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

...

+1

...

+1

--

On 5/11/2015 11:19 PM, Bill wrote:

Agree. A '+1' would seem inappropriate here :)

John

Agree. A '+1' would seem inappropriate here :)

John

John 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.

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:

3x14 = 42 4x12 = 48 5x10 = 50 6x8 = 48 7x6 = 42

We have a winner!

3x14 = 42 4x12 = 48 5x10 = 50 6x8 = 48 7x6 = 42

We have a winner!

Bill wrote:

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

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

Lew Hodgett 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.

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:

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. ; )

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:

----------------------------------------------- Too many years and too many beers.

Thanks for the input.

Lew

----------------------------------------------- Too many years and too many beers.

Thanks for the input.

Lew

Lew Hodgett 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.

Here is a fun one. How many zeros appear at the end of the product: 1

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:

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.

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:

Yes, you are absolutely correct! I had never heard of rot13 until I looked it up just now and found a decoder! Cool! : )

Yes, you are absolutely correct! I had never heard of rot13 until I looked it up just now and found a decoder! Cool! : )

wrote:

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

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

Twenty.

Bill wrote:

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: ?2 = - (Where p and q are integers.) q

p PROVE: ?2 is irrational because - cannot be expressed as a q

reducible fraction

STATEMENTS: ? REASONS: ????????????????????????????????????????????????????????????????????? p ? 1: ?2 = - ? Given q ? ????????????????????????????????????????????????????????????????????? 2: p and q are integers ? Given ????????????????????????????????????????????????????????????????????? p? ? 3: 2 = - ? Square both sides q? ? ????????????????????????????????????????????????????????????????????? 4: 2q? = p? ? Multiply by q? ????????????????????????????????????????????????????????????????????? 5: q? is an integer ? An integer (step 2) squared ? is an integer ????????????????????????????????????????????????????????????????????? 6: p? is an even number ? p? is 2x an integer (steps 4 and 5) ????????????????????????????????????????????????????????????????????? 7: p is an even number ? The square of an odd number is odd ????????????????????????????????????????????????????????????????????? p? p ? 8: p? = 2?( - ) = 4( - )? ? Factor 2? 2 ? ????????????????????????????????????????????????????????????????????? p ? Substitute for p? in step 4 9: 2q? = 4 ( - )? ? 2 ? (via step 8) ????????????????????????????????????????????????????????????????????? p ? 10: q? = 2 ( - )? ? Divide by 2 2 ? ????????????????????????????????????????????????????????????????????? p ? Half an even number (step 7) 11: - is an integer ? 2 ? is an integer ????????????????????????????????????????????????????????????????????? p ? An integer (step 11) squared 12: ( - )? is an integer ? 2 ? is an integer ????????????????????????????????????????????????????????????????????? 13: q? is an even number ? q? is 2x an integer (steps 10 and 12) ????????????????????????????????????????????????????????????????????? 14: q is an even number ? The square of an odd number is odd ????????????????????????????????????????????????????????????????????? 15: p & q have the common ? Both are even (steps 7 and 14) factor of 2 ? ????????????????????????????????????????????????????????????????????? p ? It is not reducible, as any fraction 16: - is irrational ? should be, because both denominator and q ? numerator must always be products of 2 ????????????????????????????????????????????????????????????????????? 17: ?2 is irrational ? Substitution (step 16)

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: ?2 = - (Where p and q are integers.) q

p PROVE: ?2 is irrational because - cannot be expressed as a q

reducible fraction

STATEMENTS: ? REASONS: ????????????????????????????????????????????????????????????????????? p ? 1: ?2 = - ? Given q ? ????????????????????????????????????????????????????????????????????? 2: p and q are integers ? Given ????????????????????????????????????????????????????????????????????? p? ? 3: 2 = - ? Square both sides q? ? ????????????????????????????????????????????????????????????????????? 4: 2q? = p? ? Multiply by q? ????????????????????????????????????????????????????????????????????? 5: q? is an integer ? An integer (step 2) squared ? is an integer ????????????????????????????????????????????????????????????????????? 6: p? is an even number ? p? is 2x an integer (steps 4 and 5) ????????????????????????????????????????????????????????????????????? 7: p is an even number ? The square of an odd number is odd ????????????????????????????????????????????????????????????????????? p? p ? 8: p? = 2?( - ) = 4( - )? ? Factor 2? 2 ? ????????????????????????????????????????????????????????????????????? p ? Substitute for p? in step 4 9: 2q? = 4 ( - )? ? 2 ? (via step 8) ????????????????????????????????????????????????????????????????????? p ? 10: q? = 2 ( - )? ? Divide by 2 2 ? ????????????????????????????????????????????????????????????????????? p ? Half an even number (step 7) 11: - is an integer ? 2 ? is an integer ????????????????????????????????????????????????????????????????????? p ? An integer (step 11) squared 12: ( - )? is an integer ? 2 ? is an integer ????????????????????????????????????????????????????????????????????? 13: q? is an even number ? q? is 2x an integer (steps 10 and 12) ????????????????????????????????????????????????????????????????????? 14: q is an even number ? The square of an odd number is odd ????????????????????????????????????????????????????????????????????? 15: p & q have the common ? Both are even (steps 7 and 14) factor of 2 ? ????????????????????????????????????????????????????????????????????? p ? It is not reducible, as any fraction 16: - is irrational ? should be, because both denominator and q ? numerator must always be products of 2 ????????????????????????????????????????????????????????????????????? 17: ?2 is irrational ? Substitution (step 16)

Tom Del Rosso wrote:

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

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

Bill wrote:

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

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

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:

...

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.

...

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.

--

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