[Solved] Groups Small Challenge  

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1

Calculate the following (modulus 59):

a) 7x9

b) 4x15 

Using your answers find the inverse of the following (in the set integers modulo 59- {0} under the operation x):

c) 4

d) 7

e) 28

f)  49

Is the set- integers modulo 59, under multiplication, a group?

PlanckTime - Amin

 
1

I hope this is correct.

a) 4

b) 1

c) 15

d) 17

e) 19

f) 53

 

"Computer science is no more about computers than astronomy is about telescopes."
~ Edsger W. Dijkstra

 
3

Yeah I agree with all of your answers. I attached worked solutions for anyone unsure on how you got the answers. For the last question though I meant does the set under multiplication satisfy all the requirements to be a group. They are:

  • Set is Closed under operation
  • Operation is associative 
  • There is an identity element within the set
  • Each element has an inverse in the set

 

And this therefore IS a group.

PlanckTime - Amin

Ah that's a more logical way of doing it.
By the way, is there a special notation for modulo?
As writing for example:

235 = 2(59)+17 = 17,

can look strange without context.

For example in some programming languages you'd say:

235 MOD 59 = 17

There is a notation, I forgot to add it:

7x9= 4 (mod 59) 

Chaps

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