1. **Problem statement:** Given a $k$-regular graph $G = (V, E)$ where $k \in \mathbb{N}$, prove or disprove that the number of edges $|E|$ is always a multiple of $k$ for even and odd values of $k$. 2. **Recall definitions:** A $k$-regular graph means every vertex has degree $k$. The degree sum formula states: $$\sum_{v \in V} \deg(v) = 2|E|$$ Since $G$ is $k$-regular, each vertex has degree $k$, so: $$\sum_{v \in V} \deg(v) = k|V|$$ 3. **Relate edges and vertices:** Using the degree sum formula: $$k|V| = 2|E| \implies |E| = \frac{k|V|}{2}$$ 4. **Analyze divisibility:** We want to check if $|E|$ is always a multiple of $k$. - For $|E|$ to be a multiple of $k$, there must exist an integer $m$ such that: $$|E| = km$$ - Substitute $|E|$: $$\frac{k|V|}{2} = km \implies \frac{|V|}{2} = m$$ - This means $\frac{|V|}{2}$ must be an integer, so $|V|$ must be even. 5. **Conclusion for even and odd $k$:** - The number of edges $|E|$ is a multiple of $k$ if and only if $|V|$ is even. - Since $|V|$ can be even or odd depending on the graph, $|E|$ is not always a multiple of $k$. 6. **Example to disprove:** - Let $k=3$ (odd), and $|V|=4$ (even): $$|E| = \frac{3 \times 4}{2} = 6$$ - $6$ is a multiple of $3$, so holds here. - Let $k=3$ and $|V|=3$ (odd): $$|E| = \frac{3 \times 3}{2} = \frac{9}{2} = 4.5$$ - Not an integer, so such a graph cannot exist (since edges count must be integer). - For $k$-regular graphs, $k|V|$ must be even (since sum of degrees is even), so $k|V|$ is even. - If $k$ is even, $k|V|$ is even regardless of $|V|$. - If $k$ is odd, $|V|$ must be even for $k|V|$ to be even. Therefore: - For even $k$, $|E| = \frac{k|V|}{2}$ is always an integer, but not necessarily a multiple of $k$. - For odd $k$, $|V|$ must be even, so $|E|$ is integer, but again not necessarily a multiple of $k$. 7. **Final check if $|E|$ is multiple of $k$:** - $|E| = \frac{k|V|}{2}$ - For $|E|$ to be multiple of $k$, $\frac{|V|}{2}$ must be integer. - So if $|V|$ is even, $|E|$ is multiple of $k$. - If $|V|$ is odd, $|E|$ is not integer for odd $k$, so graph cannot exist. **Summary:** - The number of edges $|E|$ in a $k$-regular graph is always an integer. - $|E|$ is a multiple of $k$ if and only if $|V|$ is even. - For odd $k$, $|V|$ must be even for the graph to exist. Hence, the statement "the number of edges in $G$ is always a multiple of $k$" is **true if and only if** $|V|$ is even, otherwise false. **Answer:** The number of edges in a $k$-regular graph is not always a multiple of $k$; it depends on the parity of the number of vertices $|V|$.