 4: Substituting,

\[(2c)^2=2b^2\Rightarrow4c^2=2b^2\Rightarrow b^2=2c^2\]

This means that 2 divides b², and so 2 divides b.

Therefore, both a and b are divisible by 2, which contradicts the fact that a and b are coprime.

Conclusion:
The contradiction arises from the assumption that \[\sqrt{2}\]​ is rational.

Consider the left hand side:

`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c))`

= `1/(1 + x^b xx x^-a + x^c xx x^-a) + 1/(1 + x^a xx x^-b + x^c xx x^-b) + 1/(1 + x^b xx x^-c + x^a xx x^-c)`   ...[∵ a^{m + n} = a^m × aⁿ]

= `1/(1 + x^b/x^a + x^c/x^a) + 1/(1 + x^a/x^b + x^c/x^b) + 1/(1 + x^b/x^c + x^a/x^c)`

= `1/((x^a + x^b + x^c)/x^a) + 1/((x^b + x^a + x^c)/x^b) + 1/((x^c + x^b + x^a)/x^c)`

= `x^a/(x^a + x^b + x^c) + x^b/(x^a + x^b + x^c) + x^c/(x^a + x^b + x^c)`

= `(x^a + x^b + x^c)/(x^a + x^b + x^c)`

= 1

Therefore left hand side is equal to the right hand side.

Hence proved.

Given `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m)`

Putting the values of x, y and z in `x^my^nz^l,` we get

`x^my^nz^l`

= `(a^(m + n))^m(a^(n + l))^n(a^(l + m))^l`

= `(a^(m^2 + nm))(a^(n^2 + ln))(a^(l^2 + lm))`

= `a^(m^2 + n^2 + l^2 + nm + ln + lm)`

Putting the values of x, y and z in `x^ny^lz^m,` we get

`x^ny^lz^m`

= `(a^(m + n))^n(a^(n + l))^l(a^(l + m))^m`

= `(a^(mn + n^2))(a^(nl + l^2))(a^(lm + m^2))`

= `a^(mn+n^2 + nl + l^2 + lm + m^2)`

So, `x^my^nz^l = x^ny^lz^m`.