Tips,tricks and formulae on H.C.F and L.C.M in PDF
... H.C.F is the highest common factor or also known as greatest common divisor, the greatest number which exactly divides all the given numbers. There are two methods to find H.C.F of given numbers, they are: i. Prime factorization method. ii. Division Method. How to find H.C.F of given numbers by prim ...
... H.C.F is the highest common factor or also known as greatest common divisor, the greatest number which exactly divides all the given numbers. There are two methods to find H.C.F of given numbers, they are: i. Prime factorization method. ii. Division Method. How to find H.C.F of given numbers by prim ...
number and number relations
... determine the possible combinations from two disjoint sets when choosing one item from each set (D-4-M) ...
... determine the possible combinations from two disjoint sets when choosing one item from each set (D-4-M) ...
Chapter 1.1 Rational and Irrational Numbers
... If you add 0 to any number you get the same or identical number, if you multiply any real number by 1 you get the same or identical number. Inverse property: For every real number a, there is a unique real number -a such that a + (-a) = 0. For any real number a, there is a unique real number 1/a suc ...
... If you add 0 to any number you get the same or identical number, if you multiply any real number by 1 you get the same or identical number. Inverse property: For every real number a, there is a unique real number -a such that a + (-a) = 0. For any real number a, there is a unique real number 1/a suc ...
Proving irrationality
... integers rn , sn and tn . If ξ = a/b is rational, then (ξ − 1)n = cn /b2 with cn an integer and again this is impossible for large enough n. Now one can extend this argument again to show that if ξ k + u1 ξ k−1 + · · · + un−1 ξ + un = 0 with the uj s integers, then if ξ isn’t an integer, then ξ must ...
... integers rn , sn and tn . If ξ = a/b is rational, then (ξ − 1)n = cn /b2 with cn an integer and again this is impossible for large enough n. Now one can extend this argument again to show that if ξ k + u1 ξ k−1 + · · · + un−1 ξ + un = 0 with the uj s integers, then if ξ isn’t an integer, then ξ must ...
Math 335 Homework Set 4
... Exercise 3.2.9. The one-to-one correspondence is x 7→ 6x going from N to 6N and y 7→ y/6 going in reverse from 6N to N. Exercise 3.2.30. Yes, the collection of individual candy bars has the same cardinality as the set of natural numbers. We can send package k’s contents to the numbers 2k − 1 and 2k. ...
... Exercise 3.2.9. The one-to-one correspondence is x 7→ 6x going from N to 6N and y 7→ y/6 going in reverse from 6N to N. Exercise 3.2.30. Yes, the collection of individual candy bars has the same cardinality as the set of natural numbers. We can send package k’s contents to the numbers 2k − 1 and 2k. ...
Fibonacci numbers
... Problem: For all the given numbers x0 , x1 , . . . , xn≠1 , such that 1 ˛ xi ˛ m ˛ 1 000 000, check whether they may be presented as the sum of two Fibonacci numbers. Solution: Notice that only a few tens of Fibonacci numbers are smaller than the maximal m (exactly 31). We consider all the pairs. If ...
... Problem: For all the given numbers x0 , x1 , . . . , xn≠1 , such that 1 ˛ xi ˛ m ˛ 1 000 000, check whether they may be presented as the sum of two Fibonacci numbers. Solution: Notice that only a few tens of Fibonacci numbers are smaller than the maximal m (exactly 31). We consider all the pairs. If ...
UNIT 3: Divisibility in Natural Numbers 3.1 Relationship of divisibility
... Method 2 To work out the H.C.F. of several numbers, first write them as a product of their primes factors and then take only the common factors with the least exponent Example: Find the H.C.F of 40 and 60 ...
... Method 2 To work out the H.C.F. of several numbers, first write them as a product of their primes factors and then take only the common factors with the least exponent Example: Find the H.C.F of 40 and 60 ...