Mathematics problem 投稿者: Smith Park
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You have the group A which contains natural number, a, b, c, d, e and the group B which contains natural number, i, j, k, l, m (except 0 for both).
When the sum of group A elements is identical to the sum of group B elements and the product of group A elements is identical to the product of group B elements, is the group A (a, b, c, d, e) the group B (i, j, k, l, m)?
(abcde, ijklm)
[Edited at 2003-10-10 04:55]
| | | | Csaba Ban 
ハンガリー
Local time: 14:26
2002に入会
英語 から ハンガリー語
+ ...
I assume that by \"natural numbers\" you mean positive integers and each of the two sets has five elements.
I have an idea how to prove it (indirectly):
You have to consider i, j, etc. as a sum of a+a\' , b+b\' , etc. (making such pairs that i is equal to or greater than a, j is equal to or greater than b, etc.) (a\' , b\' ... are either zeros or positive integers)
If the two sets are different, then at least one of a\' , b\' etc has to be other than ze... See more I assume that by \"natural numbers\" you mean positive integers and each of the two sets has five elements.
I have an idea how to prove it (indirectly):
You have to consider i, j, etc. as a sum of a+a\' , b+b\' , etc. (making such pairs that i is equal to or greater than a, j is equal to or greater than b, etc.) (a\' , b\' ... are either zeros or positive integers)
If the two sets are different, then at least one of a\' , b\' etc has to be other than zero.
You will have to then write down the product of these five elements (ab + ab\' + ac\' + ... + a\'b\' ... ).
You delete the products that appear on both sides of the equation (ab + ac + ad + ... de). All that remains is:
a\'b\' + a\'c\' + ... d\'e\' = 0
Since our precondition was that at least one of a\' , b\' ... is other than zero, those four products where the non-zero element is present have to be other than zero, hence their sum cannot be zero.
Indirectly we proved that all five of a\' , b\' ... have to be zero, i.e. the set of a, b, ... has to be identical with i, j, ...
Q.E.D
[ This Message was edited by: on 2002-12-11 11:15 ]
[ This Message was edited by: on 2002-12-11 11:16 ] ▲ Collapse
| | | | | Not necessarily. | Dec 11, 2002 |
A = {a,b,c,d,e}
B = {i,j,k,l,m}
a+b+c+d+e = i+j+k+l+m
a.b.c.d.e = i.j.k.l.m,
where . denotes multiplication.
This does not imply that A and B should be indentical sets.
Consider the counterexample:
A = {3,3,10,n,n}
B = {2,5,9,n,n}
The sum of the elements of set A is 2n+16, the product is 90.n.n - just like for set B.
You can choose any n.
... See more A = {a,b,c,d,e}
B = {i,j,k,l,m}
a+b+c+d+e = i+j+k+l+m
a.b.c.d.e = i.j.k.l.m,
where . denotes multiplication.
This does not imply that A and B should be indentical sets.
Consider the counterexample:
A = {3,3,10,n,n}
B = {2,5,9,n,n}
The sum of the elements of set A is 2n+16, the product is 90.n.n - just like for set B.
You can choose any n.
Of course,
A = {3,3,10,n,m}
B = {2,5,9,n,m}
works, and you can find infinitely many soultions. ▲ Collapse
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[ This Message was edited by: on 2002-12-11 22:36 ]
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| Awww poop... I knew I was wrong | Dec 11, 2002 |
excellent counterexample!
| | | | Indojin
Local time: 17:56
英語 から ヒンディー語
+ ...
a, b, c, d, e
are distinct numbers, i.e.,
abcde 0
where = is not equal to
and
i,j,k,l, m are also distinct numbers and
ijklm0
The order of the numbers does not matter.. but in this case do a,b,c,d,e and i,j,k,l,m have to be same???
What about the case when a,b,c,d,e are one-digit numbers?
| | | | Indojin
Local time: 17:56
英語 から ヒンディー語
+ ...
| Found the flaw in Ban Csaba's proof | Dec 13, 2002 |
I think the way you provew a,b,c,d,e have to be equal to i,j,k,l,m was very impressive. I was also pretty much sure of your proof.
But the flaw I think is in, when you assume that
i=a+a`
Here a` can be either positive or negative. It does not necessarily has to be positive.
When a` becomes negative, then in the case of equation -
a.b.c.d.e = (a+a`).(b+b`).(c+c`).(d+d`).(e+e`)
after cancelling the same ... See more I think the way you provew a,b,c,d,e have to be equal to i,j,k,l,m was very impressive. I was also pretty much sure of your proof.
But the flaw I think is in, when you assume that
i=a+a`
Here a` can be either positive or negative. It does not necessarily has to be positive.
When a` becomes negative, then in the case of equation -
a.b.c.d.e = (a+a`).(b+b`).(c+c`).(d+d`).(e+e`)
after cancelling the same terms on both sides, we get zero on LHS, but on the RHS, we can even get negative terms, since a` etc. can also be negative numbers.
Therefore, it is not necessary that a` etc. have to be zero.
Therefore, it is not necessary, that abcde and ijklm have to be equal.
I hope I am able to make myself clear.
Quote:
On 2002-12-11 10:52, Ban Csaba wrote:
I assume that by \"natural numbers\" you mean positive integers and each of the two sets has five elements.
I have an idea how to prove it (indirectly):
You have to consider i, j, etc. as a sum of a+a\' , b+b\' , etc. (making such pairs that i is equal to or greater than a, j is equal to or greater than b, etc.) (a\' , b\' ... are either zeros or positive integers)
If the two sets are different, then at least one of a\' , b\' etc has to be other than zero.
You will have to then write down the product of these five elements (ab + ab\' + ac\' + ... + a\'b\' ... ).
You delete the products that appear on both sides of the equation (ab + ac + ad + ... de). All that remains is:
a\'b\' + a\'c\' + ... d\'e\' = 0
Since our precondition was that at least one of a\' , b\' ... is other than zero, those four products where the non-zero element is present have to be other than zero, hence their sum cannot be zero.
Indirectly we proved that all five of a\' , b\' ... have to be zero, i.e. the set of a, b, ... has to be identical with i, j, ...
Q.E.D
[ This Message was edited by: on 2002-12-11 11:15 ]
[ This Message was edited by: on 2002-12-11 11:16 ]
[ This Message was edited by: on 2002-12-13 03:59 ] ▲ Collapse
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