Question No 02 | Exercise 2.2 | Divisibility Theory | Number Theory
Question No 02 | Exercise 2.2 | Divisibility Theory | Number Theory
Question No 02 | Exercise 2.2 | Chapter 2 | Divisibility Theory | Number Theory
Book : Elementary Number Theory
Edition : 6 & 7
Chapter Number : 02
Chapter Name : Divisibility Theory in the integers.
Lecture Number : 03
Topic Name : The division Algorithm.
Problems ( Exercise ) : 2.2
Question Number : 02
Part Number : 00
Question :
Prove that if a and b are integers, with o≤b then there exist unique integers q and r satisfying a = qb + r where 2b ≤ r ≤ 3b.
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Question No 03 | Part c | Exercise 2.2 | Divisibility Theory | Number Theory
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Question No: 02
Show that any integer of the form “ 6k + 5 ” is also of the form “ 3j + 2 ” , but not conversely.
Sol:
We want to prove that any integer of the form “6𝑘+5” is also of the form“3𝑗+2” but not conversely.
If 𝑎=6𝑘+5
=6𝑘+3+2
=3(2𝑘+1)+2 (i)
Let 2𝑘+1=𝑗
Put the 𝑗=2𝑘+1 in the equation (i).
=3𝑗+2
The integer “6𝑘+5” is also of the form “3𝑗+2”.
To disprove that any integer of the form “3𝑗+2” is also of the form “6𝑘+5”.
If 𝑏=3𝑗+2 (ii)
When 𝑗 is odd and even.
Case 1: 𝒋 is Odd:
Then 𝑗=2𝑟+1
Put the value 𝑗 in equation (ii). 𝑏=3𝑗+2
𝑏=3(2𝑟+1)+2
𝑏=6𝑟+3+2
𝑏=6𝑟+5
If 𝑗 is odd the integer “3𝑗+2” is also of the form “6𝑘+5”.
Case 2: 𝒋 is Even:
Then 𝑗=2𝑟
Put the value 𝑗 in equation (ii). 𝑏=3𝑗+2
𝑏=3(2𝑟)+2
𝑏=6𝑟+2
𝑏=6𝑟+2
If 𝑗 is even the integer “3𝑗+2” is not also of the form “6𝑘+5”.
Hence prove any integer of the form “6𝑘+5” is also of the form “3𝑗+2”, but not conversely.