User talk:Etmetalakret
AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.
Contents
Proof 1: The Pythagorean Theorem
Prove the Pythagorean Theorem for a right triangle such that .
Explanation
I showed this proof in the study group one time. We let be an altitude of and hunt for triangle similarity. See the following diagram:
We let , , and .
Examine the triangles and . They both share and a right angle, so AA Similarity guarantees that . Similarly, . We thus get the following ratios: We can solve for and as follows: But why is this useful? It's because . Using this fact, we have that Multiplying this equation by yields the desired .
Bad Proof
Note that and . We thus have by AA Similarity that . Similarly, . Therefore, We can solve for and as follows: Then the following sequence of equations holds: Multiplying this equation by yields the desired .
Why is this proof bad?
- No Diagram: You ALWAYS need a diagram in geometry proofs to help the grader remain oriented in dense notation.
- Terms have not been defined: What are , , and ? What about ? You can still use these these instead of writing out , , , and , but you need to define them.
- Not Enough Space: GIVE THESE WALLS OF EQUATIONS THEIR OWN LINES!
- Degrees Not Specified: The proof references , not whatever means.
Good Proof
Let be the point on such that , as shown in the following diagram:
Note that and . We thus have by AA Similarity that . Similarly, . Therefore, We can solve for and as follows: Then the following sequence of equations holds: Multiplying this equation by yields the desired .
Proof 2: Inequalities
The well-known Trivial Inequality states that if is a real number, then . Prove that if and are nonnegative real numbers, then
Explanation
I found the proof by working backwards; I started with the desired result, and connected it to something true. Here is the wall of equations on my page (sadly I can't get them aligned): Because the left-hand side of this equation is a perfect square, this is actually the Trivial Inequality in disguise. The desired inequality is therefore implied by a true result. We can now write a proof:
Bad Proof
I start out with Multiply the inequality by and square it, . Letting our algebra go on autopilot, and , so . This is true by Trivial Inequality, which completes the proof.
Why is this proof bad?
- Written Backwards: We must always write proofs like: true result desired result. However, the proof is written backwards so that the desired result true result. The Trivial Inequality should be at the start, not the end.
- Informal Word Choice: Please don't use the phrase "algebra autopilot" in a proof, and don't write sentences with no verbs (see the "Multiply the inequality by and square it"). Also, don't use "I," although "we" is totally acceptable.
- Not Enough Space: A little more space would make this proof easier to read. Important equations should have their own line.
Good Proof 1
By the Trivial Inequality, we have that Factoring this inequality returns . We add to both sides and factor to get . Note that because and are nonnegative, both sides are nonnegative; we may therefore take the square root of the inequality, which yields Finally, dividing both sides by gives , which completes the proof.
Good Proof 2
By the Trivial Inequality, we have that Then the following sequence of inequalities holds: Note that because and are nonnegative, both sides of this final inequality are nonnegative; we may therefore take the square root of both sides, which yields Finally, dividing the inequality by gives , which completes the proof.