Author 
Message 
David R Tribble science forum Guru
Joined: 21 Jul 2005
Posts: 1005

Posted: Mon Jun 12, 2006 9:55 pm Post subject:
Re: Funny wrong proofs



tomhcmi wrote:
Quote:  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.

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. 

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Alexmcw science forum beginner
Joined: 03 Mar 2006
Posts: 7

Posted: Mon Jun 12, 2006 9:37 pm Post subject:
Re: Funny wrong proofs



Not enough info given to solve the problem. What is the cousin's name?
Amcwill
<h.tracy@gmail.com> wrote in message
news:1150137872.443491.146390@i40g2000cwc.googlegroups.com...
Quote:  Just to beat an already dead horse, here's another silly analogy:
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
Dave Seaman wrote:
On Sat, 10 Jun 2006 19:53:53 0400, Mark Spahn wrote:
Woops. That line "= (f'(b)f'(a))/(ba)." should be
= (f'(b)f'(a))/(ba), if f' is linear."
"Mark Spahn" <mspahn@localnet.com> wrote in message
news:128mm4ufv64a8ef@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



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h.tracy@gmail.com science forum beginner
Joined: 10 Dec 2005
Posts: 17

Posted: Mon Jun 12, 2006 6:44 pm Post subject:
Re: Funny wrong proofs



Just to beat an already dead horse, here's another silly analogy:
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
Dave Seaman wrote:
Quote:  On Sat, 10 Jun 2006 19:53:53 0400, Mark Spahn wrote:
Woops. That line "= (f'(b)f'(a))/(ba)." should be
= (f'(b)f'(a))/(ba), if f' is linear."
"Mark Spahn" <mspahn@localnet.com> wrote in message
news:128mm4ufv64a8ef@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 


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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jun 11, 2006 4:04 am Post subject:
Re: Funny wrong proofs



h.tracy@gmail.com wrote:
Quote:  What's the average depth of the Grand Canyon, rim to rim? Well, the
depth at the south rim is 0, and the depth at the north rim is 0, so
the average depth is (0+0)/2 = 0.

No, that's the average change in height from one side to the other.
Quote:  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.
Quote: 
Mark Spahn wrote:
I don't get the joke. I am unfamiliar with the term
"average rate of change of a function over an interval",
but the answer given seems perfectly reasonable.
At x=1 the function is changing at the rate of 5,
it changes at the rate of 7 and x=2, and the value
of the rate of change, the derivative of the function, is 2x+3,
which is linear. Hence the average value of 2x+3
over interval [1,2] is 6. What am I misunderstanding?
What am I doing wrong?
 Mark Spahn
"Proginoskes" <CCHeckman@gmail.com> wrote in message news:1149810835.039279.180240@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 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.
=_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:CCHeckman@gmail.com">CCHeckman@gmail.com</A>> wrote in message
A
href="news:1149810835.039279.180240@y43g2000cwc.googlegroups.com">news:1149810835.039279.180240@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

Don't post the HTML part here; just include Plain Text in the future.
(This applies to h.tracy and Mark Spahn.
 Christopher Heckman 

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Dave Seaman science forum Guru
Joined: 24 Mar 2005
Posts: 527

Posted: Sun Jun 11, 2006 3:14 am Post subject:
Re: Funny wrong proofs



On Sat, 10 Jun 2006 19:53:53 0400, Mark Spahn wrote:
Quote:  Woops. That line "= (f'(b)f'(a))/(ba)." should be
= (f'(b)f'(a))/(ba), if f' is linear."
"Mark Spahn" <mspahn@localnet.com> wrote in message
news:128mm4ufv64a8ef@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).
Quote:  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/> 

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Mark Spahn science forum addict
Joined: 07 Jul 2005
Posts: 62

Posted: Sat Jun 10, 2006 11:53 pm Post subject:
Re: Funny wrong proofs



Woops. That line "= (f'(b)f'(a))/(ba)." should be
= (f'(b)f'(a))/(ba), if f' is linear."
"Mark Spahn" <mspahn@localnet.com> wrote in message
news:128mm4ufv64a8ef@corp.supernews.com...
Quote:  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).
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_lawrence_petry@yahoo.com> wrote in message
news:1149979688.049145.282320@i40g2000cwc.googlegroups.com...
Proginoskes wrote:
Quote:  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 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.

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! 

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Mark Spahn science forum addict
Joined: 07 Jul 2005
Posts: 62

Posted: Sat Jun 10, 2006 11:41 pm Post subject:
Re: Funny wrong proofs



Quote:  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).
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_lawrence_petry@yahoo.com> wrote in message
news:1149979688.049145.282320@i40g2000cwc.googlegroups.com...
Proginoskes wrote:
Quote:  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 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.

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! 

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david petry science forum Guru
Joined: 18 May 2005
Posts: 503

Posted: Sat Jun 10, 2006 10:48 pm Post subject:
Re: Funny wrong proofs



Proginoskes wrote:
Quote:  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 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.

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! 

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h.tracy@gmail.com science forum beginner
Joined: 10 Dec 2005
Posts: 17

Posted: Sat Jun 10, 2006 10:14 pm Post subject:
Re: Funny wrong proofs



What's the average depth of the Grand Canyon, rim to rim? Well, the
depth at the south rim is 0, and the depth at the north rim is 0, so
the average depth is (0+0)/2 = 0.
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.
Mark Spahn wrote:
Quote:  I don't get the joke. I am unfamiliar with the term
"average rate of change of a function over an interval",
but the answer given seems perfectly reasonable.
At x=1 the function is changing at the rate of 5,
it changes at the rate of 7 and x=2, and the value
of the rate of change, the derivative of the function, is 2x+3,
which is linear. Hence the average value of 2x+3
over interval [1,2] is 6. What am I misunderstanding?
What am I doing wrong?
 Mark Spahn
"Proginoskes" <CCHeckman@gmail.com> wrote in message news:1149810835.039279.180240@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 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
=_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:CCHeckman@gmail.com">CCHeckman@gmail.com</A>> wrote in message
A
href="news:1149810835.039279.180240@y43g2000cwc.googlegroups.com">news:1149810835.039279.180240@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 


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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Sat Jun 10, 2006 10:46 am Post subject:
Re: Funny wrong proofs



On Fri, 9 Jun 2006, Virgil wrote:
Quote:  "Pubkeybreaker" <Robert_silverman@raytheon.com> wrote:
José Carlos Santos wrote:
I would like to have examples of funny wrong proofs.
How about:
Example from class: (bonehead math for nonmajors)
lim x >8 1/(x ==> 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. 

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Oscar Lanzi III science forum Guru Wannabe
Joined: 30 Apr 2005
Posts: 176

Posted: Sat Jun 10, 2006 1:50 am Post subject:
Re: Funny wrong proofs



The classic case, IMHO, is to reduce 16/64 or 19/95 to lowest terms by
canceling the common digit.
OL 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Fri Jun 09, 2006 11:29 pm Post subject:
Re: Funny wrong proofs



Mark Spahn wrote:
Quote:  I don't get the joke. I am unfamiliar with the term
"average rate of change of a function over an interval",

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
Quote:  but the answer given seems perfectly reasonable.
At x=1 the function is changing at the rate of 5,
it changes at the rate of 7 and x=2, and the value
of the rate of change, the derivative of the function, is 2x+3,
which is linear. Hence the average value of 2x+3
over interval [1,2] is 6. What am I misunderstanding?
What am I doing wrong?
 Mark Spahn
"Proginoskes" <CCHeckman@gmail.com> wrote in message news:1149810835.039279.180240@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 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
=_NextPart_000_0015_01C68BCA.A5BEAFF0
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META httpequiv=ContentType content="text/html; charset=iso88591"
META content="MSHTML 6.00.2900.2873" name=GENERATOR
STYLE></STYLE
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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:CCHeckman@gmail.com">CCHeckman@gmail.com</A>> wrote in message
A
href="news:1149810835.039279.180240@y43g2000cwc.googlegroups.com">news:1149810835.039279.180240@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 


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Alexmcw science forum beginner
Joined: 03 Mar 2006
Posts: 7

Posted: Fri Jun 09, 2006 10:42 pm Post subject:
Re: Funny wrong proofs



"Average rate of change" versus "average change"??????????
Alex
"Mark Spahn" <mspahn@localnet.com> wrote in message
news:128jcotrj73jsc1@corp.supernews.com...
I don't get the joke. I am unfamiliar with the term
"average rate of change of a function over an interval",
but the answer given seems perfectly reasonable.
At x=1 the function is changing at the rate of 5,
it changes at the rate of 7 and x=2, and the value
of the rate of change, the derivative of the function, is 2x+3,
which is linear. Hence the average value of 2x+3
over interval [1,2] is 6. What am I misunderstanding?
What am I doing wrong?
 Mark Spahn
"Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1149810835.039279.180240@y43g2000cwc.googlegroups.com...
José Carlos Santos wrote:
Quote:  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 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 

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Kees science forum addict
Joined: 09 Sep 2005
Posts: 90

Posted: Fri Jun 09, 2006 10:21 pm Post subject:
Re: Funny wrong proofs



Quote:  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.
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

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 

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Herman Rubin science forum Guru
Joined: 25 Mar 2005
Posts: 730

Posted: Fri Jun 09, 2006 6:39 pm Post subject:
Re: Funny wrong proofs



In article <1149869586.407726.169920@i40g2000cwc.googlegroups.com>,
tomhcmi <tomhchappell@yahoo.com> wrote:
Quote:  Jos=E9 Carlos Santos wrote:
I would like to have examples of funny wrong proofs. ...
... "real", in the sense that they were made by
math students.
[snip]
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 (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
hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558 

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