Proofs Gone Wrong:Proofs Gone Wrong:

A Lecture OnA Lecture OnMathematical WritingMathematical Writing

CSC373, Winter 2007

Why care about writing?

• Right Idea, bad proof: gets a 50• Your proof is a method of persuading the reader

that something is true. • Errors in logic love to hide behind bad writing.

Toy Problem

• Prove the Full Binary Tree Theorem:

The number of leaves in a non-empty full binary tree is one more than the number of internal nodes.

Right Idea, bad proof Let T be a full binary tree. If a leaf of T is taken away, then it will have

n-1 nodes. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n -1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Who are WE???

• “The reader and I embark on a journey of proving things.”

Where did this come from? Let T be a full binary tree. If a leaf of T is taken away, then it will have

n-1 nodes. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Define all terms/variablesLet T be a full binary tree with n internal nodes. If a leaf of T is taken

away, then it will have one less node. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

What in the world are you trying to do?

Let T be a full binary tree with n internal nodes. If a leaf of T is taken away, then it will have one less node. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Always make it clear what you are trying to do.

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. If a leaf of T is taken away, then it will have one less node. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

What is this pronoun?The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. If a leaf of T is taken away, then it will have one less node. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So it has one less of them.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Logical errors love to hide behind ambiguous pronouns

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. If a leaf of T is taken away, then T will have one less node. But T will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So T has one less internal node.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Are you talking to yourself?The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. If a leaf of T is taken away, then it will have one less node. But it will not be a full-binary tree, so we need to remove leaves x and y. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So T has one less internal node.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Remove everything that is not necessary for the proof to work.

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to take away leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So T has one less internal node.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Put away the thesaurusThe proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to take away leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not touch any internal nodes, but one of them transformed into a leaf. So T has one less internal node.

We have also destroyed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be interpolated that the original T also had one more leaf than internal nodes.

Keep the Vocab SimpleThe proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to remove leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but one of them changed into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be concluded that the original T also had one more leaf than internal nodes.

Who did what now?The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to remove leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but one of them changed into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, it can be concluded that the original T also had one more leaf than internal nodes.

Avoid Passive VoiceThe proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to remove leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but we changed one into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, we can conclude that the original T also had one more leaf than internal nodes.

Avoid uncertain language.The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to remove leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but we changed one into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, we conclude that the original T also had one more leaf than internal nodes.

Logical mistakes always hide behind vagueness

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be a full binary tree with n internal nodes. We need to remove leaves x and y from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but we changed one into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, we conclude that the original T also had one more leaf than internal nodes.

Don’t be vague. If its on purpose, make it explicit

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. We need to remove I’s children from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but we changed one into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, we conclude that the original T also had one more leaf than internal nodes.

This is not programming!The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. We need to remove I’s children from T. Now T has n-1 internal nodes and n leaves. We did not remove any internal nodes, but we changed one into a leaf. So T has one less internal node.

We have also removed two leaves, but we made a new one instead, so the total decreased by one. By the induction hypothesis, the new T has one more leaf than the number of internal nodes. Since the number of internal nodes in T decreased by one and the number of leaves decreased by one as well, we conclude that the original T also had one more leaf than internal nodes.

You can’t redefine variablesThe proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. T’ has n-1 internal nodes and n leaves. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T.

We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes.

How do you know that?The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. T’ has n-1 internal nodes and n leaves. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T.

We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes.

Separate what you are trying to prove from the evidence.

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T. Thus, T’ has n-1 internal nodes.

We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes. Therefore, T has n + 1 leaves.

Random Paragraph Generator?The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T. Thus, T’ has n-1 internal nodes.

We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes. Therefore, T has n + 1 leaves.

Paragraphs are importantThe proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T. Thus, T’ has n-1 internal nodes. We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes. Therefore, T has n + 1 leaves.

Good proofs lead to shorter proofs.

The proof is by induction on the number of internal nodes in the tree. The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. We did not remove any internal nodes, but the node I becomes a leaf in T’. So T’ has one less internal node than T. Thus, T’ has n-1 internal nodes. We have also removed two leaves, but we made a new one instead, so T’ has one less leaf than T. By the induction hypothesis, T’ has n leaves. Since the number of internal nodes in T’ is one less than in T and the number of leaves is also one less than in T, we conclude that the T also had one more leaf than internal nodes. Therefore, T has n + 1 leaves.

Be concise, don’t blabber.The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

Let T be an arbitrary full binary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. We did not remove any internal nodes, but the node I becomes a leaf in T’. Thus, T’ has n-1 internal nodes, and, by the IH, n leaves. However, by the way we constructed T’, it has one less leaf than T. Therefore, T has n + 1 leaves.

Other

• Spelling, English grammar…• Math errors…

– base case?• Rules have exceptions

Correct Proof(100)The proof is by induction on the number of internal nodes in the tree.

The induction hypothesis (IH) is that a non-empty full binary tree with n-1 internal nodes has n leaves.

For the base case, we have n=0. This means that the tree has one leaf and zero internal nodes, proving the IH.

For the general case, assume the IH holds for n-1. Let T be an arbitrary tree with n internal nodes. Let I be an arbitrary internal node whose children are both leaves. Let T’ be the tree obtained by removing I’s children from T. T’ has n-1 internal nodes, and, by the IH, n leaves. However, by the way we constructed T’, it has one less leaf than T. Therefore, T has n + 1 leaves.

By induction, the IH holds for all values of n>0. The statement of the theorem follows.

Summary• Define all terms/variables.• Be concise / don’t be vague / avoid uncertain language.• Always make it clear what you are trying to do.• Remove everything that is not necessary for the proof to work• Avoid ambiguous pronouns.• Separate what you are trying to prove from the evidence.• Don’t redefine variables.• Avoid passive voice.• Use paragraphs to give your argument structure.• Keep the vocab simple.• Use present tense.• Type your proof