# Easy Money

On a table before you lie two envelopes, identical in appearance. You are told that one envelope contains twice as much money as the other, but given no indication which has more. You are then allowed to choose one of the envelopes and keep the money it contains.

You then pick one of the envelopes at random, but before you look inside you are offered the chance to exchange your envelope for the other. Should you exchange envelopes, (assuming that the more money you get, the better)?

One line of thinking is this: If the envelope you initially choose has $$x$$ dollars in it, then there is a $$50$$% chance that the other envelope has $$2x$$ dollars in it and a $$50$$% chance that it has $$\frac{x}{2}$$ dollars in it. Thus the expected amount of money you would end up with if you chose to exchange envelopes would be

$$\frac{1}{2} * 2x + \frac{1}{2} * \frac{x}{2} = \frac{5}{4} x$$ dollars.

You thus decide to exchange envelopes. But before doing so, you think to yourself, "But I could go through the same thought process with the other envelope, concluding that I should exchange my new envelope for my old one, and then go through the same thought process again, and again, ad infinitum. Now I have no idea what to do."

How can you resolve this paradox?

Edit: As mentioned in my comment below, the real "puzzle" here is to identify the flaw in the reasoning I outlined above. Or is there a flaw?

Note by Brian Charlesworth
4 years, 4 months ago

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First, look at the case when the envelope is exchanged. If you pick the envelope with $$x$$ dollars $$(P=1/2)$$ and exchange, you win $$2x$$ dollars. Instead, if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and exchange, then you win $$x$$ dollars.

$\therefore E(\text{Money when envelope is exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$

Now, look at the case when the envelope is not exchanged. If you pick the envelope with $$x$$dollars $$(P=1/2)$$ and don't exchange, you win $$x$$ dollars. But if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and don't exchange, then you win $$2x$$ dollars.

$\therefore E(\text{Money when envelope is not exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$

So it makes no difference at all whether you exchange or don't exchange your envelope.

- 4 years, 4 months ago

Yes, that is a good analysis. Since we are not given any further information prior to making a choice whether or not to exchange envelopes, we are in the same position as we were with our initial choice, i.e., our expected winnings will be $$\frac{3x}{2}$$.

But the "real" puzzle here is to find the logical miscue in the line of reasoning I laid out. It can be argued that in the two scenarios I describe, (namely that I start out with the lesser amount of money or the greater), the value $$x$$ actually represents different amounts in each case, so that I can't use the same variable $$x$$ for each of them, making my equation invalid.

However, suppose I did peek at the amount in the envelope before making a decision. Now this provides me with "information", but in reality it's useless since I have no clue as to whether it's the lesser or greater amount. But since we now can attach an actual value to $$x$$, the mathematics I applied in my initial analysis becomes re-validated, and hence so does my "conclusion". But since, as noted, the information regarding the amount in the initially chosen envelope is in reality useless, should it matter whether we know the actual amount as far as the mathematical analysis is concerned? From this perspective, no matter what amount we see inside the initial envelope we should conclude that switching envelopes is preferable.

So the puzzle of "where's the flaw in logic?" is more a question of perspective; are we dealing with unconditional or conditional probability values, and why should this difference in perspective matter given the above discussion? A surprising amount of attention has been given to this paradox, commonly referred to as the "Two Envelope Paradox", with no universally accepted resolution as of yet. I think why it does garner such attention is that it cuts to the heart of the foundation of probability theory, and as with any foundation, the desire is that it must be as secure as possible, and if it is not, it must be either fortified or torn down.

Here is an accessible discussion on the "search for the flaw" and why many find it so intriguing.

- 4 years, 4 months ago

I'm not a mathematician, just a dilettante of sorts but something tells me we're over complicating this paradox. There are only two choices and only two possibilities. The money is either greater or lesser and you either switch or stay. The fact that the sum of all probabilities is 1 is infallible. We simply assume we have knowledge we previously didn't have and compute what the probabilities are to find out what is what.

Here's another question: Are we then told that "hey, you chose the lesser or greater amount of money" after we've made our choice? I think not. So the probability will not even depend on outcome. For all intents and purposes, we need not even bother with the maths of 'x' and '2x'; and we might as well adopt an approach with x and y since the relationship between the monies simply do not matter. I might as well be a choice between winning the lottery and detonating an atomic bomb.

- 4 years ago

I think that the reasons this "paradox" has received so much attention over the years are more esoteric in nature than practical. For some it raises questions about the foundation of probability theory regarding the absolute distinction between conditional/unconditional scenarios and whether or not there might be a grey area there. However, mathematicians do have a tendency to "overcomplicate" things, and this may a case in point. :)

As for the choice between winning the lottery and detonating an atomic bomb, well, I think I would just walk away from the table; too many lives other than my own would be at stake without them having any say in the matter. However, if it were a choice between my eternal health, wealth and happiness on the one hand and my immediate extinction on the other, I would give it a bit more thought. I would probably still walk away, since it would feel like I'm making some kind of deal with the Devil, but it does make for an interesting psychological experiment. This is an entirely different question than the previous "paradox" we were dealing with; with the money in the two envelopes, no matter what choice we made we would be richer than before. With this new question, there is the possibility of a terrible outcome, so the first choice becomes whether to engage at all, and if we do then why do we think it is worth such a high risk of calamity. The more we would have to lose in our present life the less likely we would engage, but if we did not have much to live for then the risk might be worth it. However, as I said before, only the Devil would make such a proposition, so anyone, no matter what their circumstances, would be wise to just walk away. Sorry, I do ramble on, but i always do that when a good philosophical question happens by. :)

- 4 years ago

Hahaha.I do love to read you rambling on. It's more engaging than many other types of conversations. Your ending pretty much wrapped up my premise: which is that, the paradox seems to rely on the probability of making a more money outcome which is identical to that of making a lesser money. I do have some philosophical questions if you're interested, I'd love to read your views. I'm relatively new on this site so is it OK if I create a note and invite you to give your thoughts on it?

- 4 years ago

Of course; I look forward to seeing what the topic is. :)

- 4 years ago

Hi again Mr. Charlesworth

Here are two notes I have posted. I invite you to share your thoughts on them: https://brilliant.org/discussions/thread/number-bases-in-a-different-world/# https://brilliant.org/discussions/thread/the-infiniteness-of-light/?ref_id=654556

Hi @Calvin Lin. Do you think it's a good idea to find a way to incorporate Philosophy (Logic and Reasoning) on Brilliant? I think it is a very fascinating pseudoscience and it would appeal to many older people and others who are not strictly geeks and nerds like the rest of us. Just a thought

- 4 years ago

Let's incorporate Speculative Science And Math, as distinct from formal Philosophy. You know, something that would appeal to one old guy here who wants to try his hand in being a crank.

I'm reading Brian's posted scenario, and it seems to me that if it's known that one envelope contains twice the other, then the expected value in choosing either is simply $$\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } 2x=\dfrac { 3 }{ 2 } x$$ either way, as pointed out by Pratik. It's 50% percent chance that it's twice the one you've chosen, and 50% percent chance it's half that--but you can't assume the same $$x$$ dollars in your chosen envelope in both cases! That was the flaw. Maybe Brian is trying to alluding to the problem of trying to determine probabilities when information is incomplete, which is what led to Bayesian probability theory. Let me offer a slightly different scenario. You are given $$2$$ envelopes. You are told that one of them may or may not contain a valuable stock certificate. You open one up and do not find it inside. What's the probability the other one does? Extend that to $$100$$ such envelopes. This is actually what happens when you are looking for gold in the ground in a certain area when you don't have much information to go on. After so much digging, at what point in time do you give up? Are more likely to find gold, because you've already almost eliminated the places where it can be, or are you more less likely to find gold, because you've already tried to find it and failed? A lot of science can be like that.

- 4 years ago

Oh yes, that's a great idea because what I was hoping for was an avenue to wonder beyond the limits of our recorded knowledge and rub minds on what could be and what not.

I'm fascinated this "blind" probability theory. It is a little bit anti cognitive. Which reminds me of something else which defies our flawed cognizance. Consider flipping a coin and getting heads five times in a row. Flipping it a sixth time seems as though there is a higher probability to get tails. In fact, I'd bet that if you did a survey of ten mathematics professors, a higher percentage of them would choose tails for that sixth coin toss except they are just fighting against their human nature. But sometimes I think about it and really wonder if the law of averages is that subjective. If you mix two liquids together, they mix together perfectly EVERY TIME if they are of similar viscosity. I'm rambling. I have even forgotten what the point I'm trying to make is

- 4 years ago

A lot of theoretical physics was minds rambling before something more concrete came out of it. Ramble on.

Likewise, a lot of technology was preceded by science fiction. Smart phones? Didn't Star Trek popularize that idea?

- 4 years ago


# 4x4 Matrix with rank B=4 and B^2=3

#### Petrus

##### Well-known member
Hello MHB,
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$ and B also have to be a upper/lower or identity Matrix. And this Will not work.. I am wrong or can I explain this in a better way?

Regards,
$$\displaystyle |\pi\rangle$$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.

#### Petrus

##### Well-known member
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.
Hello Evgeny.Makarov,
thanks for fast respond and I meant that! And thanks for showing me this I did not know that $\det(B^2)=(\det(B))^2$ That Was what I Was looking for

Regards,
$$\displaystyle |\pi\rangle$$

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
An alternative proof (without using determinants):

Consider the linear map $B:\mathbb{R}^4\to \mathbb{R}^4,\; x\to Bx.$ As $\operatorname{rank}B=4,$ $\operatorname{nullity}B=0,$ wich implies $B$ is bijective. But the composition of bijective maps is bijective, so $\operatorname{rank}B^2=4.$

#### Deveno

##### Well-known member
MHB Math Scholar
Another formulation:

As rank(B) = 4, B is surjective, that is, B(R4) = R4 (for this is what rank means: the dimension of the image space (or column space) of B, and R4 is the ONLY 4-dimensional subspace of R4).

Consequently, B2(R4) = B(B(R4)) = B(R4) = R4, from which we conclude B2 is likewise surjective, and thus rank(B2) = 4 as well.

(I only post this to indicate one need not even invoke the rank-nullity theorem).

#### Petrus

##### Well-known member
this is impossible because $\det(B^2)=(\det(B))^2$.
Now that I think about I remember a sats that said $$\displaystyle |AB|=|A||B|$$ but in this case $$\displaystyle A=B$$ hmm I need to find the proof for this.

Regards,
$$\displaystyle |\pi\rangle$$

MHB Math Scholar

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I saw one neat proof of $\det(AB)=\det(A)\det(B)$ in Linear Algebra and Its Applications by Gilbert Strang. He defines $\det(\cdot)$ as a function satisfying three properties:

(1) $\det(I)=1$ where $I$ is the identity matrix;
(2) it changes sign when two adjacent rows are swapped;
(3) it is linear on the first row.

Signed volume in an orthonormal basis satisfies these properties, so this definition is much more intuitive than the Leibniz formula, which is derivable from (1)–(3).

Now, to prove that $\det(AB)=\det(A)\det(B)$, fix $B$ and consider $d(A)=\det(AB)/\det(B)$. It is possible to show that $d(A)$ satisfies (1)–(3), and so $d(A)=\det(A)$.

Now that I looked at the StackExchange link, this is answer #2, which is highest-ranked.


# Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$?

Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?

• Unlikely; see the related math.stackexchange.com/q/384520/73324 – vadim123 Jun 7 '13 at 18:47
• @Martin, is the sequence formed by these sums not a sequence? – Antonio Vargas Dec 1 '16 at 17:38
• @AntonioVargas The same could be said about any finite sum of the form $\sum_{k=1}^n a_n$. Of course, if you think that for some reasone the tag belongs there, you can edit it back. My understanding is that the tag-excerpt says rather clearly: "For questions on finite sums use the (summation) instead." – Martin Sleziak Dec 1 '16 at 17:41

To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that $$a(n+1)-a(n) = 1 + \sum_{k=1}^n \left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\= 1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right) = \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\= 2\sigma(n+1)-\tau(n+1).$$

It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) = \sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$ We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series $$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} = 2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$ We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$ so that by the aforementioned theorem, $$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$ In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting $$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) + \frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$ which yields $$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$ This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$

This is A222548 in the online encyclopedia of integer sequences. They don't provide a closed form, but they do give the following:

$$a(n)\approx\frac{\pi^2}{6}n^2+O(n\log n)$$

Let me complement @Marko Riedel's excellent answer by providing a low-tech approach to the leading term.

Let us denote by $\mathbf{1}\{\cdot\}$ the indicator function of the set $\{\cdot\}$. Then using the identity

$$\lfloor n/i \rfloor = \sum_{j=1}^{n} \mathbf{1}\{ij \leq n \}$$

we can rearrange the sum to find:

\begin{align*} a(n) &= \sum_{i=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \mathbf{1}\{ij \leq n\}\mathbf{1}\{ik \leq n\} = \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \sum_{i=1}^{n} \mathbf{1}\Big\{i \leq \frac{n}{\max\{j,k\}}\Big\} \\ &= \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{\max\{j,k\}} \right\rfloor = \sum_{l=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{l} \right\rfloor \mathbf{1}\{\max\{j,k\} = l\} \\ &= \sum_{l=1}^{n} (2l-1) \left\lfloor \frac{n}{l} \right\rfloor. \end{align*}

Dividing both sides by $n^2$, the RHS can be identified as Riemann sum and thus

$$\frac{a(n)}{n^2} = \sum_{l=1}^{n} \frac{2l-1}{n} \left\lfloor \frac{n}{l} \right\rfloor \frac{1}{n} \xrightarrow[\ n\to\infty \ ]{} \int_{0}^{1} 2x \left\lfloor\frac{1}{x}\right\rfloor \, dx = \zeta(2).$$

Here, the last integral can be easily computed by applying the substitution $x \mapsto 1/x$.

An Answer Without Number Theoretic Functions

Break up the sum as follows: $$\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 =\sum_{k=1}^n\frac{n^2}{k^2}-\sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right)\tag{1}$$ We can apply the Euler-Maclaurin Sum Formula $$\sum_{k=1}^n\frac{n^2}{k^2}\sim n^2\left(\frac{\pi^2}6-\frac1n+\frac1{2n^2}\right)\tag{2}$$ If we note that $\left\{\frac nk\right\}\lt1-\frac1k$, we see that \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\le\sum_{k=1}^n\left(1-\frac1k\right)2\frac nk\\[9pt] &\le2n\log(n)\tag{3} \end{align} Therefore, by $(1)$, $(2)$, and $(3)$ we have $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2=n^2\frac{\pi^2}6+O(n\log(n))}\tag{4}$$ However, if we use $\frac12{-}\frac1{2k}$ for the mean value of $\left\{\frac nk\right\}$ and $\frac13{-}\frac1{2k}{+}\frac1{6k^2}$ for the mean value of $\left\{\frac nk\right\}^2$ \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\approx\sum_{k=1}^n\left(\frac12-\frac1{2k}\right)2\frac{n}{k}-\left(\frac13-\frac1{2k}+\frac1{6k^2}\right)\\ &=n\log(n)+n\left(\gamma-\frac{\pi^2}6-\frac13\right)+O(\log(n))\tag{5} \end{align} Thus, if we look at the average behavior, we can use $(1)$, $(2)$, and approximation $(5)$ to get $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 \approx n^2\frac{\pi^2}6-n\log(n)+n\left(\frac{\pi^2}6-\gamma-\frac23\right)}\tag{6}$$

The value given by $(6)$ for $n=98$ is $15387.9230$ while the actual value is $15381$.
The value given by $(6)$ for $n=1001$ is $1641711.3692$ while the actual value is $1641772$.

The error of the approximation in $(6)$ appears to be proportional to $n$. Here is a plot of that error, for $1\le n\le3000$, divided by $n$:

The error divided by $n$ has a mean of $-0.0746805451$ and a standard deviation of $0.6513592109$.

Using number theoretic functions, Marko Riedel's answer gives a better approximation. Here is a plot of the error of that approximation, for $1\le n\le3000$, divided by $n$:
The error divided by $n$ has a mean of $0.0060900069$ and a standard deviation of $0.3674311090$. This is better than the error distribution in my answer, but it does require factoring $n$ to compute the number theoretic functions.


# Does there exist more than 3 connected open sets in the plane with the same boundary?

I've wondered about the following question, whose answer is perhaps well known (in this case I apologize in advance).

The Lakes of Wada are a famous example of three disjoint connected open sets of the plane with the counterintuitive property that they all have the same boundary (!)

My question is the following :

Can we find four disjoint connected open sets of the plane that have the same boundary?

More generally :

For each $n \geq 3$, does there exist $n$ disjoint connected open sets of the plane that have the same boundary? If not, then what is the smallest $n$ such that the answer is no?

Thank you, Malik

• The article you link to says "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, ...and so on.", so the answer seems to be "you can do it for all $n$ including $n =\aleph_0$". I'd guess (but have no idea really) that the corresponding question for other cardinalities is independent of ZFC, assuming NOT CH. Though, since $\mathbb{R}^n$ is first countable, there may be an answer... – Jason DeVito Mar 19 '12 at 15:55
• Found this in the link. It seems to answer your question. "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order ..." – Patrick Mar 19 '12 at 15:57
• Given that the basins of attraction of Newton's method for $z^3=1$ are an example for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. – lhf Mar 19 '12 at 16:49
• Jason, separability precludes finding any uncountable collection of pairwise disjoint nonempty open sets, irrespective of any conditions you put on their boundaries. – user83827 Mar 19 '12 at 17:23
• @Student73, ah, this may contain a proof: Frame, Michael; Neger, Nial. Newton's method and the Wada property: a graphical approach. College Math. J. 38 (2007), no. 3, 192–204. MR2310015 (2008e:37082) – lhf Mar 19 '12 at 17:31

Given that the basins of attraction of Newton's method for $z^3=1$ are Wada sets for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. I don't know a reference to a proof but try these:


# Thread: s4.13.5.41 relation of 2 planes

1. $\tiny{{s4}.{13}.{5}.{41}}$

$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}

$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$

2. Originally Posted by karush
$\tiny{{s4}.{13}.{5}.{41}}$

$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}

$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$
Everything you've posted is fine. You can give a plane any name you like, but I would write something like this:

\displaystyle \begin{align*} P_1 : x + z = 1 \end{align*}

That way we can see that \displaystyle \begin{align*} P_1 \end{align*} is DEFINED as the relationship "the sum of the x and z values needs to be 1", not that it is some variable that has something to do with the equation.

3. The problem asks you to do three things:
1) determine if the planes are parallel.
2) determine if the planes are perpendicular.
3) if neither of those, determine the angle of intersection of the two planes.

Yes, by using that formula to determine the angle, you can then answer all three questions but it should be simpler to determine the first two without using that formula:

The two planes are parallel if and only if the two normal vectors are parallel- if one is a multiple of the other.

The two planes are perpendicular if and only if the two normal vectors are perpendicular: if their dot product is 0.

If neither of those is true, then you can use the dot product you found as the numerator in the formula to determine the angle.

hard to see that in their examples

the next 2 problems are probably
$\displaystyle \parallel , \perp$

$\tiny{s4.854.13.5.43}$
$\textsf{Determine if the 2 given planes are perpendicular, parallel or at an angle to each other}$
\begin{align}
\displaystyle
{p_1}&:{x+4y-3z=1}
\therefore n_1=\langle 1,4,-3 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+7z=0} \therefore n_2=\langle -3,6,7 \rangle
\end{align}
\begin{align}
\displaystyle
\cos(\theta)&= \frac{n_1\cdot n_2}{|n_1||n_2|}=0\\
\therefore p_1 &\perp p_2
\end{align}

$\tiny{s4.854.13.5.45}$
\begin{align}
\displaystyle
{p_1}&:{2x+4y-2z=1}
\therefore n_1=\langle 2,4,-2 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+3z=0} \therefore n_2=\langle -3,6,7 \rangle \\
n_1&=-\frac{2}{3} n_2 \\
&\therefore n_1\parallel n_2
\end{align}

$\textit{btw what is the input string to see the 2 plane on a W|A graph?}$

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# Probability of rolling a dice 8 times before all numbers are shown.

What is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?

I don't really have a clue how to work this out.

Edit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated. So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936. Is that right? Then 1 minus this is the probability?

Edit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls

Six numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number 6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7. Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054 Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946

• Let $p$ be the probability that in $7$ rolls you see all faces at least once. We find $p$. Then the answer to the original problem is $1-p$. It should not be too hard to find $p$, by counting the "favourables" and dividing by $6^7$. – André Nicolas Aug 6 '15 at 13:34
• As a further hint to counting the "favorables" for $p$, note that if you roll 7 times you will see at least one number repeated. But if this happens more than once you will not get to see all 6 numbers. – David K Aug 6 '15 at 13:50
• The title and text of the question don't match. Please clarify whether "at least" is intended, as in the text, or "exactly", as the title appears to imply. – joriki Aug 6 '15 at 14:17
• To clarify are you asking what is the probability of needing exactly 8 rolls to see all numbers or 8 or more rolls? – Warren Hill Aug 7 '15 at 8:20

Rephrase the question:

What is the probability of not seeing all $6$ values when rolling a die $7$ times?

Find the probability of the complementary event:

What is the probability of seeing all $6$ values when rolling a die $7$ times?

Use the inclusion/exclusion principle in order to count the number of ways:

• Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\binom66\cdot6^7$
• Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\binom65\cdot5^7$
• Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\binom64\cdot4^7$
• Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\binom63\cdot3^7$
• Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\binom62\cdot2^7$
• Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\binom61\cdot1^7$

Divide the result by the total number of ways, which is $6^7$:

$$\frac{\binom66\cdot6^7-\binom65\cdot5^7+\binom64\cdot4^7-\binom63\cdot3^7+\binom62\cdot2^7-\binom61\cdot1^7}{6^7}=\frac{35}{648}$$

Calculate the probability of the original event by subtracting the result from $1$:

$$1-\frac{35}{648}=\frac{613}{648}\approx94.6\%$$

• Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 ± 0.001, compared to exact 35/648 = 0.0540 in this Answer. – BruceET Aug 7 '15 at 8:46
• @BruceTrumbo: Thanks :) – barak manos Aug 7 '15 at 8:50

(Edit: Answer is for the probability of seeing all the numbers for $n$ rolls)

For exactly $n$ rolls, the problem can be solved using a markov chain

\begin{align*} A&= \begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0\cr 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0\cr 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0\cr 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0\cr 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\cr 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix} \end{align*}

where the rows and columns are the number of faces of the die seen.

The probability of seeing all the faces in exactly 8 rolls is $(A^8)[0,6]$, which is $\dfrac{665}{5832}\approx 0.11402606310014$

For any $n$, it can be found by finding the generating function $G(z)$ and in turn finding the coefficient of $z^n$

\begin{align*} \mathbb{P}(n) = 1-\frac{20}{2^n}+15\left(\frac{2}{3}\right)^n+\frac{15}{3^n}-6\left(\frac{5}{6}\right)^n-\frac{6}{6^n} \end{align*}

• That was my original solution (should be $1-\frac{665}{5832}$, BTW)... But I then realized that "at least 8 times before you get to see all of the numbers" meant that you were allowed to see all of the numbers by the $8$th trial, so you need to replace $8$ with $7$ in your answer. – barak manos Aug 7 '15 at 12:55
• Oh, okay. I solved for the probability of seeing all the numbers for n rolls. – gar Aug 7 '15 at 13:24
• Very elegant. Clarity of presentation could be improved for quick reading (and more upvotes). I suggest to: make the first sentence bold, correcting for n=7 so that the answers match, small subtitle "general solution" at the end. – Symeof Sep 23 '15 at 21:36

Perhaps this is of some help, even if not your final answer.

If a fair die is rolled 8 times and $X$ is the number of different faces seen, then simulation of a million performances of the 8-roll experiment shows the following approximate distribution of $X$, in which the margin of simulation error is about $\pm 0.001.$

         1        2        3        4        5        6
0.000003 0.002275 0.069393 0.363681 0.450468 0.114180

Of course, $P(X = 1) = 1/6^7 = 0.000004$ to 6 places. To get $P(X = 6),$ consider that one face may be seen three times, or two different faces may each be seen twice. A straightforward computation of other probabilities in the distribution seems feasible but increasingly tedious. Perhaps there are clever combinatorial methods that make some of them easier.

However, pending clarification, I'm still not sure if $P(X = 6)$ is exactly the probability you're trying to find.

Notes:

(a) For a 7-roll experiment an analogous simulation gives $0.0544 \pm 0.001,$ which does not agree with your result $126/279936 \approx 0.00045.$ (I do not believe you are considering all possible permutations of the numbers that occur only once.)

(b) For a 6-roll experiment a simulation gives $0.015496 \pm 0.001,$ compared with the exact $6!/6^6 = 0.0154321.$ Clearly, the probability of getting all 6 faces is 7 rolls must be larger than that.

(c) A related problem is the number of rolls $W$ required before all 6 faces are seen. Then $E(W) = 6/6 + 6/5 + \dots + 6/1 = 14.7.$ For more on this, see the 'coupon collecting problem'.

Here's another way of looking at it, along the lines suggested by barak manos.

Note that 7 rolls containing all 6 values must contain a duplicate. There are $\binom{7}{2}\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by

$$1-\frac{\binom{7}{2}\cdot 6\cdot 5!}{6^7} = 1-\frac{70}{1296}=\frac{613}{648}.$$


# Garland trouble

#### grgrsanjay

##### New member
How many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?

I was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z

So, x+y+z=12
no .of ways is 14C2.....Was i Wrong somewhere??

Last edited:

#### tkhunny

##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.

W before any red
X after first red
Y after second red
Z after last red

W + X + Y + Z = 12

14C3

#### Plato

##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.

How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.

Last edited:

#### tkhunny

##### Well-known member
MHB Math Helper
I don't disagree with your solution. I suggest only that the definition of "garland" is ambiguous.

After a quick survey of those in my home, they ALL believed a "garland" to be more of a hanging rope, and not a wreath. I believe we have established significant ambiguity (at least cultural) in the problem statement.

Perhaps a picture was provided.

#### grgrsanjay

##### New member
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.

How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.

So,there must be less than 14C2 solutions??How do i eliminate them?

Ok,could you check whether i can do it like this

x + y + z = 12

Case 1:x=y=z
No.of.ways = (4,4,4) = 1

Case 2:two of x,y,z are equal
No.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6

Case 3: x,y,z are all unequal

To avoid double counting , i will take x<y<z

No.of ways =
(0,1,11),(0,2,10),(0,3,9),(0,4,8),(0,5,7)
(1,2,9),(1,3,8),(1,4,7),(1,5,6)
(2,3,7),(2,4,6)
(3,4,5)
= 12 ways

So,total no.of ways is 19

Last edited:

#### Plato

##### Well-known member
MHB Math Helper
the definition of "garland" is ambiguous.
The definition may be ambiguous here but not in textbooks on countilng. Here is a link. The subtopic on beads is the one used in counting problems.
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this
So,total no.of ways is 19
YES, 19 is correct
This sort of problem has set many arguments. If you agree on what meaning of garland we use then the following will show you why. If we have three green beads and three red beads how many different garlands are possible? Well the answer is 3. All the teds together; two reds together and one separate; then all the reds are separated. There are no other possibilities.
Three can be written as 3, 2+1, or 1+1+1.

The questions using two colors have that nice solution. They quickly become a nightmare with more than two colors.

Last edited:

#### grgrsanjay

##### New member
I think none of my case were repeated...could you conform it whether any case is left?

#### Plato

##### Well-known member
MHB Math Helper
I think none of my case were repeated...could you conform it whether any case is left?
Yes you are correct. At first I read it as if there were only 12 beads altogether.


Parametric To Rectangular With Trig

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Example 6) Finding parametric equations for a given function is easier. Arc length of a parametric curves. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Therefore, we need to use a change of variables so we can integrate using either cylindrical (polar) or spherical coordinates, or even parametric form. Find more Mathematics widgets in Wolfram|Alpha. How do you convert #r=2sin(3theta)# to rectangular form? Trigonometry The Polar System Converting Between Systems. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Pre - Calc. 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Step-by-Step Examples. what is the volume of that rectangular pyramid? asked by riza on August 6, 2012. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. The rectangular coordinates for P (5,20°) are P (4. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Parametric and Polar Equations Review Name q) 1. Parametric equation of an ellipse and a hyperbola rectangular hyperbola cartesian and parametric forms examsolutions maths tutorials parametric equations hyperbola hw 1 conic sections in polar parametric forms lesson Parametric Equation Of An Ellipse And A Hyperbola Rectangular Hyperbola Cartesian And Parametric Forms Examsolutions Maths Tutorials Parametric Equations Hyperbola Hw 1 Conic. In , the data from the parametric equations and the rectangular equation are plotted together. Historic applications of parametric equations are discussed so the use of them is realized. SOLUTION The graph of the parametric equations is given in Figure 9. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Textbook Authors: Lial, Margaret L. This is also a great Review for AP Calculus BC. When you first learned parametrics, you probably used t as your parametric variable. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. Simply let t x and then replace your y with t. It can handle horizontal and vertical tangent lines as well. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. 3+3𝑖 3√2(cos45°+𝑖sin45°) Write the complex number in the form + 𝒊. Textbook Authors: Lial, Margaret L. This problem is about converting parametric equations to rectangular form. Sketch and identify graphs using parametric equations. This is called a parameter and is usually given the letter t or θ. Find a rectangular equation for each plane curve with the given parametric equations. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos  t (x = \cos t (x = cos t and y = sin  t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. For the problems above, let x = t + 2 and find the resulting parametric equations. ACE TRIG Final EXAM REVIEW. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Algebra Review: Completing the Square. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Plot the points. Sketching the Graph of Trigonometric Parametric Equations. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. A parametric equation is where the x and y coordinates are both written in terms of another letter. Plot the following points on the polar grid at the right and label each point. Find all solutions to a trigonometric equation 10. Find new parametric equations that shift this graph to the right 3 places and down 2. The coordinates are measured in meters. Polar and Parametric Equations 3. The volume of. 2;5p 3 Give two alternate sets of coordinates for each point. 6 Plane Curves, Parametric Equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. But by recognizing the trig identity, we were able to simplify it to an ellipse, draw the ellipse. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = − = as a function of x. the process of solving one parametric equation for t so you may substitute that equation into the other parametric equation for t to create a rectangular equation where y=f(x) Example: if x=t-4 and y=¼t solve the first equation for t so t=x-4 then by substitution y=¼(x-4) or y=¼x-1. 95 per month. Thus, keep only the other equation. x: 4 sin (2t) Y : 2 cos (2t) X : 4+2 COS t. Find more Mathematics widgets in Wolfram|Alpha. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. Graph polar equations. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. x = 3t – 1, y = 2t + 1. 4 Trigonometric Functions of Any Angle 497 1-3 4. Find a Cartesian equation for the curve traced out by this function. Trig functions, complex numbers, identities, vectors, and real life applications like circuits are here to energize your classroom. Sketch and identify graphs in polar coordinates. First, we find a vector {c,d} of distance 1 having angle -θ, which is {Cos[-θ], Sin[-θ]}. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 7. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Write the equations of the circle in parametric form. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Parametric and Polar Equations Review Name 1. Challenge Sets. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate. Graphing parametric equations: The key is to plug in useful points within the speciﬁed range of t, not just any points. Show all algebraic support. The Second Fundamental Theorem of Calculus Integration Involving Powers of Trigonometric Functions. This is the a value, this is the b value. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. x = cos 2t, y = sin t, for t in 1-p, p2. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. The volume of. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Historic applications of parametric equations are discussed so the use of them is realized. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Convert the polar coordinates (5 , 2. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Plot the resulting pairs ( x,y ). Based on your work on the above problems, name another benefit of parametric equations versus rectangular functions. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Write the equations of the circle in parametric form. Use a table of values to sketch a Parametric Curve and indicated direction of motion. In parametric equations x and y are both defined in terms of a third variable. 1 Answer A. ; Hornsby, John; Schneider, David I. Polar coordinates. 244 Chapter 10 Polar Coordinates, Parametric Equations conclude that the tangent line is vertical. 7-8: 2-8 even, 9-12 all, 14-32 even. In the past, we have seen curves in two dimensions described as a statement of equality involving x and y. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Things to try. Xmin = -20 Ymin = -12 Xmax = 20 Ymax = 12 Xscl = 5 Yscl = 5. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Convert to Polar Coordinates. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Both types depend on an argument, either circular angle or hyperbolic angle. Calculus of a Single Variable. Converting from rectangular to parametric can be complicated, and requires some creativity. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Find a rectangular equation for each plane curve with the given parametric equations. Usually will stand for time. Eliminating the parameter is a method that may make graphing some curves easier. A common parameter used is time (t) or an angle (trig) (x, y) is the place, "t" is the time it is there (at that place) (at that place) Graphing Parametric Equations 2 OPTIONS: (l) Use a chart to find rectangular points. Recall the trig identity d1 Substitute x/r and y/r into the identity: Remove the parentheses: Multiply through by r 2. The curves are colored based on the quadrants. Trig Ratios and Quadrants (9:52) Inverses of Trig Functions and Inverse Trig Functions (20:18) Graphing Sine and Cosine--Amplitude and Period (11:15) Graphing Sine and Cosine--Vertical and Horizontal Shift (6:33) Graphing Sine and Cosine by Hand (29:10) Graphing Tangent by Hand (24:39) Applications (15:50) Unit 4: Triangle Applications of Trig. The parametric equations are plotted in blue; the. A curve is given by the parametric equations: #x=cos(t) , y=sin(2t)#, how do you find the cartesian equation? Calculus Parametric Functions Introduction to Parametric Equations. Homework Statement Reduce these parametric functions to a single cartesian equation:$\displaylines{ x = at^2 \cr y = 2at \cr} \$. Drag points A and B to change the size and orientation of the polygon. now the rest is a matter of simplification:. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Graphing a Parabola with vertex at (h ,k ). And then by plotting a couple of points, we were able to figure out the direction at which, if this was describing a particle in motion, the direction in which that particle was actually moving. In parametric equations x and y are both defined in terms of a third variable. It can handle horizontal and vertical tangent lines as well. We show how we can transform between these representations of the same plane. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. Plot the resulting pairs ( x,y ). (θ is normally used when the parameter is an angle, and is measured from the positive x-axis. But how do we write and solve the equation for the position of the moon when the distance from the planet,. In the past, we have been working with rectangular equations, that is equations involving only x and y so that they could be graphed on the Cartesian (rectangular) coordinate system. This calculator converts between polar and rectangular coordinates. Use power reducing formulas 9. now the rest is a matter of simplification:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Find exact values of composite functions with inverse trig functions 8. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. Trigonometric substitution. Search this site. I'm not looking for the answer here. 1 Answer Cesareo R. 1 Complex Numbers 8. Parametric curves have a direction of. x =3cosθ and y =4sinθ, 02≤θ≤ π by eliminating the parameter finding the corresponding rectangular equation. Write the equations of the circle in parametric form. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Use a table of values to sketch a Parametric Curve and indicated direction of motion. Graphing a Parabola with vertex at (h ,k ). Convert to Polar Coordinates. Let's look at each solution carefully: The first solution is correct since: r^2 = x^2 + y^2. Subtract 7 7 from both sides of the equation. position time parametric equations path rectangular equation eliminating the parameter square root function direction of motion. now the rest is a matter of simplification:. orgChapter 1. We will then introduce the polar coordinate system, which is often a preferred coordinate system over the rectangular system. Graphing parametric equations: The key is to plug in useful points within the speciﬁed range of t, not just any points. Applications. Sketch and identify graphs using parametric equations. Applications of Parametric Equations. This is the a value, this is the b value. interesting variations on the parametric equations. Linear 64) 2. Two versions. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. 5 Parametric Equations part 2. Find the rectangular equation that models its path. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. Finding Parametric Equations In Exercises 35 and 36, find two different sets of parametric equations for the rectangular equation. Graphs of trigonometric functions in polar coordinates are very distinctive. Lots of vocab in this one -- specific solutions, general solutions, 2npi, 360n -- and lots of factoring to do too. Graph Type⎮ Parametric. Identify the type of graph. Making statements based on opinion; back them up with references or personal experience. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. 2 Exercises - Page 365 36 including work step by step written by community members like you. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Finding cartesian equation from parametric trigonometric equations. This calculator converts between polar and rectangular coordinates. For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. Application: Toy Rocket The parametric equations determined by the toy rocket are Substitute from Equation 1 into equation 2: A Parabolic Path 8. Ask Question Asked 5 years, 7 months ago. Sketch and identify graphs in polar coordinates. Find a rectangular equation for each plane curve with the given parametric equations. Find more Mathematics widgets in Wolfram|Alpha. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. An alternative approach is two describe x and y separately in terms of a. 1 shows points corresponding to θ equal to 0, ±π/3, 2π/3 and 4π/3 on the graph of the function. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Then, we do this substitution into the function: x → x c - y d y → x d + y c. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos  t (x = \cos t (x = cos t and y = sin  t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. Find parametric equations for 2x 4 a) y = x 2 —2x+3 Note that there are many ways of finding parametric equations for a given function. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Example 3: Transform the equation x 2 + y 2 + 5x = 0 to polar coordinate form. Just as with non-angle parameters, when the parameter is an angle θ the plane curve can be graphed by selecting values for the angle and calculating the x- and y-values. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. This is also a great Review for AP Calculus BC. Jimmy wants to rewrite the set of parametric equations x = 1/2 T + 3 and y = 2T - 1 in rectangular form by eliminating T. Rewrite the equation as t+7 = x t + 7 = x. now expanding (x+2)² either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)² from your original parametric equation -(t²) = -x^2 - 4x - 4. Finding all arguments t in 0 <= t <= 4Pi where the parametric graph intersects. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Pre - Calc. Find the rectangular coordinates of. The Polar coordinates are in the form (r,q). Then x=f(t) and y=g(t) are called parametric equations for the curve represented by (x,y). Equations for and are plotted on the perpendicular and planes as varies from to. Graph polar equations. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Because r is a directed distance the coordinates (r, θ) and (-r, θ + π). Finding cartesian equation from parametric trigonometric equations. However, they used meters instead of feet for gravitational constant. I don't know what to do from here or if I'm going in the right direction or not. The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! If you're seeing this message, it means we're having trouble loading external resources on our website. Calculus of a Single Variable. Rotating a Curve defined by a Equation. The students will analyze and graph Sine, Cosine and Tangent and their inverses. 2sec t and y = -0. It is a parabola with a axis of symmetry along the line $$y=x$$; the vertex is at $$(0,0)$$. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. This is also a great Review for AP Calculus BC. Trig equations are problems where you're solving for X or Theta but they're hidden behind a trig function, like "2sinX-1=0" or "tan 2 X-1=0". Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Applications of Parametric Equations. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Thus, keep only the other equation. Algebra Review: Completing the Square. Trigonometric, Parametric, and Polar Graphs OBJECTIVES When you have completed this chapter you should be able to Graph the sine wave, by calculator or manually. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Two versions. hyperbolic trig identities are similar to the regular trig onesbut beware hyperbolic trig functions have seemingly little in common with regular trig functions so there's no. Finding cartesian equation from parametric trigonometric equations. Making statements based on opinion; back them up with references or personal experience. This is the parameter or a number that affects the behavior of the equation. For each graph you create, identify the specific parametric equations used and the domain for your graph. express the equation in terms of x and/or y. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. How do I convert from parametric equations to rectangular form? Write the rectangular form for the parametric equation x=cosθ ; y=4sinθ Trig, and Differential. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Applications. The Organic Chemistry Tutor 247,826 views 33:29. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Solve trigonometric equations in quadratic form 12. Basic graphing with direction to them. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Comprehensive End of Unit Review for the Polar, Parametric, and Vectors Sections of PreCalculus or Trigonometry plus Graphic Organizer. Plot the points. 1 Answer A. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Get the free "parametric to cartesian" widget for your website, blog, Wordpress, Blogger, or iGoogle. Fill in the table and sketch the parametric equation for t [-2/6] Problems 2 - 11: Eliminate the parameter to write the parametric equations as a rectangular equation. 933 13 (b) C-/cos(£)) 10. The previous section discussed a special class of parametric functions called polar functions. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. Algebra Review: Completing the Square. At any moment, the moon is located at a particular spot relative to the planet. Then, convert each polar coordinate to rectangular ntoS form (RF). For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. 2 Exercises - Page 365 36 including work step by step written by community members like you. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Textbook Authors: Lial, Margaret L. 8/14/2018 12:12 AM §10. Pre-calculus Contents C Parametric Equations Parametric and rectangular forms of equations conversions The parametric equations of a quadratic polynomial, parabola The Trigonometric functions of arcs from 0 to. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Usually will stand for time. y for values of. 2;5p 3 Give two alternate sets of coordinates for each point. ) Drawing the graphTo draw a parametric graph it is easiest to make a table and then plot the points:Example 1 Plot the graph of the. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Drag P and C to make a new circle at a new center location. 6 Plane Curves, Parametric Equations. parameter t = x – 1. Math Algebra 2 Calculus Precalculus Trigonometry Math Help Complex Numbers Pre Calculus Polar Forms Ellipse High School: Math Derivatives Polar Coordinates Sine Convert Calculus 3 Rectangular Calculus 2 Calculus 1 Polar Equation. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and. (a) Eliminate the parameter for the curve given by the parametric equations x= 2 t, y= 1 t2 for 1 P=3𝑟 N𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. y = x 2 -2. Graphing parametric equations: The key is to plug in useful points within the speciﬁed range of t, not just any points. Trig Functions sine I First, solve for 3. Plot the resulting pairs ( x,y ). It can handle horizontal and vertical tangent lines as well. The Second Fundamental Theorem of Calculus. This problem is about converting parametric equations to rectangular form. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Let A be the point where the segment OB intersects the circle, where point B lies on the line x = 2 a. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Other Parent Functions C. For the point and So, the rectangular coordinates are See Figure 10. Comments There are no comments. ACE TRIG Final EXAM REVIEW. Calculus Examples. (a) Find a polar equation of the cissoid. Write each pair of parametric equations in rectangular form. There's also a graph which shows you the meaning of what you've found. And Vector Calculus, which tends to be the last chapter for Calculus 3, deals with work on, in, and around a surface which will predominately involve polar coordinates as well. Parametric Equations and Motion Precalculus Vectors and Parametric Equations. I was trying to solve for x for some reason. Use MathJax to format equations. y = x 2 -2. t over the interval for which the functions are defined. Given a point in polar coordinates, rectangular coordinates are given by Given a point in rectangular coordinates,. Trigonometry (MindTap Course List). In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. Exchanging x and y 4. The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Both types depend on an argument, either circular angle or hyperbolic angle. To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas that we’ve developed require that functions be in one of these two forms. Finding Parametric Equations for a Graph (Page 799) Describe how to find a set of parametric equations for a given graph. 95 per month. Write each pair of parametric equations in rectangular form. Trigonometric substitution. Example for parametric equations are {eq}\displaystyle x=\sqrt{t},\:y=t-5 {/eq}, what we need to do first is to find x. I'm not looking for the answer here. (a) y = x -3 This is the equation of a line. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. So our parametric equation is x = t and y = 4t.


# Proof of infinitely many prime numbers

Here's the proof from the book I'm reading that proves there are infinitely many primes:

We want to show that it is not the case that there only finitely many primes. Suppose there are finitely many primes. We shall show that this assumption leads to a contradiction. Let $p_1, p_2,...,p_n$ be all the primes there are. Let $x = p_1...p_n$ be their product and let $y=x+1$. Then $y \in \mathbb{N}$ and $y \neq 1$, so there is a prime $q$ such that $q \mid y$. Now $q$ must be one of $p_1,...,p_n$ since these are all primes that there are. Hence $q \mid x$. Since $q \mid y$ and $q \mid x$, $q \mid (y-x)$. But $y-x=1$. Thus $q \mid 1$. But since $q$ is prime, $q \geq 2$. Hence $q$ does not divide 1. Thus we have reached a contradiction. Hence our assumption that there are only finitely many primes must be wrong. Therefore there must be infinitely many primes.

I have a couple of questions/comments regarding this proof. I will use a simple example to help illustrate my questions:

Suppose only 6 primes exist: $2, 3, 5, 7, 11, 13$

Let $x = p_1p_2p_3p_4p_5p_6=30,030$

Let $y = x + 1 = 30,030+1 = 30,031$

1. The proof states there is a prime $q$ such that $q \mid y$ and that $q$ must be either $p_1, p_2, p_3, p_4, p_5,$ or $p_6$. However, none of the 6 primes listed, $(2,3,5,7,11,13)$, divides $30,031$. In fact, the only divisors for $30,031$ are $1, 59, 509$ and $30,031$. Doesn't the proof then break down here since there is no prime $q$ that divides $y$?

2. The prime factorization of $30,031$ is $59 \times 509$. These two numbers are factors of $30,031$ and are in fact primes themselves since they are only divisible by themselves or by 1. Have I shown that there exists $\gt 6$ primes? If so, what can I conclude now that I have shown this?

3. I don't understand why the contradiction $q$ divides $1$ and $q$ does not divide $1$ lead us to the assumption that the finitely many primes must be wrong. I understand how we reached the contradiction. I don't understand why contradiction leads us to the conclusion shows that are assumption that there is only finitely many primes is wrong.

My apologies for the long post. Thanks for any and all help.

• unique factorization requires a unique prime decomposition. If no $p$ exists, there must be a new prime... – Eleven-Eleven Oct 8 '16 at 23:50
• The point of the argument is that there must be some prime not in the list that divides the number you produce. Your example confirms this! You found two such primes $59,509$ neither of which is on the list. Thus, your list can't have been complete. – lulu Oct 8 '16 at 23:51
• The proof makes an assumption that there are finitely many primes, But it then goes on to show, given the conditions, this actually can't be the case. Therefore, the flaw is in your assumption, since all arguments in the proof are mathematically valid. – Eleven-Eleven Oct 8 '16 at 23:53
• As stated above, all of your mathematical arguments are correct when you started from your original assumption. Yet you are led to a point in the proof that is impossible. If the mathematical arguments are sound and correct, what is left for there to be wrong...? The original assumption must therefore be incorrect. – Eleven-Eleven Oct 8 '16 at 23:55
• 2) In this case that there are at least 7 primes. But if you generalize this, that p1p2...pn+1 is never divisible by p1,...,pn then p1,.....pn can not be a list of all primes. No such list can exist. So there can not be a finite number of primes. – fleablood Oct 9 '16 at 0:33

This is an excellent example of a proof that is traditionally phrased as a proof by contradiction but is much better understood constructively. From a constructive viewpoint, the proof shows that given any list of primes $p_1, \ldots, p_n$ there is a prime $q$ (any prime divisor of $p_1p_2\ldots p_n + 1$) that is distinct from each $p_i$. So given any finite set of primes we can find a prime that is not in that set.

• +1 for emphasizing the constructive aspect of this proof. – P Vanchinathan Oct 9 '16 at 0:44
• Except that this process doesn't actually construct a new prime (the example $y = 30031$ being composite). – Daniel R. Collins Oct 9 '16 at 2:20
• @Rob: I think Daniel’s point is that the proof constructs an integer $y>1$ that is not divisible by $p_1$, $p_2$, …, $p_n$, and then it asserts that there must be a prime $q$ such that $q \mid y$, but it doesn’t construct $q$. – Scott Oct 9 '16 at 4:10
• @Scott But there's a simple algorithm to generate a prime divisor of any given integer. – Jack M Oct 9 '16 at 8:05
• @djechlin This is a constructive proof and you're utterly wrong. The set of divisors of a positive integer is finite. Now just take the minimal element greater than $1$ in the set of divisors; this is obviously prime. Would you say that finding the minimal element in a finite set of integers is not constructive? By the way, the proof by contradiction seems to imply that $p_1p_2\dots p_n+1$ is prime, which is the main source for confusion in newbies. – egreg Oct 9 '16 at 11:20
1. The proof does break down in a sense. You have reached a contradiction, which means the hypothesis that there are finitely many primes can not possibly be true.

2. So you thought you had 6 primes, you found 2 extra. Can you just add these two to your list and get all of the primes? If you repeat the process on these 8 primes, you will find that you will have even MORE primes by considering the product + 1. You can keep going and going and you will never run out of primes. This is the idea of the proof. It uses contradiction because you can do it all in one step and avoid the potential issue of doing the process infinitely many times.

Suppose only 6 primes exist: 2,3,5,7,11,13

However, none of the 6 primes listed, (2,3,5,7,11,13), divides 30,031.

Then we already have a contradiction. Since there are only 6 primes (we supposed that at the beginning) and none of them divide 30,031, then 30,031 must be prime. However, 30,031 is not one of the only 6 primes that exist. So 30,031 cannot be prime, yet it must be prime.

So the proof works. In fact, it works precisely the same regardless of what set of numbers we suppose are the only primes that exist. Thus no finite set of numbers can include all the primes that exist. Thus there are an infinite number of primes.

• I think you mean "... no finite set of numbers ..." – origimbo Oct 10 '16 at 10:48

You ask in #1 if the proof breaks down. No. What breaks down is the assumption that there are no more primes. You assumed you had a complete list of primes. Then you constructed a number that is not divisible by any of the primes in your list. Only two possibilities remain: Either the number you constructed is prime, or it is divisible by a number that is not in your list. Either way your list is not complete, which contradicts your initial assumption.

Simpy put, you assumed you had a complete list and by using that assumption you proved that the list was not complete. The assumption must therefore be wrong.

The answer to #3 is basically the same. By finding a contradiction when you made an assumption, you proved the assumption to be incorrect.

The answer to #2 is no, you did not prove you anything by noting that 59 and 509 are both prime. You are trying to prove that there is a finite list of primes. If you choose a particular set of primes as you did {2, 3, 5, 7, 11, 13} and show that that particular set doesn't hold all the primes, a skeptic would just say that you need to add more primes (like 59 and 509) because you didn't make your set big enough. The proof has to be more general which is why it doesn't say how many primes are in the finite set. The proof is written so that it works no matter how big the finite set is.

Point 1: It's a theorem that any natural number $n>1$ has a prime factor. The proof is easy: for any number $n>1$, the smallest natural number $a>1$ which divides $n$ is prime (if it were not prime, it would not be the smallest).

Point 2: Yes, you have proved there are more than six primes. So what? The proof by contradiction doesn't suppose there are only six, but that there are a finite number of them.

Point 3: Actually, it's not really a proof by contradiction stricto sensu. It is proved that any finite list of primes in incomplete.

Here is a non-mathematical approach to the logic behind the proof by contradiction...

Suppose your assumption is that a suspect in a brutal murder is innocent because he said he couldn't have been at the location the person was murdered. You feel as though he is possibly telling the truth, but what you do know is that there are 5 other suspects. You know with 100% certainty due to the detail of the case there can't be anymore than 5 other suspects via proven information. But, the police can verify using obvious physical evidence such as video cameras that the 5 suspects could not possibly be guilty. Then there is either another suspect or the one assumed innocent is guilty. But we said that it was impossible for any other suspects to be considered. Therefore our original suspect is guilty.

This is exactly what is happening in a proof by contradiction.

• This is good. But I think a better analogy is that you have 6 suspects and you assume that these are the only six possible suspects in the world. Then cameras show each and every one is innocent. Thus our assumption that we have all the suspects is wrong, and therefore there is a seventh possible person who could have (and did) do it. – fleablood Oct 9 '16 at 0:58
• Damn.... you are right. It made logical sense when i wrote it... but yours is better. – Eleven-Eleven Oct 9 '16 at 1:02
• Yours makes sense. And is a proof by contradiction. I just thought for this particular proof this is a closer analogy. I wouldn't say yours was wrong or anything. – fleablood Oct 9 '16 at 1:05
• That was my original goal; understand why contradition works. But yours does that AND is analogous to the problem at hand. Thanks – Eleven-Eleven Oct 9 '16 at 1:07
• I will concede. I get it. – Eleven-Eleven Oct 9 '16 at 23:01

The proof is essentially overexplaining. The point of contradiction would have been much clearer if the author had:

1. Reached it earlier.
2. Introduced less notation.

Since $p_1\cdots p_n+1$ cannot have any of $p_1,...,p_n$ as a prime factor we already have a contradiction.

The rest is just explaining to death why $p_1,...,p_n$ cannot be prime factors of $y=p_1\cdots p_n+1$ which should be clear since $y$ is $1$ off from being a multiple of either of those primes.

• I think the gif dosn't really fit this site, though I can't say if there is a specific rule against such things – Yuriy S Oct 10 '16 at 11:27
• @YuriyS: Hmm ... I am sorry to learn that people may find it inappropriate! That was by no means my intention! Do you think it would work better by removing the gif an leave the link and the figure of speach, or are all of those opfuscating the points I tried to make? – String Oct 10 '16 at 12:21
• @String: your link to a cartoon movie of a dead animal being bludgeoned is disproportionate and offensive. – Rob Arthan Mar 12 '17 at 1:27
• @RobArthan: Sorry, in some parts of the world such a cartoon would be considered merely a funny way to illustrate the saying about beating a dead horse. No offense intended, only a light tone. I cannot help that people do take offense, so I have removed it. Still it puzzles me how a cartoon matching the content of a saying would offend. I am from Denmark, after all. – String Mar 12 '17 at 8:11
• @String: on MSE it's easier just to use neutral language. As I am English (after all), may I point out that the cliched phrase is actually "flogging a dead horse" and it doesn't have the connotations you think it does (it's not "explaining to death", it's using a tired old argument that has lost all interest or relevance). Your cartoon doesn't help with that. – Rob Arthan Mar 13 '17 at 1:55
1. Well... yes... it does break down. You assumed that there are only 6 primes and reached a contradiction. You've successfully proved that there aren't only 6 primes.
2. Under the assumption that there are only 6 primes 30,031 isn't factorizable.
3. In proof by contradiction you prove proposition A by assuming A is not true, and through a series of logical steps reach an impossibility, thus proving that A must be true. Concurrently, in this proof, we assumed there is a finite number of primes. After a series of logical inferences we reached a contradiction. Since the only assumption we made in the process is that there are a finite number of primes, that assumption must be wrong.
1. Either $6$ primes you considered are not all the primes, or $30031$ is a new prime. Either way assumption is false. Proof by contradiction.
2. No, you haven't shown that two more primes exist. In this example yes, but not in general
3. Again, based on your assumption you reached something impossible that $q$ divides $1$. There is no such $q$ except $1$.

Start with $2,3,5$ being the only prime numbers. Then you would have $31$ as the new number. $2,3,5$ do not divide $31$. But unlike your case, $31$ is a prime number. So the assumption is wrong.