I would like to have examples of funny wrong proofs. Here are two
examples. They are "real", in the sense that they were made by
math students.
1) Proof of the fact that, in a field, you have
(a^2  b^2)/(a + b) = a  b
Proof: (a^2  b^2)/(a + b) = a^2/a + (/+) + b^2/b = a  b.
2) Proof of the CayleyHamilton theorem: in order to prove that a matrix
M is a zero of its characteristic polynomial, all you have to do is to
notice that this polynomila is det(x.Id  M) and that it is obvious that
det(M.Id  M) = det(0) = 0.
Best regards,
Jose Carlos Santos
> I would like to have examples of funny wrong proofs. Here are two
> examples. They are "real", in the sense that they were made by
> math students.
>
> 1) Proof of the fact that, in a field, you have
>
> (a^2  b^2)/(a + b) = a  b
>
Fantastic corollary: x is the unknown.
Let a = b = x/2. Thus x = 0/0.
> I would like to have examples of funny wrong proofs. Here are two
> examples. They are "real", in the sense that they were made by
> math students.
>
> 1) Proof of the fact that, in a field, you have
>
> (a^2  b^2)/(a + b) = a  b
>
Fantastic corollary: x is the unknown.
Let a = b = x/2. Thus x = 0/0.
which = 0, right?
That reminds me that in Yosida's Functional Analysis (sixth edition),
the author "proves" that a certain subset A of a metric space M has a
nonempty interior by proving that it is homeomorphic to another subset
B of M whose interior is nonempty.
> 1) Proof of the fact that, in a field, you have
>
> (a^2  b^2)/(a + b) = a  b
>
> Proof: (a^2  b^2)/(a + b) = a^2/a + (/+) + b^2/b = a  b.
Wow. I love this one. Lots of students would do something like:
(a^2  b^2)/(a + b) = a^2/a  b^2/b = a  b.
That's obvious. But dividing the minus sign by the plus sign: that's
insight, baby!

Jesse F. Hughes
"I think the burden is on those people who think he didn't have
weapons of mass destruction to tell the world where they are."
 White House spokesman Ari Fleischer
Indeed
x = sqr 2 = pi
because
0 = 0(sqr 2) = 0.pi
There you see, we don't know what x is, it can be anything.
However we do notice that x is a thing for it can't be
two places at once, say for example at pi and at sqr 2.
Oh no no no, that is not possible, that pi = x = sqr 2, that a mere
computably algebraic number dare rise unto transcendental eminence.
Hmm. Was this a general metric space?
Of course the sketch above is invalid in general, but for
example it's valid in R^n, by "Invariance of Domain".
Which raises an obvious question... hmm, no it's not valid
even in a Hilbert space (for example consider onesided
l^2, the unit ball and the image of the unit ball under
a right shift.)
>Best regards,
>
>Jose Carlos Santos
************************
David C. Ullrich
This is actually a valid proof, but still rather amusing. A problem on
a calculus exam:
Let F(x) = [integral from t=1 to t=2 of sin(t^2) dt] .
Compute the derivative F'(x).
Solution given by one student:
Break up the integral as a sum:
F(x) = [integral from t=1 to t=x of sin(t^2) dt] + [integral from t=x
to t=2 of sin(t^2) dt]
Switch the limits of integration on the second integral:
F(x) = [integral from t=1 to t=x of sin(t^2) dt]  [integral from t=2
to t=x of sin(t^2) dt]
Differentiate using the 2nd fundamental theorem of calculus:
F'(x) = sin(x^2)  sin(x^2) = 0.
Wrong proofs yield wrong answers (most of the time). Your
examples fit my revised subject better, IMO.
In statistics, there is a wellknown example of the "pool variance"
of the difference between two sample means, as in a Ttest, is given by
[(n1) u^2 + (m1)v^2]/(n+m2)
where u^2 and v^2 are the unbiased estimates of the variance
of x1bar and x2bar respectively.
A correct result will be obtained by those students who take
this as the correct algebraic formula:
a/b + c/d = (a+b)/(c+d),
using the numerator of u^2 as a, and the numerator of v^2 as c,
and (c+d) = sum of the denominators = (n1)+(m1) = n+m2.
 Bob.
>>> It's not exactly what you're looking for, by years ago I saw a comical
>>> gaffe in a book written by the one of the bestknown mathematicians in
>>> the game. The fellow sets out to prove a certain theorem by induction,
>>> proves it for n=1, and then moves on to the next theorem!
>> That reminds me that in Yosida's Functional Analysis (sixth edition),
>> the author "proves" that a certain subset A of a metric space M has a
>> nonempty interior by proving that it is homeomorphic to another subset
>> B of M whose interior is nonempty.
>
> Hmm. Was this a general metric space?
No, but it was a general Banach space.
How about a wrong calculation (by a nonmath student)?
PROBLEM. Find the average rate of change of the function
f(x) = x^2 + 3x + 8 over the interval [1,2].
"SOLUTION":
f'(x) = 2x + 3
f'(1) = 5
f'(2) = 7
(f'(1) + f'(2))/2 = (5+7)/2 = 6.
This almost turned into an American Mathematical Monthly problem, until
I found out someone else had included my wouldbe problem in their
article:
PROBLEM. Characterize all differentiable functions f(x) such that
(f'(a) + f'(b))/2 = (f(b)  f(a)) / (b  a),
for all real numbers a,b.
 Christopher Heckman
I don't think these were made by any math students;
but
16/64 cancel the sixes and get 1/4
19/95 cancel the nines and get 1/5
26/65 cancel the sixes and get 2/5
49/98 cancel the nines and get 4/8
I don't know whether there are more like those.
Tom
How about:
Example from class: (bonehead math for nonmajors)
lim x >8 1/(x8) ==> oo
Exam problem:
lim x > 5 1/(x5)
The answer given on the exam was:
___
__ 
When the answer was marked wrong, the student complained
that from the example given in class, she wasn't sure
whether the answer should be
___
__ 
or:
__
 ___
"Proginoskes" <CCHe...@gmail.com> wrote in message news:1149810835.0...@y43g2000cwc.googlegroups.com...
> José Carlos Santos wrote:
> > Hi all:
> >
> > I would like to have examples of funny wrong proofs. Here are two
> > examples. They are "real", in the sense that they were made by
> > math students.
>
>
> How about:
>
> Example from class: (bonehead math for nonmajors)
>
> lim x >8 1/(x8) ==> oo
>
> Exam problem:
>
>
> lim x > 5 1/(x5)
>
> The answer given on the exam was:
>
>
>   
>
> When the answer was marked wrong, the student complained
> that from the example given in class, she wasn't sure
> whether the answer should be
>
>
>   
>
>
> or:
>
>
>   
Don't you mean something like
_
_/ \
?

>Tom
How about (x^2  1)/(x  1)? Cancel x^2 against
x, leaving x in the numerator and nothing in the
denominator, and cancel  against  leaving +.
Hence one gets (x + 1)/(+ 1) = x + 1.

This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558
Here is one I encounterd myself when I was cheking homework assignments in a elementary combinatorics course. Proof that in a graph the distance between vertices satisfies the triangle inequality.
Proof (actually given by more then one student): For two vertices u,v let uv be the graph distance between u and v. We now want to prove that uw\le uv+vw which holds for the triangle inequality for . QED.
Cheers,
Kees
Alex
"Mark Spahn" <msp...@localnet.com> wrote in message
news:128jcot...@corp.supernews.com...
The average rate of change of f(x) over [a,b] is
(f(b)  f(a)) / (b  a),
and this is the formula that the student was _supposed_ to use. (It's
the basis of the derivative formula.) Perhaps you learned it with
another name.
 Christopher Heckman
> =_NextPart_000_0015_01C68BCA.A5BEAFF0
> ContentType: text/html; charset=iso88591
> ContentTransferEncoding: quotedprintable
> XGoogleAttachSize: 2311
>
> <!DOCTYPE HTML PUBLIC "//W3C//DTD HTML 4.0 Transitional//EN">
> <HTML><HEAD>
> <META httpequiv=ContentType content="text/html; charset=iso88591">
> <META content="MSHTML 6.00.2900.2873" name=GENERATOR>
> <STYLE></STYLE>
> </HEAD>
> <BODY bgColor=#ffffff>
> <DIV>I don't get the joke. I am unfamiliar with the term</DIV>
> <DIV>"average rate of change of a function over an interval",</DIV>
> <DIV>but the answer given seems perfectly reasonable.</DIV>
> <DIV>At x=1 the function is changing at the rate of 5,</DIV>
> <DIV>it changes at the rate of 7 and x=2, and the value</DIV>
> <DIV>of the rate of change, the derivative of the function, is 2x+3,</DIV>
> <DIV>which is linear. Hence the average value of 2x+3</DIV>
> <DIV>over interval [1,2] is 6. What am I misunderstanding?</DIV>
> <DIV>What am I doing wrong?</DIV>
> <DIV> </DIV>
> <DIV> Mark Spahn</DIV>
> <DIV> </DIV>
> <BLOCKQUOTE
> style="PADDINGRIGHT: 0px; PADDINGLEFT: 5px; MARGINLEFT: 5px; BORDERLEFT: #000000 2px solid; MARGINRIGHT: 0px">
> <DIV>"Proginoskes" <<A
> href="mailto:CCHe...@gmail.com">CCHe...@gmail.com</A>> wrote in message
> <A
> href="news:1149810835.0...@y43g2000cwc.googlegroups.com">news:1149810835.0...@y43g2000cwc.googlegroups.com</A>...</DIV><BR>José
> Carlos Santos wrote:<BR>> Hi all:<BR>><BR>> I would like to have
> examples of funny wrong proofs. Here are two<BR>> examples. They are
> "real", in the sense that they were made by<BR>> math students.<BR><BR>How
> about a wrong calculation (by a nonmath student)?<BR><BR>PROBLEM. Find the
> average rate of change of the function<BR>f(x) = x^2 + 3x + 8 over the
> interval [1,2].<BR><BR>"SOLUTION":<BR>f'(x) = 2x + 3<BR>f'(1) = 5<BR>f'(2) =
> 7<BR>(f'(1) + f'(2))/2 = (5+7)/2 = 6.<BR><BR>This almost turned into an
> American Mathematical Monthly problem, until<BR>I found out someone else had
> included my wouldbe problem in their<BR>article:<BR><BR>PROBLEM. Characterize
> all differentiable functions f(x) such that<BR><BR>(f'(a) + f'(b))/2 = (f(b) 
> f(a)) / (b  a),<BR><BR>for all real numbers
> a,b.<BR><BR>  Christopher
> Heckman<BR></BLOCKQUOTE></BODY></HTML>
>
> =_NextPart_000_0015_01C68BCA.A5BEAFF0
OL
> > How about:
> > Example from class: (bonehead math for nonmajors)
> >
> > lim x >8 1/(x8) ==> oo
> >
> > Exam problem:
> >
> > lim x > 5 1/(x5)
> >
> > The answer given on the exam was:
> >
> >
> >   
> >
> > When the answer was marked wrong, the student complained
> > that from the example given in class, she wasn't sure
> > whether the answer should be
> >
> >
> >   
> >
> >
> > or:
> >
> >
> >   
>
> Don't you mean something like
> _
> _/ \
>
Yes, however the multiple _ underscores didn't come thru your browser
for you to see. For you to see
__
__ 
with four underscores, two below and two above.
Actually, it's a perfectly valid calculation as long as you first
observe that the second derivative of f is constant, so the rate of
change itself is a linear function, and average of the endpoints always
gives the correct overall average if (and only if) a function is
linear. (If it were a straight line between the north and south rims,
the average depth would indeed be 0). The average rate of change means
the average over the entire intervalyou could have a function which
happened to have extremely steep rates of change at both endpoints, but
which on average went downhill instead of up. You take the average
over an entire interval by computing the integral and dividing by the
length, but in this case you're taking the integral of a derivative and
you wind up just evaluating the function itself.
If the student had offered an explanation for what he was doing, it
would have been entirely correct.
Since f'(x) is a linear function in x, **and we know that the average
value of a linear function is equal to the average value of its
endpoints**, we can proceed as above!
Actually, this question (from a math test, I presume)
does not test mathematical knowledge, but rather tests
the conventions of what various mathematical objects
are called in English.
By convention, "the average rate of change of function f
over interval [a,b]" is the wording that is used to describe
(f(b)f(a))/(ba). But I interpreted this wording to mean
"the average OF THE rate of change of function f over
interval [a,b]", which would mean
"the average of the derivative of f over interval [a,b]"
= "the average of function f' over interval [a,b]"
= (f'(b)f'(a))/(ba).
In my opinion, this is not quite fair as a math question,
but maybe it is fair as a question of math nomenclature.
The meaning of this question is too ambiguous to be
on an important test (even for people whose native
language is English).
 Mark Spahn
"david petry" <david_lawr...@yahoo.com> wrote in message
news:1149979688.0...@i40g2000cwc.googlegroups.com...
"Mark Spahn" <msp...@localnet.com> wrote in message
news:128mm4u...@corp.supernews.com...
> "Mark Spahn" <msp...@localnet.com> wrote in message
> news:128mm4u...@corp.supernews.com...
>> PROBLEM. Find the average rate of change of the function
>> f(x) = x^2 + 3x + 8 over the interval [1,2].
> Actually, this question (from a math test, I presume)
> does not test mathematical knowledge, but rather tests
> the conventions of what various mathematical objects
> are called in English.
> By convention, "the average rate of change of function f
> over interval [a,b]" is the wording that is used to describe
> (f(b)f(a))/(ba). But I interpreted this wording to mean
> "the average OF THE rate of change of function f over
> interval [a,b]", which would mean
> "the average of the derivative of f over interval [a,b]"
>= "the average of function f' over interval [a,b]"
>= (f'(b)f'(a))/(ba).
Funny, that's how I interpret it as well. But, how do you compute the
average value of f' over [a,b]. Well, f' is a function, and in general,
the way to find the average value of a function over an interval is to
integrate the function over that integral, and divide by the length of
the interval.
So if the function is f', we get an average value of (int_a^b f'(x) dx) /
(ba).
But, by the fundamental theorem of calculus, this is identical to (f(b) 
f(a)) / (ba).
> In my opinion, this is not quite fair as a math question,
> but maybe it is fair as a question of math nomenclature.
> The meaning of this question is too ambiguous to be
> on an important test (even for people whose native
> language is English).
I see no ambguity whatsoever. The meaning is the same either way.

Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia AbuJamal.
<http://www.mumia2000.org/>
No, that's the average change in height from one side to the other.
> Actually, it's a perfectly valid calculation as long as you first
> observe that the second derivative of f is constant, so the rate of
> change itself is a linear function, and average of the endpoints always
> gives the correct overall average if (and only if) a function is
> linear. (If it were a straight line between the north and south rims,
> the average depth would indeed be 0). The average rate of change means
> the average over the entire intervalyou could have a function which
> happened to have extremely steep rates of change at both endpoints, but
> which on average went downhill instead of up. You take the average
> over an entire interval by computing the integral and dividing by the
> length, but in this case you're taking the integral of a derivative and
> you wind up just evaluating the function itself.
Yes, or you could solve the differential equation.
Don't post the HTML part here; just include Plain Text in the future.
(This applies to h.tracy and Mark Spahn.
 Christopher Heckman
Andrew pulls out of his driveway in Berkeley at five miles per hour,
headed for Southern California. Five hours and three hundred and fifty
miles later, he pulls into his cousin's driveway at three miles per
hour. "Wow," says the cousin, "you must not have hit any traffic at
all. How fast did you average, anyway?"
"Well," says Andrew, "I started at five miles per hour, and just now I
finished at three miles per hour, so I figure over the whole trip I
must have averaged four miles per hour."
Tracy
Amcwill
<h.t...@gmail.com> wrote in message
news:1150137872.4...@i40g2000cwc.googlegroups.com...
Find p/q = n/d, where n = 10p+h and d = h10^i+q, i = floor(log10(q))+1,
for h=1,...,9; i.e., where pd = qn.
A quickie Java program for all 4digit numerators and denominators
gave me these:
16/64 = 1/4
19/95 = 1/5
11/110 = 1/10
13/325 = 1/25
16/640 = 1/40
19/950 = 1/50
11/1100 = 1/100
13/3250 = 1/250
16/6400 = 1/400
19/9500 = 1/500
26/65 = 2/5
22/220 = 2/20
26/650 = 2/50
27/756 = 2/56
22/2200 = 2/200
26/6500 = 2/500
27/7560 = 2/560
33/330 = 3/30
39/975 = 3/75
33/3300 = 3/300
39/9750 = 3/750
49/98 = 4/8
44/440 = 4/40
49/980 = 4/80
44/4400 = 4/400
49/9800 = 4/800
55/550 = 5/50
55/5500 = 5/500
66/660 = 6/60
66/6600 = 6/600
77/770 = 7/70
77/7700 = 7/700
79/9875 = 7/875
83/332 = 8/32
88/880 = 8/80
83/3320 = 8/320
88/8800 = 8/800
99/990 = 9/90
99/9900 = 9/900
106/6625 = 10/625
111/1110 = 11/110
133/3325 = 13/325
138/8832 = 13/832
166/664 = 16/64
166/6640 = 16/640
199/995 = 19/95
199/9950 = 19/950
217/775 = 21/75
217/7750 = 21/750
222/2220 = 22/220
249/996 = 24/96
249/9960 = 24/960
266/665 = 26/65
266/6650 = 26/650
277/7756 = 27/756
333/3330 = 33/330
399/9975 = 39/975
416/6656 = 41/656
444/4440 = 44/440
499/998 = 49/98
499/9980 = 49/980
555/5550 = 55/550
568/8875 = 56/875
666/6660 = 66/660
777/7770 = 77/770
833/3332 = 83/332
888/8880 = 88/880
999/9990 = 99/990
1249/9992 = 124/992
1666/6664 = 166/664
1999/9995 = 199/995
2177/7775 = 217/775
2499/9996 = 249/996
2666/6665 = 266/665
4999/9998 = 499/998
The ones with trailing zeroes seem to be especially boring.