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O/T: What is wrong here?
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 --------------------------------------------------------------------------------------- |
O/T: What is wrong here?
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)? |
O/T: What is wrong here?
"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. . |
O/T: What is wrong here?
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. -- |
O/T: What is wrong here?
On 05/11/2015 10:19 PM, Bill wrote:
.... Illegal to divide by 0 in going from (3) to (4). .... +1 -- |
O/T: What is wrong here?
"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 --------------------------------------------------------------------------------------- |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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! |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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? |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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. ; ) |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
"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 |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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! : ) |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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O/T: What is wrong here?
Just Wondering writes:
Is that why so many people vote Democrat? asshole. |
O/T: What is wrong here?
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) |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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 |
O/T: What is wrong here?
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. |
O/T: What is wrong here?
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. |
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