Paradox puzzle
Poll
 | 1 vote (8.33%) | ||
| | 1 vote (8.33%) | ||
| | 1 vote (8.33%) | ||
| | 2 votes (16.66%) | ||
| No votes (0%) | |||
| No votes (0%) | |||
| | 6 votes (50%) | ||
| No votes (0%) | |||
| No votes (0%) | |||
| | 1 vote (8.33%) |
12 members have voted
October 2nd, 2015 at 4:11:34 PM
permalink
Try to find the flaw in the following argument:
(1) 1+x+x^2+x^3+x^4 = 0
(2) Putting the x^4 term on the right side: 1+x+x^2+x^3= -x^4
(3) Rearranging equation (1): 1+x*(1+x+x^2+x^3) = 0
(4) Substituting equation (2) for the part in parenthesis in (3): 1 + x*(-x^4) = 0
(5) Solving equation (4): 1 - x^5 = 0
(6) x^5 = 1
(7) x=1
(8) Substituting x=1 into equation (1): 1+1+1^2 + 1^3 + 1^4 = 0
(9) 5=0
The question for the poll is which is the first equation to be in error?
(1) 1+x+x^2+x^3+x^4 = 0
(2) Putting the x^4 term on the right side: 1+x+x^2+x^3= -x^4
(3) Rearranging equation (1): 1+x*(1+x+x^2+x^3) = 0
(4) Substituting equation (2) for the part in parenthesis in (3): 1 + x*(-x^4) = 0
(5) Solving equation (4): 1 - x^5 = 0
(6) x^5 = 1
(7) x=1
(8) Substituting x=1 into equation (1): 1+1+1^2 + 1^3 + 1^4 = 0
(9) 5=0
Have a nice day.
The question for the poll is which is the first equation to be in error?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 2nd, 2015 at 4:19:16 PM
permalink
I think it's because this function has no real zeros just complex ones
So Wizard, still no basic strategy for strip poker huh?
October 2nd, 2015 at 4:34:36 PM
permalink
Quote: Wonko33top of my head.I think it's because this function has no real zeros just complex ones
What would be a complex solution? Shouldn't the algebra lead to a complex solution if there is one?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 2nd, 2015 at 4:40:59 PM
permalink
I'm working on it
1 is an extraneous root, im trying to figure out where I missed it
So Wizard, still no basic strategy for strip poker huh?
October 2nd, 2015 at 4:44:43 PM
permalink
x^n - y^n is always divisible by x - y, so x^5 - 1 has x - 1 as a factor.
x^5-1/(x-1)=x^4+x^3+x^2+x+1 Long division.
(x-1)(x^4+x^3+x^2+x+1)
http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.390481.html
More roots than x=1.
x^5-1/(x-1)=x^4+x^3+x^2+x+1 Long division.
(x-1)(x^4+x^3+x^2+x+1)
http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.390481.html
More roots than x=1.
So much bullshit; so little time!
October 2nd, 2015 at 5:23:36 PM
permalink
Quote: Wizard
(4) Substituting equation (2) for the part in parenthesis in (3): 1 + x*(-x^4) = 0
(5) Solving equation (4): 1 - x^5 = 0
Equation four is 1 + [(x)(-x)(-x)(-x)(-x)] = 0
four negatives multiplied together equal a positive. So (6) should be should be x^5 = -1
x=-1
1 + (-1) + 1 +(-1) + 1 = 0
which is still wrong
October 2nd, 2015 at 5:25:03 PM
permalink
You introduced a root in step #4. That root is 1, and not valid.
Try it with 1+x=0.
1. 1+x=0
2. 1=-x
3. 1+x*(1)=0
4. 1+x*(-x)=0
5. 1-x^2=0
6. 1=x^2
7. X=1
8. 1+1=0
9. 2=0
Try it with 1+x=0.
1. 1+x=0
2. 1=-x
3. 1+x*(1)=0
4. 1+x*(-x)=0
5. 1-x^2=0
6. 1=x^2
7. X=1
8. 1+1=0
9. 2=0
October 2nd, 2015 at 5:29:50 PM
permalink
Quote: TomG
Equation four is 1 + [(x)(-x)(-x)(-x)(-x)] = 0
four negatives multiplied together equal a positive. So (6) should be should be x^5 = -1
x=-1
1 + (-1) + 1 +(-1) + 1 = 0
which is still wrong
Here is equation (4):
1 + x*(-x^4) = 0
That is saying 1+x*(-(x^4)) = 0-->
1 + x * -1 * x^4 = 0-->
1 - x^5 = 0
In other words, the minus sign is outside of the exponential function.
1 + x*(-x^4) = 0
That is saying 1+x*(-(x^4)) = 0-->
1 + x * -1 * x^4 = 0-->
1 - x^5 = 0
In other words, the minus sign is outside of the exponential function.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 2nd, 2015 at 5:35:02 PM
permalink
Quote: ChampagneFireballYou introduced a root in step #4. That root is 1, and not valid.
Try it with 1+x=0.
1. 1+x=0
2. 1=-x
3. 1+x*(1)=0
4. 1+x*(-x)=0
5. 1-x^2=0
6. 1=x^2
7. X=1
8. 1+1=0
9. 2=0
Good answer. However, why can't I introduce that root? It seems to follow from earlier steps.
I do like your mini version.
Step 6 can also lead to x=-1. However, that leads to the question of why it can't be both. We could simplify your point as follows:
(1) x=1
(2) Square both sides: x^2=1
(3) Take the square root of both sides: x=+/- 1
However, in the original equation x was just 1. I think that saying x=+/-1 means the answer must one or the other but not necessarily both. However, I'm not entirely satisfied with that answer. If anybody can do better, please do.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 2nd, 2015 at 5:49:19 PM
permalink
Quote: WizardQuote: ChampagneFireballYou introduced a root in step #4. That root is 1, and not valid.
Try it with 1+x=0.
1. 1+x=0
2. 1=-x
3. 1+x*(1)=0
4. 1+x*(-x)=0
5. 1-x^2=0
6. 1=x^2
7. X=1
8. 1+1=0
9. 2=0
Good answer. However, why can't I introduce that root? It seems to follow from earlier steps.
I do like your mini version.
Step 6 can also lead to x=-1. However, that leads to the question of why it can't be both. Let me think about the proper response.
You can introduce a root but it won't be the solution to the original equation.
Very simple example
1. x=1
2. multiply both sides by x: x^2=x,
The equation in (2) is correct given (1), but 0 is also an answer to (2) but not an answer to (1).
October 2nd, 2015 at 5:55:49 PM
permalink
Quote: ChampagneFireball
You can introduce a root but it won't be the solution to the original equation.
Very simple example
1. x=1
2. multiply both sides by x: x^2=x,
The equation in (2) is correct given (1), but 0 is also an answer to (2) but not an answer to (1).
Can we say that if we introduce a root that not necessarily all answers will satisfy the original equation? I'm looking for a rule of mathematics engraved in granite we can quote to properly explain this.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 2nd, 2015 at 7:29:55 PM
permalink
x5 = 1 does not necessarily mean x = 1.
Otherwise, you get something like this:
Let y = -1, and x = y
x2 = (-1)2 = 1
x = 1
Since x = y, this means -1 = 1
October 2nd, 2015 at 7:40:12 PM
permalink
Quote: ThatDonGuy
x5 = 1 does not necessarily mean x = 1.
Otherwise, you get something like this:
Let y = -1, and x = y
x2 = (-1)2 = 1
x = 1
Since x = y, this means -1 = 1
There is another discussion of this argument above in spoiler tags. I would say there that if x^2=1 means that x is either to -1 and/or 1. In this case, just -1.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 3rd, 2015 at 12:58:50 AM
permalink
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
October 3rd, 2015 at 4:53:44 PM
permalink
In my opinion, your (real) math is sound, Wizard. This is a proof by falsification.
since 1+x+x^2+x^3+x^4 = 0 leads to 5=0 it follows
1+x+x^2+x^3+x^4 != 0 (There is no real x for which the left term equals 0)
It has, however, several complex solutions.
I can also say that equation number 4 is where things go wrong from a complex point of view. Your new form now has a real solution which is not true for equations 1 through 3. You mustn't introduce a new exponent (^5).
October 3rd, 2015 at 9:16:03 PM
permalink
Equation 2: Usually initial solving would set x+x^2+x^3+x^4=-1
Thus the "1" is placed on the right side of the equation. If this is done then x=-1, and further, x=1 is not a correct answer.
Thus the "1" is placed on the right side of the equation. If this is done then x=-1, and further, x=1 is not a correct answer.
Some people need to reimagine their thinking.
October 4th, 2015 at 9:36:14 AM
permalink
@98Clubs:
If I test the candidate solution of -1 in that equation, I get:
(-1) + (-1)^2 + (-1)^3 + (-1)^4 = -1
-1 + 1 - 1 + 1 = -1
0 = -1
Not quite a standard equality.
Uh, I don't think so.Quote: 98ClubsEquation 2: Usually initial solving would set x+x^2+x^3+x^4=-1
Thus the "1" is placed on the right side of the equation. If this is done then x=-1, and further, x=1 is not a correct answer.
If I test the candidate solution of -1 in that equation, I get:
(-1) + (-1)^2 + (-1)^3 + (-1)^4 = -1
-1 + 1 - 1 + 1 = -1
0 = -1
Not quite a standard equality.
October 4th, 2015 at 5:26:47 PM
permalink
Doc thats true, I should not have attempted a solve. But my actual answer as to where the 1st error is located remains unchanged. Somehow my poynty lil head got some zeros in there. ;o)
Some people need to reimagine their thinking.
October 4th, 2015 at 8:46:38 PM
permalink
Quote: 98ClubsBut my actual answer as to where the 1st error is located remains unchanged.
It has been a very long time since I studied algebra or worked with it extensively. I am quite rusty. Anyway here are my thoughts on the original question as to the flaw in the original derivation and the first equation in error.
Are we still using spoiler tags?
I think the earlier discussions are correct that the original equation only has complex roots and that the x=1 solution is an extraneous root introduced in the derivation.
I do not see any problem in going from (1) to (2); you should be able to subtract x^4 from each side of an equality, provided x is finite.
I think there may be an error in going from (2) to (3). The "rearranging" of equation (1) involves multiplying a portion of the left side of the equation by x/x, and assumes this ratio is equal to 1. That is not necessarily the case if x=0, so this may have introduced the extraneous root. I sometimes have reservations about a derivation that involves multiplying and/or dividing by an unknown.
I also think there could be an error in going from (3) to (4). The "substitution" involves multiplying part of the left side of equation (3) by (-x^4)/(1+x+x^2+x^3) which is assumed to equal 1, based on equation (2). However, if either x^4=0 or 1+x+x^2+x^3=0 or both, this "substitution" would present a problem.
For now, I will go with equation (3) as being the first one to be in error.
I do not see any problem in going from (1) to (2); you should be able to subtract x^4 from each side of an equality, provided x is finite.
I think there may be an error in going from (2) to (3). The "rearranging" of equation (1) involves multiplying a portion of the left side of the equation by x/x, and assumes this ratio is equal to 1. That is not necessarily the case if x=0, so this may have introduced the extraneous root. I sometimes have reservations about a derivation that involves multiplying and/or dividing by an unknown.
I also think there could be an error in going from (3) to (4). The "substitution" involves multiplying part of the left side of equation (3) by (-x^4)/(1+x+x^2+x^3) which is assumed to equal 1, based on equation (2). However, if either x^4=0 or 1+x+x^2+x^3=0 or both, this "substitution" would present a problem.
For now, I will go with equation (3) as being the first one to be in error.
October 4th, 2015 at 9:25:46 PM
permalink
I think we can quit using spoiler tags as the answer clearly isn't obvious.
To throw my hat in the ring, I agree with ThatDonGuy above.
To throw my hat in the ring, I agree with ThatDonGuy above.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 5th, 2015 at 7:21:31 AM
permalink
Think I have it.
You can solve it using Euler's Formula. The problem is the line x^5=1 => x=1 isn't true. x^5=1 has 5 roots, only 1 is x=1
x^5=1
x^5=e^(2*pi*i*y) where y=0,1,2,3,4
This gives 5 solutions.
x=e^(2pi/5*i)
x=e^(4pi/5*i)
x=e^(6pi/5*i)
x=e^(8pi/5*i)
x=1
The first 4 complex roots satisfy the first equation, (1+x+x^2+x^3+x^4=0), the last one does not.
You can solve it using Euler's Formula. The problem is the line x^5=1 => x=1 isn't true. x^5=1 has 5 roots, only 1 is x=1
x^5=1
x^5=e^(2*pi*i*y) where y=0,1,2,3,4
This gives 5 solutions.
x=e^(2pi/5*i)
x=e^(4pi/5*i)
x=e^(6pi/5*i)
x=e^(8pi/5*i)
x=1
The first 4 complex roots satisfy the first equation, (1+x+x^2+x^3+x^4=0), the last one does not.
October 5th, 2015 at 8:08:18 AM
permalink
Mike,
You have a polynomial, who cares what it is, call it f(x). You are solving f(x) = 0. To do this, you are simply multiplying both sides by x - 1, to get (x - 1)*f(x) = 0. You then solve that, saying that the root is x = 1. You make the claim that f(x) = 0 has no roots (you might want to review the Fundamental Theorem of Algebra).. Hence you conclude that x = 1 must be a solution to the original polynomial f(x).
This is typical of the "multiply both sides by 0" puzzles. It is not a paradox.
If you want to understand your argument better, start with f(x) = x + 1.
x + 1 = 0
(x + 1)*(x - 1) = 0*(x - 1) = 0
x^2 - 1 = 0
x^2 = 1
Solving, x = 1 (do you see your error now?)
So, plugging x = 1 into the original equation, 1 + 1 = 0
2 = 0
Climate Casino: https://climatecasino.net/climate-casino/
October 5th, 2015 at 8:31:45 AM
permalink
Quote: teliot
Mike,
You have a polynomial, who cares what it is, call it f(x). You are solving f(x) = 0. To do this, you are simply multiplying both sides by x - 1, to get (x - 1)*f(x) = 0. You then solve that, saying that the root is x = 1. You make the claim that f(x) = 0 has no roots (you might want to review the Fundamental Theorem of Algebra).. Hence you conclude that x = 1 must be a solution to the original polynomial f(x).
This is typical of the "multiply both sides by 0" puzzles. It is not a paradox.
If you want to understand your argument better, start with f(x) = x + 1.
x + 1 = 0
(x + 1)*(x - 1) = 0*(x - 1) = 0
x^2 - 1 = 0
x^2 = 1
Solving, x = 1 (do you see your error now?)
So, plugging x = 1 into the original equation, 1 + 1 = 0
2 = 0
Thank you for your time on this problem.
First let me me say that I am not an idiot. I never claimed that x=1. You can see the first sentence in the original post was "Try to find the flaw in the following argument:"
I also don't see where I multiplied anything by zero at any step.
Now, if you'll excuse me, I'm late for my beginning algebra class.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 5th, 2015 at 8:55:21 AM
permalink
Hi Mike. I apologize if my tone was condescending. Obviously, I hate these sorts of discussions. But, I replied because you asked for my input. Perhaps it's best if I just let it go next time (or you remember, "don't ask Eliot -- he hates these discussions").
You are multiplying both sides by (x - 1). You are just doing it in a roundabout way. Since you are solving for the root x = 1, you are multiplying both sides by 0. To see my point, just follow my steps with your polynomial and you will see that I get to the same place you did. "Substituting" and "Multiplying" are the same thing in some cases.
Cheers.
Climate Casino: https://climatecasino.net/climate-casino/
October 5th, 2015 at 11:17:34 AM
permalink
Let f(x) = 1 + x + x^2 + x^3 + x^4.Quote: teliotYou are solving f(x) = 0.
Rearrange to f(x) = 1 + x(1 + x + x^2 + x^3); and to 1 + x + x^2 + x^3 = f(x) - x^4.
Substitute to f(x) = 1+ x(f(x) - x^4).
Simplify to f(x) = (1 - x^5)/(1 - x) = (1 + x + x^2 + x^3 + x^4) X (1 - x)/(1 - x), and back to, the original, f(x).
________________________________________
Note that the four values of x are as well be entered into x - 1 (as into also x^4 + x^3 + x^2 + x + 1) to produce the x^5 - 1 = 0 result. A sequential re-solution of one equation instead of a simultaneous solution of more than one equation.
Note also that we can not multiply in/divide out a root factor, let alone on both sides of an equation. Such would amount to multiplication/division by zero of the selected root factor.
So much bullshit; so little time!
