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User talk:Etmetalakret

Revision as of 19:47, 1 April 2023 by Etmetalakret (talk | contribs)

AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.

Proof 1: The Pythagorean Theorem

Prove the Pythagorean Theorem for a right triangle $\triangle ABC$ such that $\angle BCA = 90^{\circ}$.

Explanation

I showed this proof in AoPS one time. We let $\overline{BP}$ be an altitude of $\triangle ABC$ and hunt for triangle similarity. See the following diagram:

Examine the triangles $\triangle CBP$ and $\triangle ABC$. They both share $\angle B$ and a right angle, so AA Similarity guarantees that $\triangle CBP \sim \triangle ABC$. Similarly, $\triangle ACP \sim \triangle ABC$. We thus get the following ratios: \[\frac{PB}{a} = \frac{a}{c} \textrm{ and } \frac{AP}{b} = \frac{b}{c}.\] We can solve for $PB$ and $AP$ as follows: \[PB = \frac{a^2}{c} \textrm{ and } AP = \frac{b^2}{c}.\] But why is this useful? It's because $AP + PB = c$. Using this fact, we have that \[c = AP + PB = \frac{a^2}{c} + \frac{b^2}{c}.\] Multiplying this equation by $c$ yields the desired $a^2 + b^2 = c^2$.

It's important to acknowledge that the above is not a well-written proof. Your goal in a proof is NOT to show how you found it; it's simply to illustrate why it's true.

Bad Proof

Note that $\angle PBC = \angle CBA = \angle B$ and $\angle BPC = BCA = 90$. We thus have by AA Similarity that $\triangle CBP \sim \triangle ABC$. Similarly, $\triangle ACP \sim \triangle ABC$. Therefore, $\frac{PB}{a} = \frac{a}{c} \textrm{ and } \frac{AP}{b} = \frac{b}{c}.$ We can solve for $PB$ and $AP$ as follows: $PB = \frac{a^2}{c} \textrm{ and } AP = \frac{b^2}{c}.$ Then the following sequence of equations holds: $c = AP + PB = \frac{a^2}{c} + \frac{b^2}{c}.$ Multiplying this equation by $c$ yields the desired $a^2 + b^2 = c^2$. $\square$

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 $a$, $b$, and $c$? What about $\angle B$? You can still use these these instead of writing out $BC$, $CA$, $AB$, and $\angle ABC$, but you need to define them.
  • Not Enough Space: GIVE THESE WALLS OF EQUATIONS THEIR OWN LINES!
  • Degrees Not Specified: The proof references $90^{\circ}$, not whatever $90$ means.

Good Proof

Let $P$ be the point on $\overline{AB}$ such that $\angle BPC = 90^{\circ}$, as shown in the following diagram:

Note that $\angle PBC = \angle CBA$ and $\angle BPC = BCA = 90^{\circ}$. We thus have by AA Similarity that $\triangle CBP \sim \triangle ABC$. Similarly, $\triangle ACP \sim \triangle ABC$. Therefore, \[\frac{PB}{BC} = \frac{CB}{BA} \textrm{ and } \frac{AP}{AC} = \frac{AC}{AB}.\] We can solve for $PB$ and $AP$ as follows: \[PB = \frac{BC^2}{AB} \textrm{ and } AP = \frac{CA^2}{AB}.\] Then the following sequence of equations holds: \[AB = AP + PB = \frac{BC^2}{AB} + \frac{CA^2}{AB} = \frac{BC^2 + CA^2}{AB}.\] Multiplying this equation by $AB$ yields the desired $BC^2 + CA^2 = AB^2$. $\square$

Proof 2: Inequalities

The well-known Trivial Inequality states that if $x$ is a real number, then $x^2 \geq 0$. Prove that if $x$ and $y$ are nonnegative real numbers, then \[\frac{x + y}{2} \geq \sqrt{xy}.\] (Sidenote: this is a very different kind of inequality problem than you're used to. In school, we find when inequalities are true; here, we're showing it's always true.)

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): \begin{align*} \frac{x + y}{2} \geq \sqrt{xy} \\ x + y \geq 2 \sqrt{xy} \\ (x + y)^2 \geq 4xy \\ x^2 + 2xy + y^2 \geq 4xy \\ x^2 - 2xy + y^2 \geq 0 \\ (x - y)^2 \geq 0. \end{align*} 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. Really understand and grasp how I derived this before you read the following proofs:

Bad Proof

I start out with $\frac{x + y}{2} \geq \sqrt{xy}.$ Multiply the inequality by $2$ and square it, $(x + y)^2 \geq 2 \sqrt{xy}$. Letting our algebra go on autopilot, $x^2 + 2xy + y^2 \geq 4xy$ and $x^2 - 2xy + y^2 \geq 0$, so $(x - y)^2 \geq 0$. This is true by Trivial Inequality, which completes the proof. $\square$

Why is this proof bad?

  • Written Backwards: We must always write proofs like: true result $\implies$ desired result. However, the proof is written backwards so that the desired result $\implies$ 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 $2$ 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

By the Trivial Inequality, we have that \[(x - y)^2 \geq 0.\] Factoring this inequality returns $x^2 - 2xy + y^2 \geq 0$. We add $4xy$ to both sides and factor to get $(x + y)^2 \geq 4xy$. Note that because $x$ and $y$ are nonnegative, both sides are nonnegative; we may therefore take the square root of the inequality, which yields \[x + y \geq 2 \sqrt{xy}.\] Finally, dividing both sides by $2$ gives $(x + y) / 2 \geq \sqrt{xy}$, which completes the proof. $\square$

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