Welcome to Edukum.com

**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, a^{2} = 9q^{2} (m = 3q^{2})

or, a^{2} = 3 \(\times\) 3q^{2}

\(\therefore\) a^{2} = 3m

Similarly, r = 1, a = 3q + 1 S.B.S . . . . . (ii)

or, a^{2} = 9q^{2} + 1 + 6q (m = 3q^{2} + 2q)

or, a^{2} = 3m + 1

where, r = 2, a = 3q + 2 S.B.S . . . . . (iii)

or, a^{2} = 9q^{2} + 16q + 4 (m = 3q^{2} + 4q + 1)

or, a^{2} = 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 = 2^{2} \(\times\) 3^{1} \(\times\) 5^{1} \(\times\) 17^{1} \(\times\) 23^{1} = 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 2

Example:

\(\frac{129}{2^2 \times 5^7 \times 7^5}\)

\(\Rightarrow\) 2

\(\frac{17}{8}\) = \(\frac{17}{2^3 \times 5^0}\)

\(\Rightarrow\) Denominator 8 is of the form 2^{n} \(\times\) 5^{m}. 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 \(\Rightarrow\) P

Since 2 is a factor of P

\(\Rightarrow\) 2 will also divide P

\(\Rightarrow\) p = 2n, where n is an integer

(S.B.S) p

or, 2p

\(\Rightarrow\) q

\(\Rightarrow\) 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.

© 2019 EDUKUM.COM ALL RIGHTS RESERVED