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Modular arithmetic

by davon806
Tags: arithmetic, modular
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davon806
#1
Jul13-14, 03:38 AM
P: 52
Please see the attached,which was quoted from the following website:
http://en.wikipedia.org/wiki/Modular_arithmetic

It said that the multiplicative property is only applicable if n is an integer.On the contrary,the addition property can be applied to all real numbers.
I don't quite understand what this sentence means.Can anyone explain it to me?Any mathematical proof provided will be appreciated.
Thanks.
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DeltaČ
#2
Jul13-14, 04:31 AM
P: 403
Well suppose all [tex]a_1,a_2,b_1,b_2,n[/tex] are real numbers and that
[tex]a_1=b_1 (mod n)[/tex]
[tex]a_2=b_2 (mod n)[/tex]
then by the definition of mod n equality we can conclude that there exist integers [tex]k_1,k_2[/tex] such that [tex]a_1-b_1=k_1n , a_2-b_2=k_2n[/tex]
so by adding the last 2 equations we have [tex](a_1+a_2)-(b_1+b_2)=(k_1+k_2)n[/tex] therefore there exists the integer [tex]k_3=k_1+k_2[/tex] such that [tex](a_1+a_2)-(b_1+b_2)=(k_3)n[/tex] hence by definition of (mod n) equality this means that [tex](a_1+a_2)=(b_1+b_2) (mod n)[/tex]. As you see in the proof we dont need anything of the a's or b's or the n to be integer.

However if we want to prove that [tex]a_1a_2=b_1b_2 (mod n) (1)[/tex] we gonna need a's and b's as well n to be integers because the proof goes like this:
[tex]a_1a_2-a_1b_2-b_1a_2+b_1b_2=k_1k_2n^2[/tex] so n has to be integer in order for [tex](a_1a_2+b_1b_2)=(a_1b_2+b_1a_2) (mod n) (2)[/tex] . We will also need a's and b's to be integers in order to prove that [tex]a_1b_2=b_1b_2 (mod n) (3.1) b_1b_2=b_1a_2 (mod n) (3.2)[/tex] hold. From (2) and (3)s we can infer the (1).
davon806
#3
Jul13-14, 10:02 AM
P: 52
Why did a1a2−a1b2−b1a2+b1b2=k1k2n2?

DeltaČ
#4
Jul13-14, 10:21 AM
P: 403
Modular arithmetic

By multiplying [tex]a_1-b_1=k_1n , a_2-b_2=k_2n[/tex] together. I have omitted some steps cause it isnt allowed here to give too much help with homework.

Eventually to prove why (1) doesnt hold if all a's and b's and n are real, you ll have to construct a counter example. I just pinpointed where the proof goes wrong if a's and b's and n arent all integers, but one could claim that there might be another proof specific made for when a's and b's and n are reals.
davon806
#5
Jul14-14, 03:27 AM
P: 52
Sorry,please see the attached,I came up with a contradiction while proving the above.If my proof is correct,then the multiplicative property can be applied to rational numbers,so there must be something wrong but I couldn't find it.Thx very much
Attached Thumbnails
IMG_1533.JPG   IMG_1534.JPG  
DeltaČ
#6
Jul14-14, 04:19 AM
P: 403
The final expression for a1a2-b1b2 should have k1 inside as well and not only k2. Anyway it seems for specific a's and b's that satisfy some extra equations, the multiplicative property could hold for those specific rationals , but you want to hold for every rational in order to be usefull.
davon806
#7
Jul14-14, 05:23 AM
P: 52
thx for your help
DeltaČ
#8
Jul14-14, 09:42 AM
P: 403
You are welcome :)


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