MTH202 Practice Questions Lectures 23-45

Solution File Lecture No 23-25

Q1. Use mathematical induction to prove that 4n 3n 4 is true for integral values of n 2

Solution: · Let P(n) = 4n 3n 4 · Let P n 4n 3n 4 · n 2 · So P 2 · is true · 4k 3k 4 · Now · 4k 1 3k 1 4 · 16 4k 1 3k.(3 1) 16 · 4k 1 3k.3 3k 16 · 4k 1 3k 1 3k 4 12 · 4k 1 3k 1 4 12 3k

by (1) for k 2.

4k 1 3k 1 4 · ignoring 12 · 3k · as if a b c then a b

Hence it is proved by mathematical induction that 4n 3n 4 is true for integral values of n 2 .

Q3. Suppose n3 n is divisible by 6 is true for any positive integer. Justify the

statement for n = k+1.

Solution: · Suppose for n=k, k 3 k is dividible by · 6 then k 3 k 6.q · we need to prove k 1 · 3 k 1 is dividible by 6 · k 1 · 3 k 1 k 3 3k 2 3k 1 k 1 · k 3 k 3k k 1 · 6.q 3k k 1 · As the · product · k k 1 of integers is even so we can take as 2r · 6.q 3k k 1 · 6.q 3.2r · 6 q r · So · k 1 · 3 k 1 is dividible by 6 · Hence by induction method, n3 n is divisible 6

Q4. Suppose 5n 1 is divisible by 4 is true for n k . Justify the statement for n k 1 .

Solution: · Let for n k , 5k 1 is divisible by 4 · then 5k 1 4.q · we need to prove that 5k 1 1 is divisible by 4 · 5k 1 1 5.5k 1 · 5k 4 1 1 · 5k.4 5k 1 · 5k.4 4.q · 4 5k q · Hence 5k 1 1 is divisible by 4

Q5. Prove that if x divides y and y divides z then x divides z.

Solution:

By our assumptions, and the definition of divisibility, there are natural numbers k1 and k2 such that y xk1 and z yk2 . Consequently, z yk2 xk1k2 .

Let k

k1k2

divides z.

Now k is a natural number and z xk , so by the definition of divisibility, x

Q6. Show that for all integers a and b, if a b is even then a b is also even.

Solution: · Take a+b = · 2k a= 2k-b · Subtracting ‘b’ from both · sides a-b =2k-b-b · a-b = 2k - · 2b a-b = 2( · k-b)

Hence a-b is also even.