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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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