Show that the square of an oddnumber is an odd number using a direct proof.
Consider that for any integer (n), (2n+1) is an odd number.
By expansion and collecting terms, (2n+1)^2 = ______.
Since _______ is an integer it follows that (2n+1)^2 is odd.
Therefore the square of any odd number is also an odd number.
any odd number can be written in the form 2n + 1 , where n is a positive integer.
(2n+1)^2 = 4n^2 + 4n + 1
= 4(n^2 + 1) + 1
since any integer multiplied by 4 is even, the term 4(n^2 + 1) is even , adding 1 to it would make it odd.
To prove that the square of an odd number is also an odd number, we need to show that for any odd number n, the square of n will also be an odd number.
Let's assume n is an odd number.
Since n is odd, we can write n as 2k + 1, where k is an integer. (This is because every odd number can be represented as 2 multiplied by some integer k, plus 1.)
Now let's find the square of n:
n^2 = (2k + 1)^2
Expanding this equation:
n^2 = 4k^2 + 4k + 1
We can rewrite this equation as:
n^2 = 2(2k^2 + 2k) + 1
Since 2k^2 + 2k is an integer, let's call it m. Therefore, we can simplify the equation as:
n^2 = 2m + 1
Since 2m is an even number (as it is divisible by 2), we can write it as 2m = 2p, where p is an integer.
Substituting back into the equation, we get:
n^2 = 2m + 1 = 2p + 1
Since 2p + 1 is an odd number (as it is composed of 2 multiplied by some integer p, plus 1), we can conclude that the square of an odd number is also an odd number.
Therefore, the square of any odd number is an odd number, as shown through this direct proof.
To show that the square of an odd number is an odd number using a direct proof, we need to start with the assumption that we have an odd number and then demonstrate that its square is also odd.
Let's begin:
1. Start with the assumption that we have an odd number, say n, where n can be represented as 2k + 1, where k is an integer.
(Note: In general, an odd number can be written in the form 2k + 1, where k is an integer.)
2. Now, let's square this odd number, n^2 = (2k + 1)^2.
3. Expanding the expression, we get: n^2 = (2k + 1)(2k + 1)
4. Distributing, we get: n^2 = 4k^2 + 4k + 1
Notice that the first two terms, 4k^2 and 4k, are both divisible by 2 (since they have a common factor of 2). Thus, they are even numbers.
5. Adding an even number (4k^2) and another even number (4k) results in an even number.
6. An even number plus 1 (1 is also an odd number) yields an odd number.
Therefore, n^2 = (4k^2 + 4k) + 1 is an odd number.
Hence, we have proven that the square of an odd number is odd using a direct proof.