Question

Does {(1,3),(3,4),(2,3)}span (Z/5Z)^2? Z=integer

Answer 1

I don't understand (again) your notation: (Z/5Z)^2 means what?

Answer 2

Z/5Z is the integers mod 5 right?

Answer 3

yeaps

Answer 4

best way to do this is to make a table of all the possible ordered pairs mod 5 and see if you can make each value in the table using only the given ordered pairs

Answer 5

if you can, then that set spans (Z/5Z)^2

Answer 6

but what does the power 2 mean though? is there any significance?

Answer 7

Z^2 is the set of all ordered pairs drawn from Z

Answer 8

(Z/5Z)^2 is the set of all ordered pairs drawn from Z mod 5 = {0, 1, 2, 3, 4}

Answer 9

more formally, \[\large \frac{\mathbb{Z}^2}{5\mathbb{Z}}=\{(a,b)|a,b\in\frac{\mathbb{Z}}{5\mathbb{Z}}\}\]

Answer 10

hmm okay. i think i get it now. thanks!

Answer 11

oh and you might see it as
\[\large \mathbb{Z}^2_{5}=\{(a,b)|a,b\in\mathbb{Z}_{5}\}\]

Answer 12

If you're still stuck, notice that

(1,3)+(3,4)=(4,7)=(4,2)


and


(4,2)+(1,3)=(5,5)=(0,0) <--- this is one key ingredient



Also,


(4,2)+(2,3)=(6,5)=(1,0) <--- another useful (very useful) element



Finally,

(3,4)+(2,3)=(5,7)=(0,2)

and

(0,2)+(0,2)+(0,2)=(0,6)=(0,1)


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Why did I do all that? Well for one thing, I've shown that (0,0) is in there. This is the additive identity.


Also, I've shown that (1,0) and (0,1) can also be generated


So we can then say that any element x can be generated by the linear combination


x = c*(1,0)+d*(0,1)


to get EVERY element thats in the desired set that we want.


So therefore, {(1,3),(3,4),(2,3)} spans (Z/5Z)^2

Answer 13

ok now that I think about it, you don't need to generate (0,0) explicitly since the formula involving x takes care of it.