Have fun.
Answer to follow.
Lew
---------------------------------------------------------------------------------------
Let: a = b,
Then: (1) a^2 = ab,
(2) a^2 - b^2 = ab - b^2,
(3) (a + b)(a - b) = b(a - b),
(4) a + b = b,
(5) 2b = b,
(6) 2 = 1
It would seem that we have proved 2 = 1.
What is wrong here?
---------------------------------------------------------------------------------------
From:
Basic Mathematics For Engineering And Science, Page 18, 1952
By:
Walter R Van Voorhis, PhD, Professor of Mathematics
Fenn College, Cleveland, Ohio
and
Elmer E Haskins, PhD, Professor of Mechanical Engineering
Fenn College, Cleveland, Ohio
---------------------------------------------------------------------------------------
"Lew Hodgett" <[email protected]> wrote:
>Have fun.
>
>Answer to follow.
>
>Lew
>---------------------------------------------------------------------------------------
>
>Let: a = b,
>Then: (1) a^2 = ab,
> (2) a^2 - b^2 = ab - b^2,
> (3) (a + b)(a - b) = b(a - b),
> (4) a + b = b,
> (5) 2b = b,
> (6) 2 = 1
>
>It would seem that we have proved 2 = 1.
>
>What is wrong here?
>
>---------------------------------------------------------------------------------------
Grrrr, math can be such a pain in the arse! LOL
http://www.facebook.com/photo.php?v=10202517290999274
On 5/15/2015 3:43 PM, Lew Hodgett wrote:
> "Just Wondering" wrote:
>
>> My exchange with Leon is an accurate description of how Obamacare
>> got passed. People like that deserve my scorn.
> ---------------------------------------------------
> The only thing that counts is that Obamacare is now the law of the
> land.
At least right now, that can be fixed a little later on.
"Lew Hodgett" <[email protected]> wrote:
> Have fun.
>
> Answer to follow.
>
> Lew
> ---------------------------------------------------------------------------------------
>
> Let: a = b,
> Then: (1) a^2 = ab,
> (2) a^2 - b^2 = ab - b^2,
> (3) (a + b)(a - b) = b(a - b),
> (4) a + b = b,
> (5) 2b = b,
> (6) 2 = 1
>
> It would seem that we have proved 2 = 1.
>
> What is wrong here?
>
> ---------------------------------------------------------------------------------------
> From:
>
> Basic Mathematics For Engineering And Science, Page 18, 1952
>
> By:
>
> Walter R Van Voorhis, PhD, Professor of Mathematics
> Fenn College, Cleveland, Ohio
>
> and
>
> Elmer E Haskins, PhD, Professor of Mechanical Engineering
> Fenn College, Cleveland, Ohio
> ---------------------------------------------------------------------------------------
What is wrong?
I think the beginning false pretense that a is equal in value to b.
You can make anything look correct if you begin with deception. .
Bill wrote:
>
> A farmer has 20 feet of fencing and wishing to build a rectangular
> pen against his barn (by adding 3 sides), maximizing the area. What
> should the dimensions be? Note: It makes the kids smile if I
> include a little animal in the diagram.
------------------------------------------------------
Not enough information to develop 3 independent equations.
You can use Taylor's theorem of approximation to determine
that a 5 ft x 10 ft pen with the 10 ft side parallel to barn wall will
yield max area.
-----------------------------------------------------------------
W = Width
L = Length
A = Area
2W + L = 20
A = WL
-------------------------------------------------------------
WHEN W = 7, THEN L = 6
A = 42
WHEN W = 6, THEN L = 8
A = 48
WHEN W = 5, THEN L = 10
A = 50
WHEN W = 4, THEN L = 12
A = 48
WHEN W = 3, THEN L = 14
A = 42
Ye gads you are making me dig.
Haven't used most of this stuff in over 50 years.
Lew
On 5/12/2015 6:43 AM, Leon wrote:
> "Lew Hodgett" <[email protected]> wrote:
>> Have fun.
>>
>> Answer to follow.
>>
>> Lew
>> ---------------------------------------------------------------------------------------
>>
>> Let: a = b,
>> Then: (1) a^2 = ab,
>> (2) a^2 - b^2 = ab - b^2,
>> (3) (a + b)(a - b) = b(a - b),
>> (4) a + b = b,
>> (5) 2b = b,
>> (6) 2 = 1
>>
>> It would seem that we have proved 2 = 1.
>>
>> What is wrong here?
>>
>> ---------------------------------------------------------------------------------------
>> From:
>>
>> Basic Mathematics For Engineering And Science, Page 18, 1952
>>
>> By:
>>
>> Walter R Van Voorhis, PhD, Professor of Mathematics
>> Fenn College, Cleveland, Ohio
>>
>> and
>>
>> Elmer E Haskins, PhD, Professor of Mechanical Engineering
>> Fenn College, Cleveland, Ohio
>> ---------------------------------------------------------------------------------------
>
> What is wrong?
> I think the beginning false pretense that a is equal in value to b.
> You can make anything look correct if you begin with deception. .
>
Is that why so many people vote Democrat?
Bill wrote:
>>
>> Instead of using 2 variables L and W, try it with just 1 (say just
>> L or W), you can write the area as a single equation/function of
>> it. You'll have a quadratic equation whose graph is a parabola...
>> This will lead you not only an answer, but the fact that the answer
>> lies at the vertex of a parabola opening downwards will justify for
>> you that it is the unique best answer.
>>
>>
> Here is a start.
> Let L denote the length. Then the width equals (20-L)/2.
>
> So the area = L(20-L)/2=-.5L^2 +10L, a quadratic equation in terms
> of the single variable L.
>
> You may find the vertex of the graph of A = -.5L^2 +10L in several
> ways (as a local max in calculus, completing the square, using it's
> symmetry about it's axis of symmetry--which is the midpoint of it's
> L-intercepts, for instance (these can be found with the quadratic
> formula or just by factoring)).
>
> Any of these will show you why the answer you already found really
> is the correct and
> only one. ; )
-----------------------------------------------
Too many years and too many beers.
Thanks for the input.
Lew
"Lew Hodgett" wrote:
> Have fun.
>
> Answer to follow.
>
> Lew
---------------------------------------------------------------------------------------
>
> Let: a = b,
> Then: (1) a^2 = ab,
> (2) a^2 - b^2 = ab - b^2,
> (3) (a + b)(a - b) = b(a - b),
> (4) a + b = b,
> (5) 2b = b,
> (6) 2 = 1
>
> It would seem that we have proved 2 = 1.
>
> What is wrong here?
>
---------------------------------------------------------------------------------------> From:>> Basic Mathematics For Engineering And Science, Page 18, 1952>> By:>> Walter R Van Voorhis, PhD, Professor of Mathematics> Fenn College, Cleveland, Ohio>> and>> Elmer E Haskins, PhD, Professor of Mechanical Engineering> Fenn College, Cleveland, Ohio----------------------------------------------------------------------------------------------------------------------------------------------------------------------------The first commandment of algebra according to "Van",Thou shall not divide by zero.After all these years, I still remember.Lew
On 5/13/2015 8:42 AM, Scott Lurndal wrote:
> Just Wondering <[email protected]> writes:
>> On 5/12/2015 6:43 AM, Leon wrote:
>>
>>> You can make anything look correct if you begin with deception.
>>
>> Is that why so many people vote Democrat?
>
> asshole.
>
Looks like I hit a nerve.
On 5/14/2015 8:01 AM, Scott Lurndal wrote:
> Just Wondering <[email protected]> writes:
>> On 5/13/2015 8:42 AM, Scott Lurndal wrote:
>>> Just Wondering <[email protected]> writes:
>>>> On 5/12/2015 6:43 AM, Leon wrote:
>>>>
>>>>> You can make anything look correct if you begin with deception.
>>>>
>>>> Is that why so many people vote Democrat?
>>>
>>> asshole.
>>>
>> Looks like I hit a nerve.
>
> Actually, find your behavior abhorent.
>
You're the one who gratuitously butted your nose into someone else's
converation, in the process using words like asshole, and misspelling
abhorrent.
> But I'm sure you are doing it as a troll,
> since you aren't proud enough of your namecalling
>
Your reading comprehension is also lacking. I obviously did no namecalling.
> and pejoratives (look it up)
>
http://www.merriam-webster.com/thesaurus/pejorative
Synonyms belittling, contemptuous, decrying, degrading, demeaning,
denigrative, denigratory, deprecatory, depreciative, depreciatory,
derisory, derogative, detractive, disdainful, disparaging, pejorative,
scornful, slighting, uncomplimentary
You're a hypocrite, butting in uninvited with pejorative namecalling,
and then falsely accusing me of doing what you did.
My exchange with Leon is an accurate description of how Obamacare got
passed. People like that deserve my scorn.
"Just Wondering" wrote:
> My exchange with Leon is an accurate description of how Obamacare
> got passed. People like that deserve my scorn.
---------------------------------------------------
The only thing that counts is that Obamacare is now the law of the
land.
As far as your scorn is concerned, who cares.
Lew
Just Wondering <[email protected]> writes:
>On 5/13/2015 8:42 AM, Scott Lurndal wrote:
>> Just Wondering <[email protected]> writes:
> >> On 5/12/2015 6:43 AM, Leon wrote:
>>>
> >>> You can make anything look correct if you begin with deception.
>>>
>>> Is that why so many people vote Democrat?
>>
>> asshole.
>>
>Looks like I hit a nerve.
Actually, find your behavior abhorent. But I'm sure you are
doing it as a troll, since you aren't proud enough of your
namecalling and pejoratives (look it up) to attach your name to it.
On 5/11/2015 11:19 PM, Bill wrote:
> Lew Hodgett wrote:
>> Have fun.
>>
>> Answer to follow.
>>
>> Lew
>> ---------------------------------------------------------------------------------------
>>
>>
>> Let: a = b,
>> Then: (1) a^2 = ab,
>> (2) a^2 - b^2 = ab - b^2,
>> (3) (a + b)(a - b) = b(a - b),
>> (4) a + b = b,
>> (5) 2b = b,
>> (6) 2 = 1
>>
>> It would seem that we have proved 2 = 1.
>>
>> What is wrong here?
>>
>> ---------------------------------------------------------------------------------------
>>
>> From:
>>
>> Basic Mathematics For Engineering And Science, Page 18, 1952
>>
>> By:
>>
>> Walter R Van Voorhis, PhD, Professor of Mathematics
>> Fenn College, Cleveland, Ohio
>>
>> and
>>
>> Elmer E Haskins, PhD, Professor of Mechanical Engineering
>> Fenn College, Cleveland, Ohio
>> ---------------------------------------------------------------------------------------
>>
>>
> Illegal to divide by 0 in going from (3) to (4).
Agree. A '+1' would seem inappropriate here :)
John
DerbyDad03 <[email protected]> wrote in
news:[email protected]:
> 2 does equal 1...just ask my wife. Our 2 salaries equal 1 amount for
> her to spend.
Lucky you. My girlfriend hasn't yet quite grasped the idea
that the amount she can spend can't be more than our two
combined salaries :-(
Hoping she gets a better job soon...
John
Bill <[email protected]> wrote in news:[email protected]:
> Lew Hodgett wrote:
>> Too many years and too many beers. Thanks for the input. Lew
>
> Here is a fun one. How many zeros appear at the end of the product:
> 1*2*3*4*5*...*100?
>
> Hint: The answer is Not 2. No credit for multiplying it out, by hand.
>
Twenty.
DerbyDad03 <[email protected]> wrote in
news:[email protected]:
> On Tuesday, May 12, 2015 at 2:16:57 PM UTC-4, John McCoy wrote:
>> DerbyDad03 <[email protected]> wrote in
>> news:[email protected]:
>>
>> > 2 does equal 1...just ask my wife. Our 2 salaries equal 1 amount
>> > for her to spend.
>>
>> Lucky you. My girlfriend hasn't yet quite grasped the idea
>> that the amount she can spend can't be more than our two
>> combined salaries :-(
>>
>> Hoping she gets a better job soon...
>>
>> John
>
> Getting a better job won't help.
>
> "I got a new job that pays 10% more than my old one. Now I can spend
> 10% more than I used to."
>
> Changing the job probably won't change the habit.
Well, she interviewed today for a job that pays substantially
more than she was making (like 2x) and starts Monday, so we
shall see...
John
On Tuesday, May 12, 2015 at 2:16:57 PM UTC-4, John McCoy wrote:
> DerbyDad03 <[email protected]> wrote in
> news:[email protected]:
>
> > 2 does equal 1...just ask my wife. Our 2 salaries equal 1 amount for
> > her to spend.
>
> Lucky you. My girlfriend hasn't yet quite grasped the idea
> that the amount she can spend can't be more than our two
> combined salaries :-(
>
> Hoping she gets a better job soon...
>
> John
Getting a better job won't help.
"I got a new job that pays 10% more than my old one. Now I can spend 10% more than I used to."
Changing the job probably won't change the habit.
>=20
> Let: a =3D b,
> Then: (1) a^2 =3D ab,
> (2) a^2 - b^2 =3D ab - b^2,
> (3) (a + b)(a - b) =3D b(a - b),
> (4) a + b =3D b,
> (5) 2b =3D b,
> (6) 2 =3D 1
>=20
> It would seem that we have proved 2 =3D 1.
>=20
> What is wrong here?
The problem is in line 3. The value of (a-b) is zero. Zero x any number is=
zero. So equation 3 results out to be 0=3D0. You can't do the following st=
eps afterwards. 4/5/6. Even if you use symbols, 2 does not equal 1.
Also to allow the equation to be balanced, you need to apply the same value=
s and operations to both sides of the equation. In #3, you have (a+b) x (a-=
b) on the left. On the right it's b x (a-b). The equation is NOT equal at t=
his point. And as I mentioned before (a-b) is zero.
Do I win something?
MJ
Just Wondering <[email protected]> writes:
>Is that why so many people vote Democrat?
asshole.
On Tuesday, May 12, 2015 at 9:28:32 PM UTC-7, Bill wrote:
> Lew Hodgett wrote:
> > Too many years and too many beers. Thanks for the input. Lew
>
> Here is a fun one. How many zeros appear at the end of the product:
> 1*2*3*4*5*...*100?
>
> Hint: The answer is Not 2. No credit for multiplying it out, by hand.
Remember rot13?
Gjragl-sbhe Gur uneq cneg, vf erzrzorevat gung zhygvcyrf bs gjragl-svir
unir gjb snpgbef bs svir. Gurer'f na rkprff bs snpgbef bs gjb.
On Monday, May 11, 2015 at 10:50:59 PM UTC-4, Lew Hodgett wrote:
> Have fun.
>
> Answer to follow.
>
> Lew
> ---------------------------------------------------------------------------------------
>
> Let: a = b,
> Then: (1) a^2 = ab,
> (2) a^2 - b^2 = ab - b^2,
> (3) (a + b)(a - b) = b(a - b),
> (4) a + b = b,
> (5) 2b = b,
> (6) 2 = 1
>
> It would seem that we have proved 2 = 1.
>
> What is wrong here?
>
There is nothing wrong.
2 does equal 1...just ask my wife. Our 2 salaries equal 1 amount for her to spend.
"Lew Hodgett" <[email protected]> writes:
>Have fun.
>
>Answer to follow.
>
>Lew
>---------------------------------------------------------------------------------------
>
>Let: a = b,
therefore:
a = a
>Then: (1) a^2 = ab,
a^2 == aa
> (2) a^2 - b^2 = ab - b^2,
a^2 - a^2 = a^2 - a^2 (reduces to 0 = 0)
> (3) (a + b)(a - b) = b(a - b),
(a+a)(a-a) = b(a-a)
Here you multiply by zero (so 0 = 0)
> (4) a + b = b,
a+a = a
Demonstrably false.
> (5) 2b = b,
> (6) 2 = 1
>
>It would seem that we have proved 2 = 1.
>
>What is wrong here?
>
>---------------------------------------------------------------------------------------
>From:
>
>Basic Mathematics For Engineering And Science, Page 18, 1952
>
>By:
>
>Walter R Van Voorhis, PhD, Professor of Mathematics
>Fenn College, Cleveland, Ohio
>
>and
>
>Elmer E Haskins, PhD, Professor of Mechanical Engineering
>Fenn College, Cleveland, Ohio
>---------------------------------------------------------------------------------------
>
>
Lew Hodgett wrote:
> Have fun.
>
> Answer to follow.
>
> Lew
> ---------------------------------------------------------------------------------------
>
> Let: a = b,
> Then: (1) a^2 = ab,
> (2) a^2 - b^2 = ab - b^2,
> (3) (a + b)(a - b) = b(a - b),
> (4) a + b = b,
> (5) 2b = b,
> (6) 2 = 1
>
> It would seem that we have proved 2 = 1.
>
> What is wrong here?
>
> ---------------------------------------------------------------------------------------
> From:
>
> Basic Mathematics For Engineering And Science, Page 18, 1952
>
> By:
>
> Walter R Van Voorhis, PhD, Professor of Mathematics
> Fenn College, Cleveland, Ohio
>
> and
>
> Elmer E Haskins, PhD, Professor of Mechanical Engineering
> Fenn College, Cleveland, Ohio
> ---------------------------------------------------------------------------------------
>
Illegal to divide by 0 in going from (3) to (4).
How about: Show that the square root of 2 is not a rational number (the
ratio of 2 integers)?
On 05/12/2015 7:43 AM, Leon wrote:
> "Lew Hodgett"<[email protected]> wrote:
>> Have fun.
>>
>> Answer to follow.
>>
>> Lew
>> ---------------------------------------------------------------------------------------
>>
>> Let: a = b,
...
> What is wrong?
> I think the beginning false pretense that a is equal in value to b.
> You can make anything look correct if you begin with deception. .
Nope, you can make any starting postulates you wish; starting with the
one that a=b is perfectly valid. The problem is in a step farther down.
--
John wrote:
> On 5/11/2015 11:19 PM, Bill wrote:
>> Lew Hodgett wrote:
>>> Have fun.
>>>
>>> Answer to follow.
>>>
>>> Lew
>>> ---------------------------------------------------------------------------------------
>>>
>>>
>>>
>>> Let: a = b,
>>> Then: (1) a^2 = ab,
>>> (2) a^2 - b^2 = ab - b^2,
>>> (3) (a + b)(a - b) = b(a - b),
>>> (4) a + b = b,
>>> (5) 2b = b,
>>> (6) 2 = 1
>>>
>>> It would seem that we have proved 2 = 1.
>>>
>>> What is wrong here?
>>>
>>> ---------------------------------------------------------------------------------------
>>>
>>>
>>> From:
>>>
>>> Basic Mathematics For Engineering And Science, Page 18, 1952
>>>
>>> By:
>>>
>>> Walter R Van Voorhis, PhD, Professor of Mathematics
>>> Fenn College, Cleveland, Ohio
>>>
>>> and
>>>
>>> Elmer E Haskins, PhD, Professor of Mechanical Engineering
>>> Fenn College, Cleveland, Ohio
>>> ---------------------------------------------------------------------------------------
>>>
>>>
>>>
>> Illegal to divide by 0 in going from (3) to (4).
>
>
> Agree. A '+1' would seem inappropriate here :)
>
> John
A farmer has 20 feet of fencing and wishing to build a rectangular pen
against his barn (by adding 3 sides), maximizing the area. What should
the dimensions be? Note: It makes the kids smile if I include a little
animal in the diagram.
On Tue, 12 May 2015 19:18:22 -0400, Bill wrote:
> A farmer has 20 feet of fencing and wishing to build a rectangular pen
> against his barn (by adding 3 sides), maximizing the area. What should
> the dimensions be? Note: It makes the kids smile if I include a little
> animal in the diagram.
3x14 = 42
4x12 = 48
5x10 = 50
6x8 = 48
7x6 = 42
We have a winner!
Lew Hodgett wrote:
> Bill wrote:
>> A farmer has 20 feet of fencing and wishing to build a rectangular
>> pen against his barn (by adding 3 sides), maximizing the area. What
>> should the dimensions be? Note: It makes the kids smile if I
>> include a little animal in the diagram.
> ------------------------------------------------------
> Not enough information to develop 3 independent equations.
>
> You can use Taylor's theorem of approximation to determine
> that a 5 ft x 10 ft pen with the 10 ft side parallel to barn wall will
> yield max area.
> -----------------------------------------------------------------
> W = Width
> L = Length
> A = Area
>
> 2W + L = 20
>
> A = WL
> -------------------------------------------------------------
> WHEN W = 7, THEN L = 6
> A = 42
>
> WHEN W = 6, THEN L = 8
> A = 48
>
> WHEN W = 5, THEN L = 10
> A = 50
>
> WHEN W = 4, THEN L = 12
> A = 48
>
> WHEN W = 3, THEN L = 14
> A = 42
>
> Ye gads you are making me dig.
>
> Haven't used most of this stuff in over 50 years.
>
> Lew
>
>
Instead of using 2 variables L and W, try it with just 1 (say just L or
W), you can write the area as a single equation/function of it. You'll
have a quadratic equation whose graph is a parabola... This will lead
you not only an answer, but the fact that the answer lies at the vertex
of a parabola opening downwards will justify for you that it is the
unique best answer.
Bill wrote:
> Lew Hodgett wrote:
>> Bill wrote:
>>> A farmer has 20 feet of fencing and wishing to build a rectangular
>>> pen against his barn (by adding 3 sides), maximizing the area. What
>>> should the dimensions be? Note: It makes the kids smile if I
>>> include a little animal in the diagram.
>> ------------------------------------------------------
>> Not enough information to develop 3 independent equations.
>>
>> You can use Taylor's theorem of approximation to determine
>> that a 5 ft x 10 ft pen with the 10 ft side parallel to barn wall will
>> yield max area.
>> -----------------------------------------------------------------
>> W = Width
>> L = Length
>> A = Area
>>
>> 2W + L = 20
>>
>> A = WL
>> -------------------------------------------------------------
>> WHEN W = 7, THEN L = 6
>> A = 42
>>
>> WHEN W = 6, THEN L = 8
>> A = 48
>>
>> WHEN W = 5, THEN L = 10
>> A = 50
>>
>> WHEN W = 4, THEN L = 12
>> A = 48
>>
>> WHEN W = 3, THEN L = 14
>> A = 42
>>
>> Ye gads you are making me dig.
>>
>> Haven't used most of this stuff in over 50 years.
>>
>> Lew
>>
>>
>
> Instead of using 2 variables L and W, try it with just 1 (say just L
> or W), you can write the area as a single equation/function of it.
> You'll have a quadratic equation whose graph is a parabola... This
> will lead you not only an answer, but the fact that the answer lies at
> the vertex of a parabola opening downwards will justify for you that
> it is the unique best answer.
>
>
Here is a start.
Let L denote the length. Then the width equals (20-L)/2.
So the area = L(20-L)/2=-.5L^2 +10L, a quadratic equation in terms of
the single variable L.
You may find the vertex of the graph of A = -.5L^2 +10L in several ways
(as a local max in calculus, completing the square, using it's symmetry
about it's axis of symmetry--which is the midpoint of it's L-intercepts,
for instance (these can be found with the quadratic formula or just by
factoring)).
Any of these will show you why the answer you already found really is
the correct and
only one. ; )
whit3rd wrote:
> On Tuesday, May 12, 2015 at 9:28:32 PM UTC-7, Bill wrote:
>>
>> Here is a fun one. How many zeros appear at the end of the product:
>> 1*2*3*4*5*...*100?
>>
>> Hint: The answer is Not 2. No credit for multiplying it out, by hand.
> Remember rot13?
>
> Gjragl-sbhe Gur uneq cneg, vf erzrzorevat gung zhygvcyrf bs gjragl-svir
> unir gjb snpgbef bs svir. Gurer'f na rkprff bs snpgbef bs gjb.
Yes, you are absolutely correct! I had never heard of rot13 until I
looked it up just now
and found a decoder! Cool! : )
On 05/12/2015 4:41 PM, [email protected] wrote:
>
>>
>> Let: a = b,
>> Then: (1) a^2 = ab,
>> (2) a^2 - b^2 = ab - b^2,
>> (3) (a + b)(a - b) = b(a - b),
>> (4) a + b = b,
>> (5) 2b = b,
>> (6) 2 = 1
>>
>> It would seem that we have proved 2 = 1.
>>
>> What is wrong here?
>
> The problem is in line 3. ...
...
> Also to allow the equation to be balanced, you need to apply the
> same values and operations to both sides of the equation. In #3, you
> have (a+b) x (a-b) on the left. On the right it's b x (a-b). The
> equation is NOT equal at this point. And as I mentioned before (a-b)
> is zero.
Nope--thru (3) everything is ok... (3) is simply factoring (2)
(a-b)*(a+b) = a^2 + ab - ab - b^2; the two ab terms of alternate sign
cancel leaving --> a^2 - b^2
The RHS is also legitimate irrespective of what a, b actually are;
multiplication is associative so taking the one 'b' out to the outside
of the factor (a-b) is ok. As you noted, substituting values here
yields 0=0 so that's ok.
The problem doesn't come until (4) when divide out the (a-b) term. ["You
can't do that!!!!"] Until then all is well...
> Do I win something?
Sorry, not this time... :)
--
Bill wrote:
> Illegal to divide by 0 in going from (3) to (4).
>
> How about: Show that the square root of 2 is not a rational number
> (the ratio of 2 integers)?
I wrote this up 25 years ago using the IBM graphics characters. If you copy
the text to an editor and switch to a font like Terminal then you can see
the square root and superscript square characters as well as the lines.
p
GIVEN: û2 = - (Where p and q are integers.)
q
p
PROVE: û2 is irrational because - cannot be expressed as a
q
reducible fraction
STATEMENTS: º REASONS:
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p º
1: û2 = - º Given
q º
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2: p and q are integers º Given
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pý º
3: 2 = - º Square both sides
qý º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
4: 2qý = pý º Multiply by qý
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
5: qý is an integer º An integer (step 2) squared
º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
6: pý is an even number º pý is 2x an integer (steps 4 and 5)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
7: p is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
pý p º
8: pý = 2ý( - ) = 4( - )ý º Factor
2ý 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Substitute for pý in step 4
9: 2qý = 4 ( - )ý º
2 º (via step 8)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º
10: qý = 2 ( - )ý º Divide by 2
2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Half an even number (step 7)
11: - is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º An integer (step 11) squared
12: ( - )ý is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
13: qý is an even number º qý is 2x an integer (steps 10 and 12)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
14: q is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
15: p & q have the common º Both are even (steps 7 and 14)
factor of 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º It is not reducible, as any fraction
16: - is irrational º should be, because both denominator and
q º numerator must always be products of 2
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
17: û2 is irrational º Substitution (step 16)
Tom Del Rosso wrote:
> Bill wrote:
>> Illegal to divide by 0 in going from (3) to (4).
>>
>> How about: Show that the square root of 2 is not a rational number
>> (the ratio of 2 integers)?
> I wrote this up 25 years ago using the IBM graphics characters. If you copy
> the text to an editor and switch to a font like Terminal then you can see
> the square root and superscript square characters as well as the lines.
>
>
>
> p
> GIVEN: û2 = - (Where p and q are integers.)
> q
>
> p
> PROVE: û2 is irrational because - cannot be expressed as a
> q
>
> reducible fraction
>
>
> STATEMENTS: º REASONS:
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º
> 1: û2 = - º Given
> q º
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 2: p and q are integers º Given
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> pý º
> 3: 2 = - º Square both sides
> qý º
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 4: 2qý = pý º Multiply by qý
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 5: qý is an integer º An integer (step 2) squared
> º is an integer
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 6: pý is an even number º pý is 2x an integer (steps 4 and 5)
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 7: p is an even number º The square of an odd number is odd
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> pý p º
> 8: pý = 2ý( - ) = 4( - )ý º Factor
> 2ý 2 º
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º Substitute for pý in step 4
> 9: 2qý = 4 ( - )ý º
> 2 º (via step 8)
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º
> 10: qý = 2 ( - )ý º Divide by 2
> 2 º
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º Half an even number (step 7)
> 11: - is an integer º
> 2 º is an integer
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º An integer (step 11) squared
> 12: ( - )ý is an integer º
> 2 º is an integer
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 13: qý is an even number º qý is 2x an integer (steps 10 and 12)
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 14: q is an even number º The square of an odd number is odd
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 15: p & q have the common º Both are even (steps 7 and 14)
> factor of 2 º
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> p º It is not reducible, as any fraction
> 16: - is irrational º should be, because both denominator and
> q º numerator must always be products of 2
> ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
> 17: û2 is irrational º Substitution (step 16)
>
>
Yes, you may also assume the fraction p/q is reduced at the beginning
and derive a contradiction (which I find a bit more direct, but
equivalent of course).
It occurred to me after I posted the problem that sqrt(2) is a zero of
the polynomial
x^2 -2.
So the "Rational Zeros Theorem" applies, and makes short work of the
problem (Rational Zeros Theorem: If a polynomial function, written in
descending order of the exponents, has integer coefficients, then any
rational zero must be of the form ± p/ q, where p is a factor of the
constant term and q is a factor of the leading coefficient).
I found Vi Hart's proof interesting. After first watching one of her
videos I watched all of them that she had posted, but I have not kept up.
Cheers,
Bill
On Sat, 16 May 2015 09:49:26 -0500, Leon wrote:
>> The only thing that counts is that Obamacare is now the law of the
>> land.
>
> At least right now, that can be fixed a little later on.
There are now over sixteen million people covered by insurance that
weren't before, and more every day. Not many politicians are willing to
incur those folks wrath at losing their insurance.
On Wed, 13 May 2015 00:27:41 -0400, Bill <[email protected]>
wrote:
>Lew Hodgett wrote:
>> Too many years and too many beers. Thanks for the input. Lew
>
>Here is a fun one. How many zeros appear at the end of the product:
>1*2*3*4*5*...*100?
>
>Hint: The answer is Not 2. No credit for multiplying it out, by hand.
It would be three, unless there is a decimal point then infinite.
On 5/16/2015 11:47 AM, Larry Blanchard wrote:
> There are now over sixteen million people covered by insurance that
> weren't before, and more every day. Not many politicians are willing to
> incur those folks wrath at losing their insurance.
Except those numbers are arguably subject to a reality check.
They are "signups" and, statistically provably with ObamaCare thus far,
only about 84% will actually "enroll".
And those numbers do not include those who previously had insurance, a
statistic admittedly not tracked by the government.
Fact: HHS Secretary admitted to congress in November that the official
enrollment figures for the previous year were inflated: With the
reported 11.4 million who "signed up", 6.7 actually enrolled, and 43% of
those were previously insured.
Makes the statement that "sixteen million people covered by insurance
that weren't before" highly suspect.
Just like the Bin Laden incident, you can't believe a damned thing
coming out either side of this political asshattery we call government.
--
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