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prove that the paired degree cannot be equal to $8k-3$

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Composition of linear and quadratic functions: find a value

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Q. prove that the paired degree cannot be equal to

8k-3

Assumed Paired Degree: Let's assume that the paired degree can be equal to $8k-3$ for some integer $k$ .
Even Number Property: A paired degree is always an even number since it's the sum of the degrees of two vertices in a graph.
Expression as $2n$ : An even number can be expressed as $2n$ , where $n$ is an integer.
Equation Rearrangement: If the paired degree is $8k-3$ , then we can write it as $2n = 8k-3$ .
Solving for n: Rearrange the equation to solve for n: $n = \frac{8k-3}{2}$ .
Integer Property of $8k$ : Since $k$ is an integer, $8k$ is also an integer because $8$ times any integer is an integer.
Result of Subtracting $3$ : However, when we subtract $3$ from an even integer ( $8k$ ), the result is an odd integer.
Odd Integer Result: An odd integer cannot be divided by $2$ to result in an integer, which means $n$ cannot be an integer if the paired degree is $8k-3$ .
Contradiction of Initial Assumption: This contradicts our initial assumption that the paired degree, which is an even number, can be expressed as $2n$ .
Conclusion: Therefore, the paired degree cannot be equal to $8k-3$ because it would not result in an even number.

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