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**Amartyanil****Member**- From: Universe
- Registered: 2013-05-27
- Posts: 80

hi;

Suppose not all four integers a, b, c, d are equal. Start with

and repeatedly replace by . Then at least one number of the quadruple will eventually become arbitrarily large.The solution to this problem is:

Let

From the equation for

However I saw this result, can anyone explain how is it so?

Explain how this is so:

And explain what this statement means:

The distance of the points

How is it necessary that one component would become large if the distance increases?

Note: This is Example 5 of first chapter of Arthur Engel's Problem Solving Strategies.

*Last edited by Amartyanil (2019-02-04 02:16:11)*

"Every place is the center of the universe. And every moment is the most important moment. And everything is the meaning of life." ~ Dan Harmon

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

The problem and the deduction is very simple. How did you come across this problem? Anyway the explanation of this problem is stated below:

It can be shown that the distance of any point (a, b, c, d) in the Euclidean four-dimensional space from the origin (using the Cartesian coordinate system) is given by:

N.B. This formula of the distance in the Euclidean space can be either obtained by definition or deduced using Pythagoras theorem. Now it goes without saying that if one of the integers a, b, c and d becomes arbitrarily large the distance of the corresponding point (according to the last formula) will be arbitrarily large from the origin. However, the point P=(a, b, c, d) (where all of the components are integers and at least two of them unequal) is defined recursively as follows:

Therefore to prove the proclaimed statement (that the distance of these points from the origin gets arbitrarily large), it is sufficient to show that at least one of the numbers a, b, c and d must get arbitrarily large. So the last task can be proven as follows. Suppose that after n iterations we get:

however from the definition of these points (stated above) we have:

Now from the first assumption we have:

and hence

The last inequality holds since all of the terms are squared. Therefore, we get:

however, we have

and hence substituting in the first inequality we get:

so continuing in this process recursively we get at last:

Therefore since by definition at least one of the integers in the right of the last inequality in not zero and all of the terms are squared, the proclaimed statement is established and the related proposition is true. I hope that answers your question.

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**Amartyanil****Member**- From: Universe
- Registered: 2013-05-27
- Posts: 80

hi Grantingriver;

Is

?*Last edited by Amartyanil (2019-02-08 02:32:07)*

"Every place is the center of the universe. And every moment is the most important moment. And everything is the meaning of life." ~ Dan Harmon

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

Dear Amartyanil, yes it is greater than or equal to zero. Any sum of square numbers is greater than or equal to zero.

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**Amartyanil****Member**- From: Universe
- Registered: 2013-05-27
- Posts: 80

hi Grantingriver;

What if they are complex?

*Last edited by Amartyanil (2019-02-13 01:51:57)*

"Every place is the center of the universe. And every moment is the most important moment. And everything is the meaning of life." ~ Dan Harmon

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

If they are complex then they can be negative or positive numbers. However, in your problem they assume that the numbers are integers and any integer is a real number.

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**Amartyanil****Member**- From: Universe
- Registered: 2013-05-27
- Posts: 80

hi Grantimgriver;

I am sorry for not paying attention to the question. I am thus removing my post.

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