Where Algebra Fails

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Sun Nov 01, 2015 6:48 pm

Paidion wrote:
Now for the really impressive part! Ask your friend to take use the calculator to divide his number in line 10 by his number in line 9. In this case it will be 1995 ÷ 1233. Then ask him to write down the first three digits of his answer. Now have him turn the paper over and look at the other side. He will see the very number he wrote down, that is, 1.61
I found this last puzzle very interesting. The most intriguing aspect to me was multiplying line seven to get the total. I tried it but it didn't work - until I checked my calculations! By the time you get to the seventh number you have 8 of the first number and 5 of the second. At the sum of all to line 10 you have 88 of the first and 55 of the second, so 11 times the 7th line total = the grand total of all the numbers. Incredible.

As soon as I saw that the answer to the question was going to be 1.61 I figured that this was going to be a Fibonacci sequence or the 'Golden ratio', which it is. I never cease to be amazed at how this pattern of numbers shows up in nature, architecture, human anatomy and even ancient religious sites such as Stonehenge.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Sun Nov 01, 2015 9:21 pm

Graeme, you are more skilled in mathematics than you know!

I think you have solved the line 7 thing. Just to make the proof of it more formal, let x and y represent the numbers in lines 1 and 2. Thus we have:

1) x
2) y
3) x+y
4) x+2y
5) 2x+3y
6) 3x+5y
7) 5x+8y
8) 8x+13y
9) 13x+21y
10)21x+34y
And 55x+88y is the total. It just so happens that 11(5x+8y)=55x+88y
Thus the proof that line 7 multiplied by 11 is the total becomes clearly obvious.

I am more fascinated by the 1.61 thing myself. Yes, you are right! This number is the first three digits of φ or "the golden ratio." But why does the first three digits of line 10 divided by line 9 ALWAYS work out to 1.61 no matter what two numbers are chosen for lines 1 and 2? Can you think of a proof that this is the case?
Paidion

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Mon Nov 02, 2015 12:15 am

Paidion wrote:Graeme, you are more skilled in mathematics than you know!

I think you have solved the line 7 thing. Just to make the proof of it more formal, let x and y represent the numbers in lines 1 and 2. Thus we have:

1) x
2) y
3) x+y
4) x+2y
5) 2x+3y
6) 3x+5y
7) 5x+8y
8) 8x+13y
9) 13x+21y
10)21x+34y
And 55x+88y is the total. It just so happens that 11(5x+8y)=55x+88y
Thus the proof that line 7 multiplied by 11 is the total becomes clearly obvious.

I am more fascinated by the 1.61 thing myself. Yes, you are right! This number is the first three digits of φ or "the golden ratio." But why does the first three digits of line 10 divided by line 9 ALWAYS work out to 1.61 no matter what two numbers are chosen for lines 1 and 2? Can you think of a proof that this is the case?
I had set out a table similar to what you have shown above to convince myself that line 7 was not just a fluke.

The proof of 1.61 as the ratio line 10:line 9 is illustrated in the pattern you have listed above.

Suppose that you let y = 0 and x = 1000. Then the ratio becomes 21000/13000 which gives 1.61538461538. If you let x = 0 and y = 1000 then you will get 34000/21000 = 1.61904761904. These two values give the range of value to the final answer. Even if the y value was in the millions (e.g. 34,000,000/21,000,000), the y result would always be 1.619..., and similarly the x result will always be 1.615.... The actual values of x and y will blend (how much of x and how much of y) to give an answer within these ranges. So you are safe to use 1.61.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Mon Nov 02, 2015 4:39 pm

Yes, you've shown that the first three digits will be 1.61 for y=0 and x=1000, and for x=0 and y=1000. But how do you determine that this will be true for ALL values of x>0 and y<1000? Can you produce a rigid proof?
Paidion

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Mon Nov 02, 2015 11:55 pm

Paidion wrote:Yes, you've shown that the first three digits will be 1.61 for y=0 and x=1000, and for x=0 and y=1000. But how do you determine that this will be true for ALL values of x>0 and y<1000? Can you produce a rigid proof?
I don't think I can call it rigid proof, but here's my reasoning. Where we made x=0 we eliminated any influence of x on y and found the result to be simply 34/21 for whatever positive value of y. Similarly where y=0 we have eliminated any influence of y on x as long is has a positive value. So the absolute range of the answer must always fall between the value of 21/13 (1.615385) and 34/21 (1.619048). Any two positive numbers, then of whatever value,must provide a result that is greater than 21/13 but less than 34/21, as x and y provide the ratio in which these values are weighted.

On my spreadsheet I observed these results:
x=0, y=any number, result = 1.6190476190 = maximum value
y=0, x=any number, result = 1.6153846154 = minimum value
x=1, y=1 result = 1.6176470588 = somewhere in the middle
x=1 y=1,000,000 result = 1.6190476168 a shade less than where x=0
y=1, x=1,000,000 result = 1.6153846213 a shade more than where y=0
where I had a very large number (10^20), opposite 1, the answer never exceeded the 0 values.

I did not try to reduce the formula and I have long since forgotten what calculus I learned. The above is proof enough for me.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Tue Nov 03, 2015 1:07 am

Okay Graeme, your reasoning may well qualify as a proof.

Here is a formal proof:
First consider the fact that you can always find a fraction that is between two fractions by adding the numerators and denominators. This has been proved, and you are probably already be aware of it. For example to find a fraction between 2/3 and 3/4, by adding the numerators and denominators, you would obtain the fraction 5/7. This fraction is NOT half-way between, but still it is greater than 2/3 and less than 3/4. I will use this fact in the following formal proof.

When we used x and y for lines 1 and 2, then line 9 was 13x + 21y and line 10 was 21x + 34y. So line 10 divided by line 9 would be:

21x + 34y / 13x + 21y (Let's hereafter call this number A since it is the Answer to line 10 divided by line 9

But using the "fraction between" trick in the first paragraph, that would mean that

21x/13x < A < 34y/21y

But by dividing the numerator and denominator by x, the fraction 21x/13x = 21/13. Similarly 34y/21y = 31/21.
Therefore 21/13 < A < 34/21, that is, the answer is between 21/13 and 34/21.
Now 21/13 = 1.61538461538...
And 34/21 = 1.61904760904...

I see that these lower and upper limits are very close to the ones you calculated (and you did mention that your second answer was "a shade more," and the first one is, as well.) Congratulations!
Last edited by Paidion on Tue Nov 03, 2015 1:28 am, edited 1 time in total.
Paidion

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Tue Nov 03, 2015 1:26 am

Oh, I just went back and re-examined the first paragraph of your post. You had gotten those very fraction limits! 21/13 and 34/21
Paidion

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SamIam
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Re: Where Algebra Fails

Post by SamIam » Mon Nov 09, 2015 9:29 pm

I did not bother to read all of the posts, so this might have been covered ...

The series

n = 3/2 + 9/4 + 27/8 + ...

which is the summation

n = (summation over j = 1 to infinity) 1.5^j

does not converge, so the algebra in the first post is not valid.

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Tue Nov 10, 2015 12:31 pm

Yes, Sam, as you say, (and this has been stated in some of the posts; indeed, I stated it myself) if the series does not converge, the algebra is not valid. But why not? That is what I would like to know.
Paidion

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SamIam
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Re: Where Algebra Fails

Post by SamIam » Tue Nov 10, 2015 3:07 pm

Since the series

n = (summation over j = 1 to infinity) 1.5^j

does not converge, the statement is meaningless. You can't make a variable equal to something that has no finite value. You are trying to violate the meaning of the word "equal."

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