## Favourite Erroneous "Proofs"

For the discussion of math. Duh.

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schrodingersduck
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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
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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.

bonder
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### Re: Favourite Erroneous "Proofs"

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

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
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quintopia
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### Re: Favourite Erroneous "Proofs"

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
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### Re: Favourite Erroneous "Proofs"

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.

Geekthras
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### Re: Favourite Erroneous "Proofs"

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?!

SimonM
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### Re: Favourite Erroneous "Proofs"

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.
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gmalivuk
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### Re: Favourite Erroneous "Proofs"

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.)
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SimonM
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### Re: Favourite Erroneous "Proofs"

And why is that? It because x^x^x^.^.^. only converges for e-e<x<e1/e
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Droooo
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### Re: Favourite Erroneous "Proofs"

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.

Macbi
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### Re: Favourite Erroneous "Proofs"

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.
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Stanford
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### Re: Favourite Erroneous "Proofs"

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.

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### Re: Favourite Erroneous "Proofs"

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.
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rrwoods
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### Re: Favourite Erroneous "Proofs"

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 +/-.)
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Stanford
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### Re: Favourite Erroneous "Proofs"

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
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### Re: Favourite Erroneous "Proofs"

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).
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### Re: Favourite Erroneous "Proofs"

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.
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Alpha Omicron
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### Re: Favourite Erroneous "Proofs"

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.
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### Re: Favourite Erroneous "Proofs"

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.
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btilly
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### Re: Favourite Erroneous "Proofs"

I've always liked the inflation theorem.

Code: Select all

1$= 100 c = (10c)^2 = (1/10$)^2   = 1/100$= 1c Some of us exist to find out what can and can't be done. Others exist to hold the beer. quintopia Posts: 2906 Joined: Fri Nov 17, 2006 2:53 am UTC Location: atlanta, ga ### Re: Favourite Erroneous "Proofs" btilly wrote:I've always liked the inflation theorem. Code: Select all 1$ = 100 c   = (10c)^2   = (1/10$)^2 = 1/100$   = 1c

I C Wut U Did There.

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

the tree
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### Re: Favourite Erroneous "Proofs"

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
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### Re: Favourite Erroneous "Proofs"

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
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### Re: Favourite Erroneous "Proofs"

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.

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### Re: Favourite Erroneous "Proofs"

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.

Alpha Omicron
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### Re: Favourite Erroneous "Proofs"

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.
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2ndFloor
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### Re: Favourite Erroneous "Proofs"

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.

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### Re: Favourite Erroneous "Proofs"

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.

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) = 1Then integrate again: f(x) = x + C (don't forget your constant of integration!).Consider another function, g(x) = x + b, such that a <> bDifferentiate and then integrate to get g(x) = x + C. x + C = x + Cf(x) = g(x)x + a = x + ba = b

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

GuyWithFunnyHat
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### Re: Favourite Erroneous "Proofs"

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
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### Re: Favourite Erroneous "Proofs"

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.
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### Re: Favourite Erroneous "Proofs"

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.
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### Re: Favourite Erroneous "Proofs"

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.

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.

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.

Blatm
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### Re: Favourite Erroneous "Proofs"

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

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### Re: Favourite Erroneous "Proofs"

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

antonfire
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### Re: Favourite Erroneous "Proofs"

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)
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### Re: Favourite Erroneous "Proofs"

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.
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### Re: Favourite Erroneous "Proofs"

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.

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### Re: Favourite Erroneous "Proofs"

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.]

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### Re: Favourite Erroneous "Proofs"

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.

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### Re: Favourite Erroneous "Proofs"

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).
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