Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5; where q is some integer.
Here’s a detailed and easy-to-understand explanation for your query:
Proof: Any Positive Odd Integer is of the Form 6q + 1, 6q + 3, or 6q + 5
To prove that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is an integer, let’s start with the concept of divisibility and remainders.
Step 1: Represent Positive Integers in Terms of Divisibility by 6
Let n be any positive integer. When n is divided by 6, it will leave one of the following remainders:
[ 0, 1, 2, 3, 4, \text{or } 5 ]
This is because the remainder when dividing by 6 is always less than 6.
Thus, any positive integer n can be expressed in the following form:
[
n = 6q + r
]
Where:
- ( q ) is an integer (quotient), and
- ( r ) is the remainder, such that ( r = 0, 1, 2, 3, 4, \text{or } 5 ).
Step 2: Identify Which Forms Represent Odd Integers
For an integer to be odd, it must not be divisible by 2. Let’s analyze the forms based on the value of ( r ):
- Case 1: r = 0
[
n = 6q + 0
]
Here, ( n ) is divisible by 2 (as it’s a multiple of 6). Hence, ( n ) is even and does not qualify as an odd integer. - Case 2: r = 1
[
n = 6q + 1
]
This is an odd number, as ( (6q + 1) \mod 2 = 1 ). - Case 3: r = 2
[
n = 6q + 2
]
This is even, as ( (6q + 2) \mod 2 = 0 ). - Case 4: r = 3
[
n = 6q + 3
]
This is odd, as ( (6q + 3) \mod 2 = 1 ). - Case 5: r = 4
[
n = 6q + 4
]
This is even, as ( (6q + 4) \mod 2 = 0 ). - Case 6: r = 5
[
n = 6q + 5
]
This is odd, as ( (6q + 5) \mod 2 = 1 ).
Step 3: Conclude the Forms of Positive Odd Integers
From the cases above, the odd integers occur only when ( r = 1, 3, \text{or } 5 ).
Therefore, any positive odd integer can be written as:
[
n = 6q + 1, \, 6q + 3, \text{or } 6q + 5
]
where ( q ) is an integer.
Conclusion
We have shown that any positive odd integer is of the form ( 6q + 1, 6q + 3, \text{or } 6q + 5 ), as required.
