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Write whether every positive integer can be of the form $4q + 2$ where \[q\] an integer is. Justify your answer.

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Last updated date: 17th Jun 2024
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Answer
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Hint: The form $4q + 2$ can be written by $4q + 2 = 2(2q + 1)$.Try to substitute \[q\] for different values of integers. Integer is a set of numbers in the form $\{ - \infty ,...... - 2, - 1,0,1,2,3,4,5,.....\infty \} $ it contains both positive and negative numbers and also 0.Here is the set of positive numbers that are $\{ 1,2,3,4,5,.....\infty \} $
In the above set of positive numbers, if we can substitute \[q\] for different integers, then we get a positive integer, and also we say that the form $4q + 2$ be a positive integer.

Complete step-by-step answer:
It is given that, \[q\] is an integer.
Let us consider \[y = 4q + 2\]
We can split into \[y = 2(2q + 1)\]
Substitute the value of $q = \{ - \infty ,...... - 2, - 1,0,1,2,3,4,5,.....\infty \} $ in \[y\], we get
Let us substitute the values randomly,
Consider \[q = - 1\] in \[y\], then
\[y = 2(2( - 1) + 1)\]
\[y = 2( - 2 + 1)\]
 \[y = 2( - 1)\]
  \[y = - 2\]
Also we consider \[q = 0\] in \[y\], then
\[y = 2(2(0) + 1)\]
\[y = 2(0 + 1)\]
\[y = 2(1)\]
\[y = 2\]
Now we consider \[q = 1\] in \[y\], then
\[y = 2(2(1) + 1)\]
\[y = 2(2 + 1)\]
\[y = 2(3)\]
\[y = 6\]
Also we consider, \[q = 2\] in \[y\], then
\[y = 2(2(2) + 1)\]
\[y = 2(4 + 1)\]
\[y = 2(5)\]
\[y = 10\]
Hence we get the value of
\[y = - 2,y = 2,y = 6,y = 10\]
As we observe the above substitution for \[q\] and result of \[y\], it is given that only even numbers.
If we take \[q\] is negative values, it gives result as \[y\] is negative integer,
If we take \[q\] is zero, it gives result as \[y\] is positive even integer,
If we take \[q\] as positive values, it gives a result as \[y\] is positive even integer.
Hence, we can justify our answer, this form does not give positive integer for every integer because positive integer has positive odd integer, positive even integer.
This form gives only positive even integers but our discussion is about every positive integer has the form $4q + 2$.
Hence we justified our answer.

Note: We are discussing the form positive integers, actually such a form exists with the condition.
E.g. If \[\{ n\} _{n = 1}^\infty \] for all positive integers of \[n\], then the sequence gives a positive integer.
If \[\{ n\} _{n = - 1}^{ - \infty }\] for all the negative integers of \[n\], then the sequence gives a negative integer also for zero it is zero integer.
It is followed by only certain conditions.
The form \[4q + 2\] for any values of integer, odd integer is missing that odd number is missing. So the form does not satisfy the condition of positive integer.