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# Real Numbers

Euclid's Division Lemma
a = bq + r, where a and b are positive integers
$\Rightarrow$ 0 £ r < b
Example:
a = 13, b = 3 find q and r.
a = bq + r      0 £ r < 3
13 = 3 $\times$ 4 + 1
q = 4 and r = 1 are required  numbers.

Question:
Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, q is an integer.
Solution:
a be some integer, b = 2
According to Division Lemma a = bq + r
0 £ r < 2         i.e r = 0                          r = 1
or, a = 2q + 0                or, a = 2q + 1
If a = 2q then a is an even integer.
Since, we know that an integer can be either even or odd.
Therefore, any +ve odd integer is of the form 2q + 1.

For showing the square or cube of any positive integer is either in the form of:
Square – 3m, 3m + 1
Cube – 9m, 9m + 1 or 9m + 8
Just square or cube the equation while putting value of r in Euclid Division Lemma.
Example:
a be any positive integer
b = 3, 0 £ r < 3 and r = 0
or, a = 3q + 0 . . . . . (i) S.B.S
or, a2 = 9q2 (m = 3q2)
or, a2 = 3 $\times$ 3q2
$\therefore$ a2 = 3m
Similarly, r = 1, a = 3q + 1   S.B.S . . . . . (ii)
or, a2 = 9q2 + 1 + 6q  (m = 3q2 + 2q)
or, a2 = 3m + 1
where, r = 2, a = 3q + 2  S.B.S . . . . . (iii)
or, a2 = 9q2 + 16q + 4   (m = 3q2 + 4q + 1)
or, a2 = 3m + 1
Hence, square of any positive integer is in the form of 3m, 3m + 1.

Find HCF by using Euclid's Division Algorithm; find the HCF of 56, 96 and 404.
Solution:
96 > 50, so first dividend is 96.
or, 96 = 56 $\times$ 1 + 40
or, 56 = 40 $\times$ 1 + 16
or, 40 = 16 $\times$ 2 + 8
or, 16 = 8 $\times$ 2 + 0
$\Rightarrow$ 8 is a divisor
Now,
404 > 8, start with 404
or, 404 = 8 $\times$ 10 + 4
or, 8 = 4 $\times$ 2 + 0
So, HCF of 96, 56, 404 is 4.

Relationship between LCM and HCF
LCM (a, b) $\times$ HCF (a, b) = Product (a, b)
Example:

HCF = 2
LCM = 22 $\times$ 31 $\times$ 51 $\times$ 171 $\times$ 231 = 23460
Verification:
HCF $\times$ LCM = a $\times$ b
or, 2 $\times$ 23460 = 92 $\times$ 510
$\therefore$ 46920 = 46920

Theorems
Let 'a' be a rational number whose decimal expansion terminates. Then we can express a = $\frac{P}{q}$, where, p and q are co – primes and the prime factorization of q is of the form 2n $\times$ 5m, where m and n.
Example:
$\frac{129}{2^2 \times 5^7 \times 7^5}$
$\Rightarrow$ 2n $\times$ 57 $\times$ 75 is not of the form 2n $\times$ 5m. So, the decimal is non – terminating repeating.

$\frac{17}{8}$ = $\frac{17}{2^3 \times 5^0}$
$\Rightarrow$ Denominator 8 is of the form 2n $\times$ 5m. Hence, $\frac{17}{8}$ has terminating decimal expansion.

Question:
Show that $\sqrt{2}$ is not a rational number. By using method of contradiction.
Solution:
Suppose $\sqrt{}$ is not an irrational number.
$\therefore$ It is rational number
Simplest form of $\sqrt{2}$ is $\frac{p}{q}$
p, q are integers having no common factor other than 1 and q $\neq$ 0
or, $\frac{p}{q}$ = $\sqrt{2}$ (S.B.S)
or, $\frac{p^2}{q^2}$ = 2 $\Rightarrow$ P2 = 2q2
Since 2 is a factor of P2
$\Rightarrow$ 2 will also divide P
$\Rightarrow$ p = 2n, where n is an integer
(S.B.S) p2 = 4n2
or, 2p2 = 4n2
$\Rightarrow$ q2 = 2n2
$\Rightarrow$ 2 will also divide q
p and q both are divisible by 2 and this is in contradiction that p and q have no common factor. Our supposition is wrong.
Hence, $\sqrt{2}$ is an irrational number.

5 – 2$\sqrt{3}$ is an irrational number.
Solution:
Suppose 5 – 2$\sqrt{3}$ is rational number.
5 – 2$\sqrt{3}$ = $\frac{p}{q}$, q $\neq$ 0, p and q are co-prime integers.
or, 5 - $\frac{p}{q}$ = 2$\sqrt{3}$
or, $\sqrt{3}$ = $\frac{5q – p}{2q}$
Now,
$\frac{5q – p}{2q}$ is a rational number and $\sqrt{3}$ is an irrational number.
This is not possible as a rational number cannot be equal to an irrational number.
Our supposition is wrong.
Hence, 5 – 2$\sqrt{3}$ is irrational number.

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