Prove that every positive odd integer is of the form \[(4q+1)\] or \[(4q+3)\], where q is some integer.

Hint: Every positive integer is either even or odd. Every positive integer can be written in the form of other smaller positive integers.

Complete step-by-step answer:

We know that we can express every positive integer of the form \[4q\] or \[(4q+1)\] or \[(4q+2)\] or \[(4q+3)\] where q is some positive integer.
Let x be a positive integer where \[x=4q\].
Now, \[x=4q=2(2q)\]. Here, x can be written as a multiple of 2. Hence,\[x\] is divisible by 2.
We know from the definition of even numbers that they are divisible by 2. So \[x\] is even.
Now, \[x+1=4q+1\]
Now, \[x+1\] is the just next number of \[x\]. We know that even and odd numbers are placed alternatively one by one.
So as \[x\] is even, then \[x+1\] is odd. So \[4q+1\] is odd.
Similarly, \[x+2=(4q+2)=2(2q+1)\]. Here, x+2 can be written as a multiple of 2. Hence,\[(4q+2)\] is even.
Finally, \[x+3=(4q+3)=(4q+2)+1\].
So as \[x+2\] is even, then \[x+3\] is odd. So \[4q+3\] is odd.
Now, as every number is expressible in the above-mentioned forms and q is an arbitrary positive integer, then it is proved that every odd integers are of the form \[(4q+1)\] or \[(4q+3)\].

Note: If n be an positive integer and q be any arbitrary integer, then every positive integer can be written of the form \[\text{nq},\text{ nq}+1,\text{ nq}+2,\text{ nq}+3,\text{ }........\text{ },\text{ }nq+n-1\]. Here, according to the problem n=4. Therefore, it only has 4 types of representations. Two of them for events and the rest of two are for odds. You can check any natural number by substitution and recheck whether it is odd.