Show that any positive integer is of the form 4q, 4q+2, where q is any positive integer.
Did you mean to say 'positive even integer' ?
Just like in your previous post, the even integers 2 and 4 cannot be obtained using your expressions.
I think you should change both statements to say " where q is any whole number "
To show that any positive integer is of the form 4q or 4q+2, where q is any positive integer, we can consider two cases:
Case 1: The positive integer is even.
If the positive integer is even, it can be expressed in the form 2p, where p is also a positive integer. Now, we need to show that 2p can be rewritten as either 4q or 4q+2 for some positive integer q.
Since p is positive, we can express it as q + r, where q is a non-negative integer and r is a remainder (0 ≤ r < 2). This is a result of dividing p by 2.
Substituting q + r for p, we have:
2p = 2(q + r) = 4q + 2r
Now, we have two possibilities for the remainder r:
1) If r = 0, then 2r = 0. Hence, 2p = 4q, where q = (q + r).
2) If r = 1, then 2r = 2. Hence, 2p = 4q + 2, where q = (q + r).
Therefore, any positive even integer can be expressed as either 4q or 4q+2, where q is a positive integer.
Case 2: The positive integer is odd.
If the positive integer is odd, it can be expressed in the form 2p + 1, where p is a positive integer. Now, we need to show that 2p + 1 can be rewritten as either 4q or 4q+2 for some positive integer q.
Again, we express p as q + r, where q is a non-negative integer and r is a remainder (0 ≤ r < 2).
Substituting q + r for p, we have:
2p + 1 = 2(q + r) + 1 = 4q + 2r + 1
Now, we have two possibilities for the remainder r:
1) If r = 0, then 2r = 0. Hence, 2p + 1 = 4q + 1, where q = (q + r).
2) If r = 1, then 2r = 2. Hence, 2p + 1 = 4q + 3, where q = (q + r - 1).
In both cases, we can express any positive odd integer as either 4q or 4q+2, where q is a positive integer.
In conclusion, for any positive integer, we have shown that it can be expressed in the form 4q or 4q+2, where q is any positive integer.