7-7=13-13
7(1-1)=13(1-1)
By dividing on (1-1) at both the two sides
then 7=13
2006-11-13 18:51:02
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answer #1
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answered by Anonymous
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In the binary system: 0, 1, 10, 11 stands for 0, 1, 2, 3
In the quinary system (of 4): 0, 1, 2, 3, 10, 11, 12, 13 represents
0, 1, 2, 3, 4, 5, 6, 7 respectively. Therefore, 13 = 7 from the conversion.
2006-11-13 18:50:18
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answer #2
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answered by dreamofyz 2
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7 * x = 13 * x if x= 0
2006-11-13 18:34:32
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answer #3
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answered by indoragincajun 1
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if 7 and 13 are naturale numbers than there is no way prove this. If you are interested in the construction of the set of naturales take a look at Peano axioms. You will find there a function called the succesor function that works like this: s(0)=1, s(1)=2 and so on.
now supose that 7=13. ie s(12)=13 and s(12)=7, this is in contradiction with the first Peano axiom(every natural has a unique successor)
2006-11-13 19:02:25
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answer #4
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answered by Dao R 1
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they're equal in mod 6
mod 6 is the base of the number system. our decimal system is in base 10, or mod 10
7 = 1 in mod 6 as the remainder of 7/6 is 1
13 = 1 in mod 6 as the remainder of 13/6 is 1
thus, 7=13. QED
2006-11-13 18:43:41
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answer #5
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answered by mahnamahna 2
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Nope can you?
i can prove 1 = -1 by using Phytagorean Theorem but not 7 = 13.... seems to be impossible!!!
2006-11-13 18:31:12
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answer #6
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answered by bugi 6
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I can prove 7=13 if you can prove that earth has a sequared shape!!!
2006-11-13 19:39:38
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answer #7
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answered by Ibraheem G 2
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7(base 10) = 13(base 4)
Rob S, you cannot say (-1^2)^(1/2) = -1
because (-1^2)^(1/2) = (-1)^(1/2) which is an imaginary number
is that the subtle error your talking about?
2006-11-13 18:58:57
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answer #8
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answered by Jason D 2
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Practically what u r asking is impossible, but theortically if u want to prove it u can use any trial and error method to prove it.
2006-11-13 19:18:17
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answer #9
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answered by Napster 2
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since that your question is can you prove that 7=13?
then my answer is yes i can
and i think that is the correct answer ?
2006-11-13 22:56:49
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answer #10
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answered by alaa_cancer 3
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