Favourite Erroneous "Proofs"

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

User avatar
schrodingersduck
Posts: 130
Joined: Mon Aug 13, 2007 2:20 pm UTC
Location: People's Democratic Republic of Leodensia
Contact:

Postby schrodingersduck » Wed Aug 29, 2007 5:12 pm UTC

BlochWave wrote:How fast would it have to average on the second half to bring its total average to 120mph?


Speed of light; assuming that the speed is being measured from the car, as time would seem to stop and the rest of the course could be finished instantaneously

eattre
Posts: 7
Joined: Fri Sep 07, 2007 5:05 am UTC

Postby eattre » Fri Sep 07, 2007 6:20 am UTC


I thought that 7 and 11 were the most subtle ones; I wonder whether it would be a good idea to post the solutions for all of these. :?

User avatar
bonder
Posts: 168
Joined: Mon Jul 16, 2007 6:41 am UTC
Location: /home/bonder
Contact:

Re: Favourite Erroneous "Proofs"

Postby bonder » Tue Oct 23, 2007 3:31 am UTC

I just came across this one and thought I'd share.

start with:
x = (pi + 3)/2 .................................. multiply both sides by 2 to get:
2x = pi + 3 ..................................... multiply both sides by (pi - 3) to get:
2x(pi - 3) = (pi + 3)(pi - 3) ................... distribute to get:
2*pi*x - 6x = pi^2 - 9 ......................... add 9 - 2*pi*x to both sides to get:
9 - 6x = pi^2 - 2*pi*x ......................... add x^2 to both sides to get:
9 - 6x + x^2 = pi^2 - 2*pi*x + x^2 .......... factor both sides to get:
(3 - x)^2 = (pi - x)^2 ......................... take square roots of both sides to get:
3 - x = pi - x ................................... add x to both sides to get:
3 = pi
I've never made anyone's life easier and you know it.

User avatar
quintopia
Posts: 2906
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

Re: Favourite Erroneous "Proofs"

Postby quintopia » Tue Oct 23, 2007 4:47 am UTC

I didn't find 7 to be all that subtle. It was a very straightforward misapplication of the rule. A simple check of the rule's statement gives lie the error.

And 11 makes its error obvious by stating the error outright (by literally writing the limit of a series and one of its partial sums on opposite sides of an equals sign).

In fact, I must say those are the 2 most obvious ones on there (to me).

I like the ones that involve a catch that is not a straightforward misapplication of a rule or technique. 4, for example, seems particularly tricky.

scarletmanuka
Posts: 533
Joined: Wed Oct 17, 2007 4:29 am UTC
Location: Perth, Western Australia

Re: Favourite Erroneous "Proofs"

Postby scarletmanuka » Mon Nov 05, 2007 2:33 am UTC

Weird. I thought #4 was one of the more obvious ones - like most of them, it was just another case of "left out the constant of integration". I thought #6 was about the best, though I also liked #3.

User avatar
Geekthras
3) What if it's delicious?
Posts: 529
Joined: Wed Oct 03, 2007 4:23 am UTC
Location: Around Boston, MA

Re: Favourite Erroneous "Proofs"

Postby Geekthras » Mon Nov 05, 2007 2:59 am UTC

I liked the car one. I reasoned it out like this:
Let's say the racetrack is 120 miles long.
So he goes the fist half (60 mi) in one hour.
In order to go 120 mph on the whole track, he must do the entire thing in 1 hour.
The end.


For things like this, it's easy to just pick a random number and see what happens.
Wait. With a SPOON?!

User avatar
SimonM
Posts: 280
Joined: Sat Jul 21, 2007 4:49 pm UTC
Location: Guernsey, CI
Contact:

Re: Favourite Erroneous "Proofs"

Postby SimonM » Mon Nov 05, 2007 8:12 am UTC

a^a^a^.^.^. = 2

Find a

a = \sqrt 2

b^b^b^b^.^.^. = 4

b = \sqrt 2

Therefore

a^a^a^.^.^. b^b^b^b^.^.^. = 2 = 4

Therefore 2=4

I like that one

I also like

Assume 2=1
Be reversal 1=2
By Euclid's Second Common Notion, we may add the the two identities side by side: 3=3

Since the latter is true the former is true
Last edited by SimonM on Mon Nov 05, 2007 8:25 pm UTC, edited 1 time in total.
mosc wrote:How did you LEARN, exactly, to suck?

User avatar
gmalivuk
GNU Terry Pratchett
Posts: 26818
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:

Re: Favourite Erroneous "Proofs"

Postby gmalivuk » Mon Nov 05, 2007 8:00 pm UTC

SimonM wrote:aa[sub]a[sub]a[sub].[sub].[sub].[/sub][/sub][/sub][/sub][/sub] = 2

Find a

a = \sqrt 2

bb[sub]b[sub]b[sub].[sub].[sub].[/sub][/sub][/sub][/sub][/sub] = 4

b = \sqrt 2

Therefore

aa[sub]a[sub]a[sub].[sub].[sub].[/sub][/sub][/sub][/sub][/sub] = bb[sub]b[sub]b[sub].[sub].[sub].[/sub][/sub][/sub][/sub][/sub] = 2 = 4

Therefore 2=4

I like that one

I'd like it even more if you got your tags right. (Powers are sup, not sub.) Also, since you can see they don't nest, better to just use the old-fashioned method for this.

(The error in that one, of course, is that there is no number you can keep raising to itself to ever approach four.)
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

(he/him/his)

User avatar
SimonM
Posts: 280
Joined: Sat Jul 21, 2007 4:49 pm UTC
Location: Guernsey, CI
Contact:

Re: Favourite Erroneous "Proofs"

Postby SimonM » Mon Nov 05, 2007 8:28 pm UTC

And why is that? It because x^x^x^.^.^. only converges for e-e<x<e1/e
mosc wrote:How did you LEARN, exactly, to suck?

User avatar
Droooo
Posts: 98
Joined: Tue Aug 14, 2007 7:40 pm UTC
Location: Tomorrow, Scotland
Contact:

Re: Favourite Erroneous "Proofs"

Postby Droooo » Tue Nov 06, 2007 7:27 pm UTC

i=i

sqrt(-1)=sqrt(-1)

sqrt(-1/1)=sqrt(1/-1)

sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)

i/1=1/i

i2=1

Explain.
Like Drooo but 10x better.

User avatar
Macbi
Posts: 941
Joined: Mon Apr 09, 2007 8:32 am UTC
Location: UKvia

Re: Favourite Erroneous "Proofs"

Postby Macbi » Tue Nov 06, 2007 8:07 pm UTC

sqrt(a/b)=sqrt(a)/sqrt(b)
only when a and b are posititve (did I mean Real?)

Edit: This might be false, see post below for a better explanation.
Last edited by Macbi on Tue Nov 06, 2007 8:21 pm UTC, edited 1 time in total.
    Indigo is a lie.
    Which idiot decided that websites can't go within 4cm of the edge of the screen?
    There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

Stanford
Posts: 38
Joined: Fri Nov 02, 2007 6:52 pm UTC

Re: Favourite Erroneous "Proofs"

Postby Stanford » Tue Nov 06, 2007 8:19 pm UTC

Droooo wrote:i=i

sqrt(-1)=sqrt(-1)

sqrt(-1/1)=sqrt(1/-1)

sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)

i/1=1/i

i2=1

Explain.

sqrt is multivalued.

User avatar
Govalant
Posts: 249
Joined: Mon Sep 17, 2007 2:50 am UTC
Location: Rosario, Argentina
Contact:

Re: Favourite Erroneous "Proofs"

Postby Govalant » Tue Nov 06, 2007 8:52 pm UTC

Macbi wrote:sqrt(a/b)=sqrt(a)/sqrt(b)
only when a and b are posititve (did I mean Real?)

Edit: This might be false, see post below for a better explanation.


Nope, you got it right.

sqrt(a*b) = sqrt(a) * sqrt(b) Only for real positive values of a and b.
Now these points of data make a beautiful line.

How's things?
-Entropy is winning.

User avatar
rrwoods
Posts: 1509
Joined: Mon Sep 24, 2007 5:57 pm UTC
Location: US

Re: Favourite Erroneous "Proofs"

Postby rrwoods » Tue Nov 06, 2007 8:52 pm UTC

Stanford wrote:
Droooo wrote:i=i

sqrt(-1)=sqrt(-1)

sqrt(-1/1)=sqrt(1/-1)

sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)

i/1=1/i

i2=1

Explain.

sqrt is multivalued.

This also applies to the pi = 3 proof above. (Note that the sqrt function itself isn't multivalued, but when you use it to remove a ^2 from each side of an equation, you aren't correct until you add a +/-.)
31/M/taken/US
age/gender/interest/country

Belial wrote:The sex card is tournament legal. And I am tapping it for, like, six mana.

Stanford
Posts: 38
Joined: Fri Nov 02, 2007 6:52 pm UTC

Re: Favourite Erroneous "Proofs"

Postby Stanford » Tue Nov 06, 2007 10:23 pm UTC

rrwoods wrote:Note that the sqrt function itself isn't multivalued, but when you use it to remove a ^2 from each side of an equation, you aren't correct until you add a +/-.

The +/- is a way to allow sqrt to be single-valued, by choosing a single branch. It's really a matter of notation, whether you define sqrt to be multivalued, or single-valued but requiring the +/-. I just prefer to call it a multivalued function, because that's the way most of my professors teach exponents, but either is valid.

Token
Posts: 1481
Joined: Fri Dec 01, 2006 5:07 pm UTC
Location: London

Re: Favourite Erroneous "Proofs"

Postby Token » Tue Nov 06, 2007 10:33 pm UTC

Stanford wrote:
rrwoods wrote:Note that the sqrt function itself isn't multivalued, but when you use it to remove a ^2 from each side of an equation, you aren't correct until you add a +/-.

The +/- is a way to allow sqrt to be single-valued, by choosing a single branch. It's really a matter of notation, whether you define sqrt to be multivalued, or single-valued but requiring the +/-. I just prefer to call it a multivalued function, because that's the way most of my professors teach exponents, but either is valid.

Sqrt can be single-valued without requiring a ±. The ± becomes necessary if you want to set up an iff in a chain of implications - for example, x^2 = 4 if x=sqrt(4), but x^2 = 4 iff x = ±sqrt(4).
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

User avatar
gmalivuk
GNU Terry Pratchett
Posts: 26818
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:

Re: Favourite Erroneous "Proofs"

Postby gmalivuk » Tue Nov 06, 2007 11:24 pm UTC

Stanford wrote:
rrwoods wrote:Note that the sqrt function itself isn't multivalued, but when you use it to remove a ^2 from each side of an equation, you aren't correct until you add a +/-.

The +/- is a way to allow sqrt to be single-valued, by choosing a single branch. It's really a matter of notation, whether you define sqrt to be multivalued, or single-valued but requiring the +/-. I just prefer to call it a multivalued function, because that's the way most of my professors teach exponents, but either is valid.

This topic has come up before, I believe, though it was ages ago. The general consensus is that Sqrt(), which is equivalent to the radical symbol, is defined to be a function (single-valued) which takes only the positive real or imaginary number that, when squared, gives the argument. This isn't to say that "square root" is also only that single value, since it makes perfect sense to talk about the square roots of a number.

So in other words, Sqrt is a single-valued function that gives the positive square root of a number.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

(he/him/his)

User avatar
Alpha Omicron
Posts: 2765
Joined: Thu May 10, 2007 1:07 pm UTC

Re: Favourite Erroneous "Proofs"

Postby Alpha Omicron » Wed Nov 07, 2007 11:56 pm UTC

gmalivuk wrote:So in other words, Sqrt is a single-valued function that gives the positive square root of a number.

my Grade 11 Math teacher wrote:The square root of a number is understood to be positive.
Here is a link to a page which leverages aggregation of my tweetbook social blogomedia.

User avatar
gmalivuk
GNU Terry Pratchett
Posts: 26818
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:

Re: Favourite Erroneous "Proofs"

Postby gmalivuk » Thu Nov 08, 2007 12:31 am UTC

Alpha Omicron wrote:
gmalivuk wrote:So in other words, Sqrt is a single-valued function that gives the positive square root of a number.

my Grade 11 Math teacher wrote:The square root of a number is understood to be positive.

In high school, you were also probably told that negative numbers don't have square roots. At the very least, you probably thought that 1 only had one cube root.

However, it is fairly common to talk about the (three) cube roots or all n of the nth roots of unity. So I don't think it's unreasonable to stick square roots in this category, and accept that there are two of them.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

(he/him/his)

btilly
Posts: 1877
Joined: Tue Nov 06, 2007 7:08 pm UTC

Re: Favourite Erroneous "Proofs"

Postby btilly » Thu Nov 08, 2007 12:31 am UTC

I've always liked the inflation theorem.

Code: Select all

1$ = 100 c
   = (10c)^2
   = (1/10$)^2
   = 1/100$
   = 1c
Some of us exist to find out what can and can't be done.

Others exist to hold the beer.

User avatar
quintopia
Posts: 2906
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

Re: Favourite Erroneous "Proofs"

Postby quintopia » Thu Nov 08, 2007 12:57 am UTC

btilly wrote:I've always liked the inflation theorem.

Code: Select all

1$ = 100 c
   = (10c)^2
   = (1/10$)^2
   = 1/100$
   = 1c


I C Wut U Did There.

Spoiler:
$^2 != $ and c^2 != c

User avatar
the tree
Posts: 801
Joined: Mon Apr 02, 2007 6:23 pm UTC
Location: Behind you

Re: Favourite Erroneous "Proofs"

Postby the tree » Fri Nov 23, 2007 1:30 am UTC

If a relation is symmetric and transitive, then it is also reflexive. (I came up with this one myself during a lecture today, I was quite proud)

aRb, bRc -> aRc that is transitivity.

let c=a. aRb, bRa -> aRa

since the first half of that proposition is true for a symmetric R, we can be sure of the second half so aRa

QED

PaulT
Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

Re: Favourite Erroneous "Proofs"

Postby PaulT » Fri Nov 23, 2007 1:51 am UTC

the tree wrote:If a relation is symmetric and transitive, then it is also reflexive. (I came up with this one myself during a lecture today, I was quite proud)

aRb, bRc -> aRc that is transitivity.

let c=a. aRb, bRa -> aRa

since the first half of that proposition is true for a symmetric R, we can be sure of the second half so aRa

QED

I think I was asked this as a 'find the error' question as an undergrad. It's quite clever.

aguacate
Posts: 209
Joined: Fri Feb 16, 2007 10:29 pm UTC

Re: Favourite Erroneous "Proofs"

Postby aguacate » Fri Nov 23, 2007 6:29 am UTC

PaulT wrote:
the tree wrote:If a relation is symmetric and transitive, then it is also reflexive. (I came up with this one myself during a lecture today, I was quite proud)

aRb, bRc -> aRc that is transitivity.

let c=a. aRb, bRa -> aRa

since the first half of that proposition is true for a symmetric R, we can be sure of the second half so aRa

QED

I think I was asked this as a 'find the error' question as an undergrad. It's quite clever.


There was another one like this on metrics that I can't remember.

Spoiler:
Symmety+transitivity implies reflexivity only when 'a' belongs to an equivalence class that includes more than just 'a'. If there is no 'b' such that aRb, then reflexivity implies that aRa, while symmetry+transitivity does not.


I like this one.
Image

User avatar
the tree
Posts: 801
Joined: Mon Apr 02, 2007 6:23 pm UTC
Location: Behind you

Re: Favourite Erroneous "Proofs"

Postby the tree » Fri Nov 23, 2007 10:17 am UTC

aguacate wrote:
Spoiler:
Symmety+transitivity implies reflexivity only when 'a' belongs to an equivalence class that includes more than just 'a'. If there is no 'b' such that aRb, then reflexivity implies that aRa, while symmetry+transitivity does not.
Spoiler:
Since you could have a=b, I think the error in fact is the assumption that for every a, there exists something in relation with it.

User avatar
Alpha Omicron
Posts: 2765
Joined: Thu May 10, 2007 1:07 pm UTC

Re: Favourite Erroneous "Proofs"

Postby Alpha Omicron » Fri Nov 23, 2007 7:01 pm UTC

gmalivuk wrote:
Alpha Omicron wrote:
gmalivuk wrote:So in other words, Sqrt is a single-valued function that gives the positive square root of a number.

my Grade 11 Math teacher wrote:The square root of a number is understood to be positive.

In high school, you were also probably told that negative numbers don't have square roots. At the very least, you probably thought that 1 only had one cube root.

However, it is fairly common to talk about the (three) cube roots or all n of the nth roots of unity. So I don't think it's unreasonable to stick square roots in this category, and accept that there are two of them.

I know. I wasn't being cereal.
Here is a link to a page which leverages aggregation of my tweetbook social blogomedia.

2ndFloor
Posts: 5
Joined: Thu Nov 22, 2007 8:42 am UTC

Re: Favourite Erroneous "Proofs"

Postby 2ndFloor » Tue Nov 27, 2007 9:16 pm UTC

A friend of mine came up with a proof that stated that all numbers were equal. It actually made our calculus teacher thing for a while. Then she ignored the kid for a while. :-P

User avatar
schrodingersduck
Posts: 130
Joined: Mon Aug 13, 2007 2:20 pm UTC
Location: People's Democratic Republic of Leodensia
Contact:

Re: Favourite Erroneous "Proofs"

Postby schrodingersduck » Wed Nov 28, 2007 3:19 pm UTC

2ndFloor wrote:A friend of mine came up with a proof that stated that all numbers were equal. It actually made our calculus teacher thing for a while. Then she ignored the kid for a while. :-P


When you say calculus, did the proof go something like this:

Code: Select all

Take a function (let's use f(x) = x + a for example's sake):
Differentiate it: f'(x) = 1
Then integrate again: f(x) = x + C (don't forget your constant of integration!).
Consider another function, g(x) = x + b, such that a <> b
Differentiate and then integrate to get g(x) = x + C.
x + C = x + C
f(x) = g(x)
x + a = x + b
a = b


So any two different numbers are actually the same! Of course, the error in this one should be blindingly obvious...

GuyWithFunnyHat
Posts: 14
Joined: Mon Dec 03, 2007 10:06 pm UTC

Re: Favourite Erroneous "Proofs"

Postby GuyWithFunnyHat » Wed Dec 05, 2007 2:32 am UTC

here's one I came up with last year due to a misunderstanding of how calculus works

0/3=x
3x=0
3=0/x
3=0

edit: and yes, I do know this proof fails for many reasons, the least of which is 0/0 being undefined
-there are no stupid questions, only stupid people
-never take a blind date to a silent film
-an optimist is a person whom, on noticing that a rose smells better than a cabbage, concludes that it will make a better soup

User avatar
duckshirt
Posts: 567
Joined: Thu Feb 15, 2007 1:41 am UTC
Location: Pacific Northwest

Re: Favourite Erroneous "Proofs"

Postby duckshirt » Sat Dec 08, 2007 2:25 am UTC

Droooo wrote:i=i

sqrt(-1)=sqrt(-1)

sqrt(-1/1)=sqrt(1/-1)

sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)

i/1=1/i

i2=1

Explain.
The version I heard was -1 = i^2 = i*i = sqrt(-1)*sqrt(-1) = sqrt(-1*-1) = sqrt(1) = 1.
i is wrong!

Also, the calculus thing on the first page was funny.
lol everything matters
-Ed

User avatar
jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

Re: Favourite Erroneous "Proofs"

Postby jestingrabbit » Sat Dec 08, 2007 3:44 am UTC

duckshirt wrote:i is wrong!


Nah, thinking of it as the squareroot of -1 is wrong. Correctly you think about it as solving x2+1 = 0. Mathematics: splitting hairs since forever.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Ranorith
Posts: 16
Joined: Sat Dec 22, 2007 6:17 am UTC

Re: Favourite Erroneous "Proofs"

Postby Ranorith » Sat Dec 22, 2007 6:36 am UTC

I am new to the forums, hopefully my first post won't disappoint.

This one was given to my class by my first-year University maths teacher.


Take a triangle with a hypotenuse going for a length of sqrt(2), and with the two other sides of length 1 above the hypotenuse.

Image

Define the cut operation by travelling up from the base line of the triangle along all lines leaving it, half-way to the first vertex, then drawing a line coming down from that point making angle with the first line equal to the angle between the upper lines of the original triangle.

This process can be thought of as cutting new triangles into the old 1/2 way up the old edges, and the first step illustrated below.

Image

The second step would contain another 4 smaller triangles embedded in the two new ones above, with side lengths 1/4 (and a hypotenuse of another length), etc.

Now the path along Pn is given by taking the path along the cuts after n steps, and we use Pn also to represent that path's length. P(1) is shown above.

Inspection shows that Pn = 2 for all n, as you always double the number of lines you travel along, and halve the length of each one (and the original length was 1+1 = 2)
Hence limn->infinity Pn= 2
However this is clearly equal to the length of the hypotenuse, as the limiting Path is the hypotenuse.

Hence sqrt(2) = 2.

User avatar
Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

Re: Favourite Erroneous "Proofs"

Postby Blatm » Sun Dec 23, 2007 12:54 am UTC

There was a thread about that a while back. You can find it here.

User avatar
quintopia
Posts: 2906
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

Re: Favourite Erroneous "Proofs"

Postby quintopia » Mon Dec 24, 2007 3:28 am UTC

I was disappointed by that first post, in that it continues to be the first and only post.

User avatar
antonfire
Posts: 1772
Joined: Thu Apr 05, 2007 7:31 pm UTC

Re: Favourite Erroneous "Proofs"

Postby antonfire » Sat Jan 19, 2008 10:44 pm UTC

Recall that RPn is the set Rn+1\{0} (which we will denote Rn+1-1), with all points on each line through the origin identified. Each line through the origin is R-1, so we can write RPn = (Rn+1-1)/(R-1). Also, we can get RPn by gluing RPn-1 to Rn at the boundary in the appropriate way, so we will write RPn = Rn+RPn-1, which gives us RPn=Rn+Rn-1+...+R+1.

So, Rn+Rn-1+...+R+1 = (Rn+1-1)/(R-1)
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Favourite Erroneous "Proofs"

Postby skeptical scientist » Sat Jan 19, 2008 11:18 pm UTC

antonfire wrote:So, Rn+Rn-1+...+R+1 = (Rn+1-1)/(R-1)

Nice. I think I'll put that up on the blackboard in my office.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
UserGoogol
Posts: 90
Joined: Mon Jan 29, 2007 5:58 am UTC

Re: Favourite Erroneous "Proofs"

Postby UserGoogol » Sun Jan 20, 2008 3:07 am UTC

All Canadians are the same age:

Suppose that you have a group of n Canadians. If n = 1, then the group is all of the same age trivially, and if n = k + 1, then if it holds for n = k, the group can be divided into a_1, a_2 ... a_k and a_2, a_3 ... a_k, a_(k+1). The first group is all of the same age and the second group is all of the same age and the groups overlap, therefore it is true for n = k + 1. Therefore by induction, it holds for all n, including whatever the total amount of Canadians is, therefore all Canadians are of the same age.

This proof can of course be extended to any arbitrary group and property, but I find the "all Canadians have the same age" formulation particularly amusing.

Ranorith wrote:I am new to the forums, hopefully my first post won't disappoint.

This one was given to my class by my first-year University maths teacher.

[snip]

Hence sqrt(2) = 2.


When I was in elementary school something along those lines occured to me. On what I believe was Square One Television the question was being asked how to measure the length of a set of stairs without measuring each stair (in the context of some story). My initial guess was the Pythagorean theorem, but in fact the answer was simply height plus width for the reasons you say. I immediately came to the conclusion of this oddness, since the hypotenuse is surely just an infinitely fine stairway. The precise reason why this is wrong I didn't find out until last year: that being that a curve needs to approach another curve in slope for the lengths to be the same. It amuses me when stuff I found vaguely interesting in elementary school pops up in college.

User avatar
Owehn
Posts: 479
Joined: Tue Oct 09, 2007 12:49 pm UTC
Location: Cambridge, UK

Re: Favourite Erroneous "Proofs"

Postby Owehn » Sun Jan 20, 2008 3:16 am UTC

UserGoogol wrote:On what I believe was Square One Television the question was being asked how to measure the length of a set of stairs without measuring each stair (in the context of some story).


I have three disconnected memories that I think are related. One is this problem, another is some bloke saying "'Tri' means 3", and the third is a girl explaining how all of her father's vacation days are the same as those of a serial criminal. Are these all from Square One Television?
[This space intentionally left blank.]

JoshuaZ
Posts: 401
Joined: Tue Apr 24, 2007 1:18 am UTC
Contact:

Re: Favourite Erroneous "Proofs"

Postby JoshuaZ » Mon Jan 21, 2008 8:11 pm UTC

Proposition: 1 does not exist.
Proof:consider limx->oo (x^2 + sin x)/(x^2+2x + cos x). Clearly the limit of this is 1. But we have the top and bottom both going to infinity, so we could also apply L'Hospital's rule. So we get limx->oo (x^2 + sin x)/(x^2+2x + cos x) = limx->oo (2x + cos x)/(2x +2 - sin x). Again, the top and bottom are going to infinity so we can apply L'Hospital's rule. So limx->oo (2x + cos x)/(2x +2 - sin x) = limx->oo (2 - sin x)/(2 - cos x) but this limit clearly does not exist. So 1 does not exist.

User avatar
Charlie!
Posts: 2035
Joined: Sat Jan 12, 2008 8:20 pm UTC

Re: Favourite Erroneous "Proofs"

Postby Charlie! » Mon Feb 04, 2008 8:15 pm UTC

Fun math joke: prove that the double integral of sin(x) is x - sin(x).

First, take the taylor series of sin(x): x - x^3/3! + x^5/5! etc...

Then integrate once to get x^2/2! - x^4/4! + x^6/6! etc...

Integrate for the second time to get x^3/3! - x^5/5! + x^7/7! etc..., which is clearly x - sin(x).
Some people tell me I laugh too much. To them I say, "ha ha ha!"


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 8 guests