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closed as off-topic by arjafi Sep 18 '14 at 10:37

  • This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

8  
I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
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(+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
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At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
3  
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
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This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. – arjafi Sep 18 '14 at 10:37
    
PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07
    
@GraceNote Could you please also change to Community owned all of the posts in our Formatting Sandbox., including deleted posts too (I just got many pings when someone started using a deleted post). Thanks. – Bill Dubuque Dec 24 '15 at 19:50

17 Answers 17

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Common Mistakes
The following are some of the most common mistakes people make when writing mathematical proofs.
$\color{#08F}{\text{1. Arguing from examples.}}$
. . .
$\color{#08F}{\text{2. Using the same letter to mean two different things.}}$
...

$\color{#08F}{\text{3. Jumping to a conclusion.}}$
To jump to a conclusion means to allege the truth of something without giving an adequate reason. Consider the following “proof” that the sum of any two even integers is even.
Suppose m and n are any even integers. By definition of even, m = 2r and n = 2s for some integers r and s. Then m + n = 2r + 2s. So m + n is even. The problem with this “proof” is that the crucial calculation
2r + 2s = 2(r + s)
is missing. The author of the “proof” has jumped prematurely to a conclusion.

$\color{#08F}{\text{4. Circular reasoning.}}$
To engage in circular reasoning means to assume what is to be proved; it is a variation of jumping to a conclusion. As an example, consider the following “proof” of the fact that the product of any two odd integers is odd:
Suppose m and n are any odd integers. When any odd integers are multiplied, their product is odd. Hence mn is odd."

$\color{#08F}{\text{5. Confusion between what is known and what is still to be shown.}}$ A more subtle way to engage in circular reasoning occurs when the conclusion to be shown is restated using a variable.

$\color{#08F}{\text{6. Use of any rather than some.}}$
There are a few situations in which the words any and some can be used interchangeably. For instance, in starting a proof that the square of any odd integer is odd, one could correctly write “Suppose m is any odd integer” or “Suppose m is some odd integer.” In most situations, however, the words any and some are not interchangeable.
Here is the start of a “proof” that the square of any odd integer is odd, which uses any when the correct word is some:
Suppose m is a particular but arbitrarily chosen odd integer.
By definition of odd, m = 2a + 1 for any integer a.
In the second sentence it is incorrect to say that “m = 2a + 1 for any integer a” because a cannot be just “any” integer; in fact, solving m = 2a + 1 for a shows that the only possible value for a is (m − 1)/2. The correct way to finish the second sentence is, “m = 2a + 1 for some integer a” or “there exists an integer a such that m = 2a + 1.

$\color{#08F}{\text{7. Misuse of the word if.}}$
Another common error is not serious in itself, but it reflects imprecise thinking that sometimes leads to problems later in a proof. This error involves using the word if when the word because is really meant. Consider the following proof fragment:
Suppose p is a prime number. If p is prime, then p cannot be written as a product of two smaller positive integers.
The use of the word if in the second sentence is inappropriate. It suggests that the primeness of p is in doubt. But p is known to be prime by the first sentence. It cannot be written as a product of two smaller positive integers because it is prime. Here is a correct version of the fragment:
Suppose p is a prime number. Because p is prime, p cannot be written as a product of two smaller positive integers. ”

Source: Discrete Mathematics with Its Applications, Susanna Epp. p.156-157

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$$\frac{(900-3n)r^{n+1}+(3n-897)r^n-900r+897)}{r^2-2r+1}$$

$$-14100r^{5001}+14103r^{5000}-900r+897+{600000000000r^2-1200000000000r+600000000000}=0$$

$$-14100r^{5001}+14103r^{5000}-900r+600000000000r^2-1200000000900r+600000000897=0$$

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$y = f(x)$ instead of $(x,\space y)\space \in \space f$

$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+ a_0$
$g(x) = b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x+ b_0$

$(f+g)(x)=y$ instead of $(x, y) \in (f+g)$

$f(x) + g(x) = y$ instead of $(x, y) \in (f+g)$

$(f+g)(x)=(a_n +b_n)x^n + (a_{n-1} + b_{n-1})x^{n-1} + \cdots + (a_1+b_1)x+ (a_0 +b_0)$ instead of $[x,\space (a_n +b_n)x^n + (a_{n-1} + b_{n-1})x^{n-1} + \cdots + (a_1+b_1)x+ (a_0 +b_0)] \in (f+g)$

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Für Luise:

$$(a+b)^2\\=(a+b)(a+b)\\=(a+b)a+(a+b)b\\=(a^2+ab)+(ab+b^2)\\=a^2+2ab+b^2$$


$$(a+b)^3\\=(a+b)^2(a+b)\\=(a^2+2ab+b^2)(a+b)\\=(a^2+2ab+b^2)a+(a^2+2ab+b^2)b\\=(a^3+2a^2b+ab^2)+(a^2b+2ab^2+b^3)\\=a^3+3a^2b+3ab^2+b^3$$


$$(a-b)(a+b)\\=(a-b)a+(a-b)b\\=(a^2-ba)+(ab-b^2)\\=a^2-ab+ab-b^2\\=a^2-b^2$$

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available-------------------------------

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$$AP = \lambda BP$$


Notationally, $XY$ will represent the length of the segment $\overline{XY}$ and $XY^2$ means $(XY)^2$. $X-Y-Z$ means that point $Y$ is on the segment $\overline{XZ}$. We say that $Y$ is between $X$ and $Z$. This is true if and only if $XY + YZ = XZ$.

An expression like $A-B-C-D-E$ would then indicate how the points $B$, $C$, and $D$ are positioned on the segment $\overline{AE}$.

We need to show that $\Gamma_{AB}(\lambda) = \{ P \in \mathbb R^2: AP = \lambda BP\}$ is a circle.

$\Gamma_{AB}(1) = \left\{ \dfrac 12(A+B) \right\}$

Since $AP = \lambda BP \iff BP = \dfrac 1\lambda AP$, then we can assume that points $A$ and $B$ have been chosen so that $\lambda \gt 1$.

LEMMA $1$: For $\lambda > 1$, there are two points, $P_L$ and $P_R$, on the line $\overleftrightarrow{AB}$ such that $AP = \lambda BP$. They can be described by

  • $A-P_L-B-P_R$.
  • $AP_L = \dfrac{1}{\lambda+1}AB$.
  • $AP_R = \dfrac{1}{\lambda - 1}AB$.

DEFINITION $2$: Let $\Gamma$ be the circle with diameter $\overline{P_LP_R}$.

We will show that $\Gamma_{AB}(\lambda) = \Gamma$.

We will assume that the following theorem is already known.

THEOREM $3$: $P \not \in \{X, Y\}$ is on the circle with diameter $\overline{XY}$ if and only if $\angle XPY$ is a right angle.

Let's say that a solution, $P$, to $AP = \lambda BP$ that isn't on the line $\overleftrightarrow{AB}$ is a non trivial solution.

then $\Gamma_{AB}(\lambda) = \Gamma$ will follow if we can prove the following theorem.

THEOREM $4$: $P$ is a non trivial solution to $PA = \lambda PB$ if and only if $\angle P_LPP_R$ is a right angle.

PROOF. First we show that $AP = \lambda BP \implies \angle P_LPP_R$ is a right angle.

$\angle P_1PP_2$ is a right angle.

We know that $AP = \lambda BP$ and $AP_L = \lambda BP_L$. Hence $\dfrac{AP}{AP_L} = \dfrac{BP}{BP_L}$. This implies that ray $\overrightarrow{PP_L}$ bisects $\angle APB$. So let $m\angle APP_L = m\angle BPP_L = \theta$.

Let $D$ be the point on $\overline{AP}$ such that $PA::PD = \lambda::1$. Then

\begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{PA}{PB} &\implies PB = PD \\ &\implies \triangle PCB \sim \triangle PCD \\ &\implies m \angle DBP = \theta' = \dfrac{\pi}{2} - \theta \end{align}

Also \begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{P_RA}{P_RB} &\implies \triangle ADB \sim \triangle APP_R &\text{By SAS for similar triangles.}\\ &\implies \overline{BD} \parallel \overline{P_RP} \\ &\implies m\angle BPP_R = m\angle DBP = \theta'\\ &\implies m\angle P_LPB + m\angle BPP_R = \dfrac{\pi}{2} \\ &\implies \angle P_LPP_R \text{ is a right triangle.} \end{align}

Finally, we show that $\angle P_LPP_R$ is a right angle $\implies AP = \lambda BP$.

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