Question

plzzzzzzzzzzzzzzzz. ans this questions. 1. using euclid division algorithm find which of the following pairs of number are co prime: 231,396 847,2160. 2. using euclid division algorithm to find hcf of 441, 567, 693. 3. show that the square of any positive integer cannot be of the form 5m+2 or 5m+3 for some integer m. 4. show that the cube of any positive integer is the 4m, 4m+1, or 4m+3 for some integer m.  

Answer 1

• Euclid’s Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. • Euclid’s Division Algorithm to obtain the HCF of two positive integers, say c and d, c > d. Step 1 : Apply Euclid’s division lemma to c and d, to find whole numbers q and r, such that c = dq + r, 0 ≤ r < d. Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. • Fundamental Theorem of Arithmetic : Every composite number can be expressed as a product of primes, and this expression (factorisation) is unique, apart from the order in which the prime factors occur. • Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. • 2 , 3 , 5 are irrational numbers.The sum or difference of a rational and an irrational number is irrational. • The product or quotient of a non-zero rational number and an irrational number is irrational. • For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b• Let x = p q , p and q are co-prime, be a rational number whose decimal expansion terminates. Then, the prime factorisation of q is of the form 2 m .5n ; m, n are non-negative integers. • Let x = p q be a rational number such that the prime factorisation of q is not of the form 2m.5n ; m, n being non-negative integers. Then, x has a non-terminating repeating decimal expansion.

Answer 2

Step-by-step explanation:

Question : -

→ Use Euclid's Division lemma to show that the Square of any positive integer cannot be of form 5m + 2 or 5m + 3 for some integer m.

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▶ Step-by-step explanation : -

Let ‘a’ be the any positive integer .

And, b = 5 .

→ Using Euclid's division lemma :-

==> a = bq + r ; 0 ≤ r < b .

==> 0 ≤ r < 5 .

•°• Possible values of r = 0, 1, 2, 3, 4 .

→ Taking r = 0 .

Then, a = bq + r .

==> a = 5q + 0 .

==> a = ( 5q )² .

==> a = 5( 5q² ) .

•°• a = 5m . [ Where m = 5q² ] .

→ Taking r = 1 .

==> a = 5q + 1 .

==> a = ( 5q + 1 )² .

==> a = 25q² + 10q + 1 .

==> a = 5( 5q² + 2q ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 2q ] .

→ Taking r = 2 .

==> a = 5q + 2 .

==> a = ( 5q + 2 )² .

==> a = 25q² + 20q + 4 .

==> a = 5( 5q² + 4q ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 4q ] .

→ Taking r = 3 .

==> a = 5q + 3 .

==> a = ( 5q + 3 )² .

==> a = 25q² + 30q + 9 .

==> a = 25q² + 30q + 5 + 4 .

==> a = 5( 5q² + 6q + 1 ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 6q + 1 ] .

→ Taking r = 4 .

==> a = 5q + 4 .

==> a = ( 5q + 4 )² .

==> a = 25q² + 40q + 16 .

==> a = 25q² + 40q + 15 + 1 .

==> a = 5( 5q² + 8q + 3 ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 8q + 3 ] .

→ Therefore, square of any positive integer in cannot be of the form 5m + 2 or 5m + 3 .

✔✔ Hence, it is proved ✅✅.

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