Higher Level only, and the first of three sub-topics on graphs. It is almost entirely vocabulary plus one theorem, which makes it the easiest block in Topic 3 and the one most often under-taught.
Leave view 1 up and make them write a number down. Individually, on paper, before anyone speaks. Then collect the numbers: 6, 7 and 8 all appear in a normal class, and the disagreement is the hook.
Step to view 2. Degrees labelled, sum 14, so 7 edges. Step to view 3, which numbers the edges and confirms it. Two independent routes to the same answer, and only one of them can be miscounted.
If you have five minutes more, redraw the same graph on the board with no crossings at all. It is the same graph, nothing about it has changed, and the edges are now trivial to count. That is the moment the class understands that a drawing is a picture of a graph and not the graph.
| Question | Answer |
|---|---|
| 1. Degree sum of 3, 3, 2, 2, 2, 2 | 14. |
| 2. Edges | 14/2 = 7. |
| 3. Edges in K5 | 5 × 4/2 = 10. |
| 4. Edges in K6 | 6 × 5/2 = 15. |
| 5. Degrees 3, 3, 3, 2, 2, 2 | B. Impossible: the sum is 15, and every degree sum is even. |
Questions 3 and 4 are one apart on purpose. Going from K5 to K6 adds 5 edges, not 1, because the new vertex joins to every vertex already there. A class that notices that has understood the formula rather than memorised it.
1 markThe degrees, read off correctly.
1 markThe handshake lemma used or stated.
1 markThe edge count, or the impossibility with its reason.
On an impossibility question the reason carries the marks. "No" alone scores nothing. Drill the sentence: the degree sum is 15, which is odd, and every degree sum is twice the number of edges and therefore even. Three clauses, and students who write one of the three get partial credit and feel hard done by.
| They wrote | What happened |
|---|---|
| 7 on question 1 | Gave the edge count where the degree sum was asked. The two questions back to back are designed to separate them. |
| 6 on question 1 | Counted the vertices. Fast to fix and worth catching, because it means the word "degree" has not landed. |
| 13 | A miscount, and the student has the tool to detect it themselves: a degree sum cannot be odd. |
| 14 on question 2 | Forgot to halve. Ask how many ends an edge has. |
| 6 or 8 on question 2 | Counted the drawing anyway. Both appear, which is itself the argument against counting. |
| 20 on question 3 | Gave the degree sum of K5. Same omission as 14 on question 2. |
| 25 | n². They have counted every ordered pair and every vertex joined to itself. |
| 11 on question 4 | 10 + 1, treating the sixth vertex as bringing one edge. It brings five. |
| "Yes, 7.5 edges" on question 5 | Worth praising halfway. They have done the right arithmetic and not drawn the conclusion. Half an edge is the contradiction, and naming it is the answer. |
"Does it matter how I draw it?" No, and this is the single most useful thing to establish. Two drawings with different numbers of crossings are the same graph if the same pairs are joined. It is why 3.15 writes graphs as matrices: the matrix has no drawing in it at all.
"What about a loop?" It contributes 2 to the degree of its vertex, because both of its ends are there. Graphs with loops are not simple. Mention it once so the word "simple" has content, and do not build questions on it.
"Why is a tree n − 1 edges?" Build one: start with one vertex and no edges, and every new vertex needs exactly one new edge to join it without creating a cycle. After n vertices you have added n − 1 edges. That construction also shows why one extra edge forces a cycle, which is the fact 3.16 leans on.
"Is the handshake lemma on the formula booklet?" No, and it does not need to be. It is one sentence: every edge has two ends. A student who can say that can rebuild it, and a student who has memorised "sum = 2E" without it will write 2E = 15 and not blink.
| Stage | What to do |
|---|---|
| Demonstrate | Define k(n):=n*(n-1)/2 on the Nspire and tabulate K4 to K8: 6, 10, 15, 21, 28. The differences are 4, 5, 6, 7, which is the new vertex joining to everyone already there. One minute, and the formula stops being arbitrary. |
| Where they stick | Looking for a graph-theory mode. There is not one on either machine, and the Graph application plots functions. Say so at the start of the lesson or several students will spend the lesson hunting through menus instead of drawing. |
| The check | Parity. After any degree sum, ask whether it is even. If it is odd, something has been miscounted, and that is certain rather than likely. It is one-directional: an odd total always means an error, and an even total is not a clean bill of health, since two degrees each misread by 1 in the same direction shift the sum by 2 and so leave the parity intact. |
Nothing here depends on the angle mode. The machine matters from 3.15 onwards, where matrix powers do real work.
| Step | What |
|---|---|
| 1 | Figure view 1. Everyone writes a number. Collect the disagreement. |
| 2 | Degree as a word, on that drawing, vertex by vertex. |
| 3 | View 2: sum 14. Ask why it is twice the edges. "Every edge has two ends." |
| 4 | View 3 to confirm, then redraw the graph with no crossings on the board. |
| 5 | Kn, from K4 up, with the differences noticed. |
| 6 | The impossible degree sequence. This is the highlight; give it its own two minutes. |
| 7 | Vocabulary sweep: simple, weighted, directed, subgraph, tree, cycle, against the drawing already up. |
| 8 | Directed graphs, and the fact that the doubling disappears. |
Do not say "the degree is how many lines come out of the dot". It is nearly right and it makes a loop count once. Say edge-ends, which is both correct and the reason the lemma works.
Do not say "just count the edges carefully". It is the advice this whole page exists to replace. Careful counting of a tangled drawing is slow and unverifiable; the degree sum is fast and self-checking. Teach the method that tells you when you are wrong.