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| − | AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.
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| − | == Proof 1: The Pythagorean Theorem ==
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| − | Prove the Pythagorean Theorem for a right triangle <math>\triangle ABC</math> such that <math>\angle BCA = 90^{\circ}</math>.
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| − | === Explanation ===
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| − | I showed this proof in the study group one time. We let <math>\overline{BP}</math> be an altitude of <math>\triangle ABC</math> and hunt for triangle similarity. See the following diagram:
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| − | [[Image:Screenshot 2023-04-01 204908.png|thumb|none|600px]]
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| − | We let <math>BC = a</math>, <math>CA = b</math>, and <math>AB = c</math>.
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| − | Examine the triangles <math>\triangle CBP</math> and <math>\triangle ABC</math>. They both share <math>\angle B</math> and a right angle, so AA Similarity guarantees that <math>\triangle CBP \sim \triangle ABC</math>. Similarly, <math>\triangle ACP \sim \triangle ABC</math>. We thus get the following ratios: <cmath>\frac{PB}{a} = \frac{a}{c} \textrm{ and } \frac{AP}{b} = \frac{b}{c}.</cmath> We can solve for <math>PB</math> and <math>AP</math> as follows: <cmath>PB = \frac{a^2}{c} \textrm{ and } AP = \frac{b^2}{c}.</cmath> But why is this useful? It's because <math>AP + PB = c</math>. Using this fact, we have that <cmath>c = AP + PB = \frac{a^2}{c} + \frac{b^2}{c} = \frac{a^2 + b^2}{c}.</cmath> Multiplying this equation by <math>c</math> yields the desired <math>a^2 + b^2 = c^2</math>.
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| − | === Bad Proof ===
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| − | Note that <math>\angle PBC = \angle CBA = \angle B</math> and <math>\angle BPC = BCA = 90</math>. We thus have by AA Similarity that <math>\triangle CBP \sim \triangle ABC</math>. Similarly, <math>\triangle ACP \sim \triangle ABC</math>. Therefore, <math>\frac{PB}{a} = \frac{a}{c} \textrm{ and } \frac{AP}{b} = \frac{b}{c}.</math> We can solve for <math>PB</math> and <math>AP</math> as follows: <math>PB = \frac{a^2}{c} \textrm{ and } AP = \frac{b^2}{c}.</math> Then the following sequence of equations holds: <math>c = AP + PB = \frac{a^2}{c} + \frac{b^2}{c} + \frac{a^2 + b^2}{c}.</math> Multiplying this equation by <math>c</math> yields the desired <math>a^2 + b^2 = c^2</math>. <math>\square</math>
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| − | '''Why is this proof bad?'''
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| − | * '''No Diagram''': You ALWAYS need a diagram in geometry proofs to help the grader remain oriented in dense notation.
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| − | * '''Terms have not been defined''': What are <math>a</math>, <math>b</math>, and <math>c</math>? What about <math>\angle B</math>? You can still use these these instead of writing out <math>BC</math>, <math>CA</math>, <math>AB</math>, and <math>\angle ABC</math>, but you need to define them.
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| − | * '''Not Enough Space''': GIVE THESE WALLS OF EQUATIONS THEIR OWN LINES!
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| − | * '''Degrees Not Specified''': The proof references <math>90^{\circ}</math>, not whatever <math>90</math> means.
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| − | === Good Proof ===
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| − | Let <math>P</math> be the point on <math>\overline{AB}</math> such that <math>\angle BPC = 90^{\circ}</math>, as shown in the following diagram:
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| − | [[Image:Screenshot 2023-04-01 204908.png|thumb|none|600px]]
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| − | Note that <math>\angle PBC = \angle CBA</math> and <math>\angle BPC = BCA = 90^{\circ}</math>. We thus have by AA Similarity that <math>\triangle CBP \sim \triangle ABC</math>. Similarly, <math>\triangle ACP \sim \triangle ABC</math>. Therefore, <cmath>\frac{PB}{BC} = \frac{CB}{BA} \textrm{ and } \frac{AP}{AC} = \frac{AC}{AB}.</cmath> We can solve for <math>PB</math> and <math>AP</math> as follows: <cmath>PB = \frac{BC^2}{AB} \textrm{ and } AP = \frac{CA^2}{AB}.</cmath> Then the following sequence of equations holds: <cmath>AB = AP + PB = \frac{BC^2}{AB} + \frac{CA^2}{AB} = \frac{BC^2 + CA^2}{AB}.</cmath> Multiplying this equation by <math>AB</math> yields the desired <math>BC^2 + CA^2 = AB^2</math>. <math>\square</math>
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| − | == Proof 2: Inequalities ==
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| − | The well-known '''Trivial Inequality''' states that if <math>x</math> is a real number, then <math>x^2 \geq 0</math>. Prove that if <math>x</math> and <math>y</math> are nonnegative real numbers, then <cmath>\frac{x + y}{2} \geq \sqrt{xy}.</cmath>
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| − | === Explanation ===
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| − | 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):
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| − | <cmath>\begin{align*}
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| − | \frac{x + y}{2} \geq \sqrt{xy} \\
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| − | x + y \geq 2 \sqrt{xy} \\
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| − | (x + y)^2 \geq 4xy \\
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| − | x^2 + 2xy + y^2 \geq 4xy \\
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| − | x^2 - 2xy + y^2 \geq 0 \\
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| − | (x - y)^2 \geq 0.
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| − | \end{align*}</cmath>
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| − | 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:
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| − | === Bad Proof ===
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| − | I start out with <math>\frac{x + y}{2} \geq \sqrt{xy}.</math> Multiply the inequality by <math>2</math> and square it, <math>(x + y)^2 \geq 2 \sqrt{xy}</math>. Letting our algebra go on autopilot, <math>x^2 + 2xy + y^2 \geq 4xy</math> and <math>x^2 - 2xy + y^2 \geq 0</math>, so <math>(x - y)^2 \geq 0</math>. This is true by Trivial Inequality, which completes the proof. <math>\square</math>
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| − | '''Why is this proof bad?'''
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| − | * '''Written Backwards''': We must always write proofs like: true result <math>\implies</math> desired result. However, the proof is written backwards so that the desired result <math>\implies</math> true result. The Trivial Inequality should be at the ''start'', not the end.
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| − | * '''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 <math>2</math> and square it"). Also, don't use "I," although ''"we" is totally acceptable''.
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| − | * '''Not Enough Space''': A little more space would make this proof easier to read. Important equations should have their own line.
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| − | === Good Proof 1 ===
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| − | By the Trivial Inequality, we have that <cmath>(x - y)^2 \geq 0.</cmath> Factoring this inequality returns <math>x^2 - 2xy + y^2 \geq 0</math>. We add <math>4xy</math> to both sides and factor to get <math>(x + y)^2 \geq 4xy</math>. Note that because <math>x</math> and <math>y</math> are nonnegative, both sides are nonnegative; we may therefore take the square root of the inequality, which yields <cmath>x + y \geq 2 \sqrt{xy}.</cmath> Finally, dividing both sides by <math>2</math> gives <math>(x + y) / 2 \geq \sqrt{xy}</math>, which completes the proof. <math>\square</math>
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| − | === Good Proof 2 ===
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| − | By the Trivial Inequality, we have that <cmath>(x - y)^2 \geq 0.</cmath> Then the following sequence of inequalities holds:
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| − | <cmath>
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| − | \begin{align*}
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| − | x^2 - 2xy + y^2 \geq 0 \\
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| − | x^2 + 2xy + y^2 \geq 4xy \\
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| − | (x + y)^2 \geq 4xy.
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| − | \end{align*}</cmath>
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| − | Note that because <math>x</math> and <math>y</math> are nonnegative, both sides of this final inequality are nonnegative; we may therefore take the square root of both sides, which yields <cmath>x + y \geq 2 \sqrt{xy}.</cmath> Finally, dividing the inequality by <math>2</math> gives <math>(x + y) / 2 \geq \sqrt{xy}</math>, which completes the proof. <math>\square</math>
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