Lulz. I bet half the retards on here won't get this.

File: 59 KB, 540x595, 20110408.gif [View same] [iqdb] [saucenao] [google]

Anonymous Sat Apr 9 20:32:42 2011 No.2861092 [Reply] [Original]

Lulz. I bet half the retards on here won't get this.

>>	Anonymous Sat Apr 9 20:35:42 2011 No.2861109 i dont get it. please enlighten me.

>>	Anonymous Sat Apr 9 20:36:42 2011 No.2861124 >>2861109 The girl on the left is wrong because the type of fraction isn't specified.

>>	Anonymous Sat Apr 9 20:37:42 2011 No.2861126 This comic is wrong. The definiton of a rational number is that two _whole_ numbers can express the number - pi is not a whole number, and thus pi/1 does not demonstrate that pi is a rational number.

>>	Anonymous Sat Apr 9 20:37:42 2011 No.2861131 >>2861124 bingo >>2861109 retard so i was right in OP

>>	Anonymous Sat Apr 9 20:38:42 2011 No.2861135 The comic isn't wrong, one of the characters in the comic is wrong. The comic is right actually.

>>	Anonymous Sat Apr 9 20:38:42 2011 No.2861139 But it NEVER ask that The art of learn mathematics is the art of learning to troll with formalism!

>>	Anonymous Sat Apr 9 20:38:42 2011 No.2861140 >>2861126 retard, see >>2861124

>>	rec & play !!lz9pGsOAX6N Sat Apr 9 20:38:43 2011 No.2861138 >>2861126 >This comic is wrong You mean the girl on the right is wrong. Thatsthejoke.gif.zip.jpeg

>>	Anonymous Sat Apr 9 20:41:42 2011 No.2861155 Idiots. It says FRACTION, and PI/1 is clearly a fraction. Losers.

>>	Anonymous Sat Apr 9 20:41:42 2011 No.2861162 >>2861092 This is correct because female mathematicians wear glasses, have messy hair and never dress in pink.

>>	Anonymous Sat Apr 9 20:47:42 2011 No.2861197 1pi/1 = pi

>>	Anonymous Sat Apr 9 20:49:42 2011 No.2861207 ITT: OP is proven right.

>>	Anonymous Sat Apr 9 20:50:42 2011 No.2861212 >>2861162 >where*** glasses

>>	Anonymous Sat Apr 9 20:53:42 2011 No.2861227 What does π mean?

>>	Anonymous Sat Apr 9 20:54:42 2011 No.2861232 >>2861162 one of our (female) lecturer's favourite tops this term was a pink stripy thing. she did also wear glasses and have messy hair.

>>	Anonymous Sat Apr 9 20:55:42 2011 No.2861240 >>2861227 >look up acii character >requires knowing the name of the symbol in the first place :\|

>>	Anonymous Sat Apr 9 20:56:42 2011 No.2861243 >>2861240 I laughed until I realized: http://en.wikipedia.org/wiki/Greek_alphabet

>>	Anonymous Sat Apr 9 21:42:42 2011 No.2861440 Denominator is 1, so it's not a fraction. If she'd said tau/2 that'd be a different story.

>>	Anonymous Sat Apr 9 21:43:42 2011 No.2861449 >>2861440 If Pi/1 is not a fraction, so is tau/2.

>>	Anonymous Sat Apr 9 21:44:42 2011 No.2861455 >>2861449 OK, that's not a grammatical English sentence.

>>	Rainbow Dash !!pGgXu8jfKhv Sat Apr 9 21:44:42 2011 No.2861456 >>2861092 Relationship between 2 INTEGERS. Fixed that A+ math student

>>	Anonymous Sat Apr 9 21:45:42 2011 No.2861464 >>2861455 OK, that´s not related to this thread.

>>	Anonymous Sat Apr 9 21:46:42 2011 No.2861470 >>2861456 Relationship between 2 RATIONAL NUMBERS. Fixed that /sci/bro.

>>	Anonymous Sat Apr 9 21:47:42 2011 No.2861476 >>2861464 OK, it's hard to explain in reply to something that doesn't make sense.

>>	Anonymous Sat Apr 9 21:47:42 2011 No.2861480 >>2861207 u got that right.

>>	Anonymous Sat Apr 9 21:49:42 2011 No.2861489 Can "i" be expressed as a fraction?

>>	Anonymous Sat Apr 9 21:51:42 2011 No.2861504 >>2861476 It's pretty easy to see where he was going with that, unless you're a retard.

>>	Anonymous Sat Apr 9 21:54:42 2011 No.2861522 >>2861489 i/1

>>	Anonymous Sat Apr 9 21:54:42 2011 No.2861523 File: 8 KB, 283x283, wat (2).png [View same] [iqdb] [saucenao] [google] 24/7

>>	Anonymous Sat Apr 9 21:57:42 2011 No.2861536 http://www.wolframalpha.com/input/?i=ln+-1+%2F+i YOU JELLY? FIELDS MEDAL PLEASE.

>>	Anonymous Sat Apr 9 21:58:42 2011 No.2861543 File: 82 KB, 376x500, wizard.jpg [View same] [iqdb] [saucenao] [google] >>2861536

>>	Anonymous Sat Apr 9 22:01:42 2011 No.2861564 >>2861536 >logarithm of a negative number >divide by imaginary unit Shit just got imaginary, son.

>>	Anonymous Sat Apr 9 22:03:42 2011 No.2861581 >>2861536 > implying logs of negative numbers exist I thought wolfram alpha was smart.

>>	Anonymous Sat Apr 9 22:04:42 2011 No.2861582 >>2861536 Why the fuck is that pi?! Why the fuck does there exist such a relationship? Fucking mathematics.

>>	Anonymous Sat Apr 9 22:05:42 2011 No.2861593 >>2861536 >log(-1) >i*pi This is just mathematicians arbitrarily assigning shit, right?

>>	Anonymous Sat Apr 9 22:05:42 2011 No.2861595 >>2861536 Sure is Euler's Identity in here

>>	Anonymous Sat Apr 9 22:07:42 2011 No.2861601 >>2861536 wait a second. when did you learn this, from HIGH SCHOOL ALGEBRA!? LOLOLOLOLOLOL EVERYONE KNOWS THIS LULZ

>>	Anonymous Sat Apr 9 22:07:42 2011 No.2861602 >>2861582 >>2861593 >doesn't know about Euler's identity

>>	Anonymous Sat Apr 9 22:08:42 2011 No.2861607 >>2861582 because ln(-x) = i pi + ln(x)

>>	Anonymous Sat Apr 9 22:08:42 2011 No.2861611 >>2861607 AHAHAHAHA ln(-x) is undefined for real x.

>>	Anonymous Sat Apr 9 22:09:42 2011 No.2861616 > implying Euler's identity has anything to do with this.

>>	Anonymous Sat Apr 9 22:11:42 2011 No.2861630 >>2861611 No, it's what I said, moron. http://www.wolframalpha.com/input/?i=ln+x

>>	Anonymous Sat Apr 9 22:13:42 2011 No.2861640 >>2861616 Do you even know what Euler's Identity is?

Anonymous Sat Apr 9 22:14:42 2011 No.2861645

>>2861630
Wolfram alpha is wrong there.

Ln(x) is a function. As such, it must have unique solutions everywhere. According to your formula, ln(e^(i3Pi)) is also Pi.
Also, the limit does not converge for negative real values (which basically means the same).

mathematicians would be laughing hard at this.

>>	Anonymous Sat Apr 9 22:14:42 2011 No.2861648 File: 21 KB, 241x230, haah.png [View same] [iqdb] [saucenao] [google] >>2861109 Is the retard who got it wrong. >>2861124 Is the retard who got it right.

>>	Anonymous Sat Apr 9 22:15:42 2011 No.2861651 >>2861640 Do you?

>>	Anonymous Sat Apr 9 22:16:42 2011 No.2861663 >>2861651 <span class="math">e^{i \pi} + 1 = 0 <span class="math">e^{i \pi} = -1 i \pi = ln(-1) \pi = \frac{ln(-1)}{i}[/spoiler][/spoiler]

>>	Anonymous Sat Apr 9 22:17:42 2011 No.2861665 >>2861648 Actually, he didn´t get it wrong, he didn´t get it at all.

>>	Anonymous Sat Apr 9 22:18:42 2011 No.2861674 >>2861663 Fucking tags <div class="math">e^{i \pi} + 1 = 0</div> <div class="math">e^{i \pi} = -1</div> <div class="math">i \pi = ln(-1)</div> <div class="math">\pi = \frac{ln(-1)}{i}</div>

Anonymous Sat Apr 9 22:18:42 2011 No.2861676

>>2861611
No matter how you slice it, you're wrong.
1) For negative real x ln(-x) is defined trivially, as <span class="math">\int_0^{-x}\frac{1}{x}dx[/spoiler]
2) Otherwise, we can extend the definition of log to all complex numbers pretty simply. Considering finding the inverse expression to e^z = c where c is an arbitrary complex number. Write c in polar form and z in rectangular. e^x*e^(iy) = r*e^(i theta). So clearly, x = log(r), r is a positive real so the real definition of log applied. Also, e^(iy) = e^(i theta). This is of course a family of solutions, that is y = theta + 2pi*k where k is an integer, (you know e^(ix) is 2pi periodic right?). So there you go, log(c) is a set valued function such that
log(z) = log|z| + i(Arg(z) + 2pi*k) k = 0,+/-1,+/-2...

Anonymous Sat Apr 9 22:18:42 2011 No.2861680

>>2861663

Oh boy. Where do i start?

First two are correct, third line is false, therefore also the fourth, as you CAN´T use ln on negative real numbers, as explained before.

You can however specify the numbers that when exponentiated will return -1. Those will be inPi, the n being essential.

he correct equation therefore would be inPi/i which is nPi and therefore useless.

>>	Anonymous Sat Apr 9 22:19:42 2011 No.2861684 >>2861665 You get the post anyways.

>>	Anonymous Sat Apr 9 22:19:42 2011 No.2861685 >>2861616 e^(i pi) = -1 e^(i pi + x) = e^(i pi) * e^x = -e^x ln(-e^x) = x + i pi

Anonymous Sat Apr 9 22:21:42 2011 No.2861697

>>2861676

> No matter how you slice it, you're wrong.

I disagree. So does mathematics.

> 1)

wut?

> (you know e^(ix) is 2pi periodic right?)

Yes, exactly, and that´s why Ln(x) is not a function for these values, as it is not unique/the limit does not converge.

> log(c) is a set valued function

Misnomer, since it´s not a function. It´s a relation.

>>	Anonymous Sat Apr 9 22:22:42 2011 No.2861703 >>2861685 see >>2861680

>>	Anonymous Sat Apr 9 22:23:42 2011 No.2861710 >>2861645 You're a full-blown idiot. Not all functions are 1-to-1. For example y=x^2 is not 1-to-1, as x=1, and x=-1, both give y=1. ln(e^(i3Pi)) = pi i ln(-1) = pi i this does not imply that ln(x) is not a function.

>>	Anonymous Sat Apr 9 22:23:42 2011 No.2861715 File: 10 KB, 333x331, sadPikachu-Crying.jpg [View same] [iqdb] [saucenao] [google] >>2861676 > use ln as a function > then say it's not a function > mfw almost nobody on /sci/ knows how to math

Anonymous Sat Apr 9 22:25:42 2011 No.2861726

>>2861680
You're wrong. There is no magical law saying you can't take logarithms of negative numbers. The logarithms of negative numbers are complex, which is why there's an i involved.

Protip: When wolfram alpha disagrees with you, just assume you're wrong.

>>	Anonymous Sat Apr 9 22:26:42 2011 No.2861728 >>2861710 You get that wrong, sir. Look at the graph of y = x². There are no two points "on top of" each other. Now look at Ln(-x) interpreted as a "set-valued function". Infinitely many points on top of each other -> not a function.

Anonymous Sat Apr 9 22:27:42 2011 No.2861744

>>2861726

For a value to exist, the limit must converge. It does so, if it converges for coming from every possible "direction" in any case. This is not true for the logarithm, as well as for roots by the way.
Do i seriously have to write down the proof for you now? Can´t you just look at how the logarithm is defined?

>>	Anonymous Sat Apr 9 22:28:42 2011 No.2861747 PI is defined as the ratio between the Circumference and Diameter. Why is it irrational?

>>	Anonymous Sat Apr 9 22:28:42 2011 No.2861750 >>2861645 >ln(e^(i3Pi)) is also Pi. holy fuck you're an idiot. e^(i 3 pi) = -1. So you're just restating the fact that ln(-1) = pi i

>>	Anonymous Sat Apr 9 22:29:42 2011 No.2861757 >>2861747 Because there is no circle for which the circumference and diameter are both integers.

>>	Anonymous Sat Apr 9 22:31:42 2011 No.2861767 >>2861744 ln(x) is a continuous function. It's value at every point exists as a limit from both directions. Give it the fuck up.

>>	Anonymous Sat Apr 9 22:31:42 2011 No.2861770 >>2861750 Oh my god are you retarded or wut? I meant ln(e^(i3Pi)) is also -1 of course, obvious as fuck. And no, i´m saying that you are wrong, because ln(-x) is not a function for real x, as explained before.

>>	Anonymous Sat Apr 9 22:31:42 2011 No.2861774 >>2861728 What are you smoking? There are no points "on top of each other" in ln(x) or ln(-x). Each point in the domain maps to a single complex value.

Anonymous Sat Apr 9 22:35:42 2011 No.2861792

>>2861767

Sigh, guess i have to. I´ll use the example of -1. I´ll use polar coordinates (z = r*e^(i\phi)) and set r = 1 for convenience.

from the left of \phi = Pi:

c_1 = lim ln(e^(i(Pi-d))) as d -> 0 = lim iPi + d = iPi

from the right:

c_2 = lim ln(e^(-i(Pi+d))) as d -> 0 = lim -iPi + d = -iPi

c_1 != c_2

> It's value at every point exists as a limit from both directions

HURR DURR I´M SMURT

>>	Anonymous Sat Apr 9 22:36:42 2011 No.2861799 >>2861774 No you fucking idiot. The solution is inPi, which obviously all lie on the y-axis of the complex plane.

>>	Anonymous Sat Apr 9 22:37:42 2011 No.2861803 >>2861792 That should be a "lim iPi - d" in the line where c_1 is calculated.

>>	sage sage Sat Apr 9 22:40:42 2011 No.2861831 ITT: People unable to do math. The ln is only defined on the SLICED complex plane for a reason, dumbfucks.

>>	Anonymous Sat Apr 9 22:41:42 2011 No.2861842 >>2861792 y=ln(x) is a function that maps from x to y, where x is real and y is complex. You're changing it to a function from a complex x to a complex y. You're a complete idiot.

>>	Anonymous Sat Apr 9 22:43:42 2011 No.2861854 >>2861842 Now you just went full retard. Either you´re talking abour ln(x):R->R or Ln(z):C->C. Your magical "function" is non-existant, as no limit at all can be computed.

>>	Anonymous Sat Apr 9 22:44:42 2011 No.2861860 >>2861842 logs are defined for complex numbers as well

Anonymous Sat Apr 9 22:45:42 2011 No.2861868

>>2861715
You're right, there's ambiguity. But we can fix that. Some definitions:
Define ln(x) as above taking only positive reals as imput.
Define log(z) as a set-valued function http://en.wikipedia.org/wiki/Multivalued_function
taking a complex number z as an input and returning a related infinite set of complex numbers as output. You don't seem to like this idea, but there is a fix! We just need to make an arbitrary choice. So let's go with the convential choice, define Arg(z) as the angle the complex number z makes with the x axis, such that -pi<x<=pi. This choice is of course entirly arbirary but it means that our branch cut of log (let's call it Log(z)) will be discontinuous (note: not undefined) on the negative reals.
If you dont like this definition of Arg, I can make it more explicit, let's say <span class="math">Arg(x+iy)=2 arctan\frac{y}{\sqrt{x^2+y^2}+x}[/spoiler] except on the negative reals where it = pi. This gives the expected valued. So finally we have an honest to goodness function to invert e^(z)=c.

Log(z) = ln|z|+i(Arg(z))

Tada!

>>	Anonymous Sat Apr 9 22:46:42 2011 No.2861874 >>2861799 Right, that's why e^(i2pi)=-1. Oh, wait, it doesn't. Fucking moron.

>>	Anonymous Sat Apr 9 22:46:42 2011 No.2861877 >>2861842 http://en.wikipedia.org/wiki/Logarithm#Definition "For the logarithm to be defined, the base b must be a positive real number not equal to 1 and y must be a positive number." Also: http://en.wikipedia.org/wiki/Complex_logarithm

Anonymous Sat Apr 9 22:50:42 2011 No.2861898

>>2861868

> Define log(z) as a set-valued function

As said, it´s not a function then, it´s a relation. You can´t use it like a function in that case, what you did before.

I agree that undefined was the wrong word. It is still not the inverse function of exp on the negative real axis.

Chosing the principal branch doesn´t really help. I could have chosen any other branch and would have gotten a different result when evaluating "Ln(e^(-Pi))".

It does not solve our problem.

>>	Anonymous Sat Apr 9 22:51:42 2011 No.2861908 >>2861874 Goddammit you know what i mean. i (2n+1) Pi.

>>	Anonymous Sat Apr 9 22:53:42 2011 No.2861920 >>2861898 The principle branch is the branch for which the logarithm is defined. just like sqrt(4) only gives 2, ln(-1) only gives pii, not 3pi*i etc. wolfram is correct, ln(-1)/i=pi just like sqrt(4)/2=1, not -1

>>	Anonymous Sat Apr 9 22:55:42 2011 No.2861925 >>2861920 Dude, did you even read what i had written? I can chose any other branch and get a false result. Therefore, your deduction was wrong. i(2n+1)Pi/i != Pi I can chose a branch in which n = 3 for example.

>>	Anonymous Sat Apr 9 22:56:42 2011 No.2861928 >>2861877 >believes wikipedia over wolframalpha on math questions

>>	Anonymous Sat Apr 9 22:58:42 2011 No.2861936 >>2861928 I believe what i can deduce myself. I have provided the proof for your deduction being wrong. If you don´t trust mathematics, that´s not my problem.

>>	Anonymous Sat Apr 9 22:58:42 2011 No.2861941 Hey guys, what is the third root of -8, without looking it up?

>>	Anonymous Sat Apr 9 22:59:42 2011 No.2861943 >>2861925 but ln(-1) doesn't equal i(2n+1)pi you fuck it equals i*pi you don't pick whatever branch you like, you have to pick the priciple branch because thats how the logarithm is defined

>>	Anonymous Sat Apr 9 23:00:42 2011 No.2861953 >>2861943 No, that´s how YOU defined it. I am free to chose whatever branch i like to. -1 = e^(i(2n+1)Pi), NOT e^(iPi) Seriously.

>>	Anonymous Sat Apr 9 23:04:42 2011 No.2861963 >>2861941 -2 is one of them

>>	Anonymous Sat Apr 9 23:06:42 2011 No.2861971 >>2861963 lolno. You confuse this with the solution to x³ = -8. Completely different.

>>	Anonymous Sat Apr 9 23:07:42 2011 No.2861976 >>2861953 so arcsin(0) = 10*pi ? does sqrt(4) = -2 ? no they fucking don't you don't get to define the logarithm, it's already defined

>>	Anonymous Sat Apr 9 23:07:42 2011 No.2861977 >>2861971 well then what's the difference between a cubed root and a third root?

Anonymous Sat Apr 9 23:08:42 2011 No.2861981

>>2861953

Me again, just to ylear this up:

I am only arguing that this deduction:

>>2861685
>>2861674
>>2861663

is false, as explained.

What can be done though is to chose the principal branch of the logarithm to be its value at the negative real axis. In that case, one is free to write Log(-1)/i = Pi.

The deduction using Euler´s formula is still wrong.

>>	Anonymous Sat Apr 9 23:10:42 2011 No.2861986 >>2861941 3 3rd roots of -8, given by 2e^(i * theta) where theta = -pi/3, pi/3, pi

>>	Anonymous Sat Apr 9 23:10:42 2011 No.2861991 >>2861977 The solution to x³ = -8 is {-2,2\pm isqrt(3)}, whereas the third root of -8 is 1+isqrt(3).

>>	Anonymous Sat Apr 9 23:11:42 2011 No.2861993 ITT: OP wins the bet.

>>	Anonymous Sat Apr 9 23:12:42 2011 No.2861996 whenever anyone mentions complex numbers /sci/ becomes full retard

>>	Anonymous Sat Apr 9 23:13:42 2011 No.2861998 >>2861996 It´s weekend. That´s the main reason i guess.

>>	Anonymous Sat Apr 9 23:21:42 2011 No.2862028 File: 93 KB, 360x270, Math-That-looks-right-to-me.jpg [View same] [iqdb] [saucenao] [google] OP here. Retards sure didn't disappoint. I love the Internets.

>>	Anonymous Sat Apr 9 23:22:42 2011 No.2862039 Can we all agree to use x as a real variable and z as a complex one, so we don't get problems like >>2861991

>>	Anonymous Sat Apr 9 23:24:42 2011 No.2862049 >>2862039 I don´t see a problem there, but ok.

>>	Anonymous Sat Apr 9 23:27:42 2011 No.2862058 >>2862049 Just makes it absolutely clear whether or not we're looking for real answers or all answers. As far as the ln(x) stuff from before, it's clear that ln(x) is undefined for negative x and ln(z) is defined for negative z.

>>	Anonymous Sat Apr 9 23:28:42 2011 No.2862064 >>2862058 > ln(z) is defined for negative z. You didn´t read, right? ln(z) is not a function if you let z be a negative real value.

>>	Anonymous Sat Apr 9 23:30:42 2011 No.2862073 Cool story bros I still don't get why the math student thinks differently than the mathematician

>>	Anonymous Sat Apr 9 23:31:42 2011 No.2862077 >>2862064 see: >>2862039

>>	Anonymous Sat Apr 9 23:32:42 2011 No.2862084 >>2862058 >it's clear that ln(x) is undefined for negative x That's just fucking wrong. We can argue about it not being a function because it has multiple answers, but it is sure as fuck defined.

>>	Anonymous Sat Apr 9 23:32:42 2011 No.2862085 >>2862077 I don´t think you get it. z = -a+ib with b = 0 and a>0 is a complex number (on the negative real axis), for which the ln(z) is not a function.

>>	Anonymous Sat Apr 9 23:32:42 2011 No.2862088 >>2862073 The math student doesn't think. The mathematician answers the question. Also the math student is using a wrong definition of irrational.

>>	Anonymous Sat Apr 9 23:33:42 2011 No.2862091 >>2862088 And just repeating definitions he learned by heart and most likely not yet understood well.

>>	Anonymous Sat Apr 9 23:33:42 2011 No.2862094 >>2862085 Bah, fuck. I can't think on the weekends it seems. You're right...I think. Probably.

>>	Anonymous Sat Apr 9 23:33:42 2011 No.2862096 >>2862084 He obviously means the real ln in that case.

>>	Anonymous Sat Apr 9 23:35:42 2011 No.2862099 >>2862096 In that case, I should have been clearer. Disregard most of what I said, I suck cocks.

>>	Anonymous Sat Apr 9 23:38:42 2011 No.2862108 File: 121 KB, 468x349, wtf.jpg [View same] [iqdb] [saucenao] [google] How can you guys derive all those equations from that question?

>>	Anonymous Sat Apr 9 23:41:42 2011 No.2862123 >>2862108 What question?

>>	Anonymous Sat Apr 9 23:42:42 2011 No.2862125 >>2862123 >can pi be expressed as a fraction?

>>	Anonymous Sat Apr 9 23:44:42 2011 No.2862128 >>2862108 Guy says ln(-1)/i was a possibility. Some people didn´t get it. Someone then tried to deduce this from Euler´s equation, which was proven to be impossible. Then shitstorming. Now we´re here.

>>	Anonymous Sat Apr 9 23:45:42 2011 No.2862133 >>2862125 You mean all the logarithm stuff? Complex analysis. Just a bunch of things that equal π.

>>	Anonymous Sat Apr 9 23:45:42 2011 No.2862134 >>2862085 >>2862064 >>2862084 ln(x) where x is <0 is defined as a fuction though by definition you take the primary branch I though we were over this ln(z) is defined as a function for all complex z except 0

>>	Anonymous Sat Apr 9 23:47:42 2011 No.2862140 >>2862125 I don't know why that's even a hard question to answer. How about fuckin' C/d... you know, the first thing you learnt about pi.

>>	Anonymous Sat Apr 9 23:49:42 2011 No.2862143 >>2862128 tldr; it wasn't impossible. 2 people did it.

>>	Anonymous Sat Apr 9 23:52:42 2011 No.2862148 >>2862134 > ln(x) where x is <0 is defined as a fuction though by definition you take the primary branch > implying there exists an a so that e^a is negative

>>	Anonymous Sun Apr 10 00:02:42 2011 No.2862164 usually, irrationality proofs begin "suppose # can be expressed as a fraction a/b where a and b are in LOWEST TERMS..." pi/1 is not in lowest terms because 1 divides pi.

>>	Anonymous Sun Apr 10 00:03:42 2011 No.2862168 >>2862164 exactly.

>>	Anonymous Sun Apr 10 00:04:42 2011 No.2862170 >>2862164 >implying any of that matters, π/1 is still a fraction

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