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 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)?

"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
---------------------------------------------------------------------------------------

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. .

On 05/12/2015 7:43 AM, Leon wrote:
"Lew 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.

--

On 05/11/2015 10:19 PM, Bill wrote:
....

Illegal to divide by 0 in going from (3) to (4).
....

+1

--

"Lew Hodgett" writes:
Have fun.

Answer to follow.

Lew
---------------------------------------------------------------------------------------

Let: a = b,

therefo

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
---------------------------------------------------------------------------------------

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.

DerbyDad03 wrote in
:

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

On Tuesday, May 12, 2015 at 2:16:57 PM UTC-4, John McCoy wrote:
DerbyDad03 wrote in
:

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.

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. The value of (a-b) is zero. Zero x any number is zero. So equation 3 results out to be 0=0. You can't do the following steps 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 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.

Do I win something?

MJ

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

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!

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" 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?

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. ; )

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

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.

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.

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 Wed, 13 May 2015 00:27:41 -0400, 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.

It would be three, unless there is a decimal point then infinite.

On 05/12/2015 4:41 PM, 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...

--

Just Wondering writes:

Is that why so many people vote Democrat?

asshole.

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×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º
1: û2 = - º Given
q º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2: p and q are integers º Given
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
pý º
3: 2 = - º Square both sides
qý º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
4: 2qý = pý º Multiply by qý
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
5: qý is an integer º An integer (step 2) squared
º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
6: pý is an even number º pý is 2x an integer (steps 4 and 5)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
7: p is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
pý p º
8: pý = 2ý( - ) = 4( - )ý º Factor
2ý 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Substitute for pý in step 4
9: 2qý = 4 ( - )ý º
2 º (via step 8)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º
10: qý = 2 ( - )ý º Divide by 2
2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Half an even number (step 7)
11: - is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º An integer (step 11) squared
12: ( - )ý is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
13: qý is an even number º qý is 2x an integer (steps 10 and 12)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
14: q is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
15: p & q have the common º Both are even (steps 7 and 14)
factor of 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
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×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
17: û2 is irrational º Substitution (step 16)

Bill wrote:

How about: Show that the square root of 2 is not a rational number
(the ratio of 2 integers)?

See the outstandingly magnificent youtube video from Vihart about Pythagoras
for a geometrical proof. See all her math videos for that matter.

https://www.youtube.com/watch?v=X1E7I7_r3Cw

Bill wrote in :

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 wrote in
:

On Tuesday, May 12, 2015 at 2:16:57 PM UTC-4, John McCoy wrote:
DerbyDad03 wrote in
:

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

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×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º
1: û2 = - º Given
q º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2: p and q are integers º Given
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
pý º
3: 2 = - º Square both sides
qý º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
4: 2qý = pý º Multiply by qý
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
5: qý is an integer º An integer (step 2) squared
º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
6: pý is an even number º pý is 2x an integer (steps 4 and 5)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
7: p is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
pý p º
8: pý = 2ý( - ) = 4( - )ý º Factor
2ý 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Substitute for pý in step 4
9: 2qý = 4 ( - )ý º
2 º (via step 8)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º
10: qý = 2 ( - )ý º Divide by 2
2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º Half an even number (step 7)
11: - is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
p º An integer (step 11) squared
12: ( - )ý is an integer º
2 º is an integer
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
13: qý is an even number º qý is 2x an integer (steps 10 and 12)
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
14: q is an even number º The square of an odd number is odd
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
15: p & q have the common º Both are even (steps 7 and 14)
factor of 2 º
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
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×ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
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 5/13/2015 8:42 AM, Scott Lurndal wrote:
Just Wondering 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.

Just Wondering writes:
On 5/13/2015 8:42 AM, Scott Lurndal wrote:
Just Wondering 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.