Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

Yes, exactly, "triples" implies "ordered triples". (1, 2, 1) is NOT the same as (2, 1, 1).however, the answer works it out tohold.not

It goes as follows:

Setu= (u1, u2, u3), then

(k+m)u

=(k+m)(u1, u2, u3)

= ((k+m)u1,(k+m)u2,(k+m)u3)

and

ku+ mu

=(ku1, ku2, ku3) + (mu1, mu2, mu3)

= (ku3 + mu3, ku2 + mu2, ku1+ mu1)

since these are not equal, this axiom does not hold.'

Now, I don't understand why these two are not equal?

Can we not further simplify:

(ku3 + mu3, ku2 + mu2, ku1+ mu1)

= ((k+m)u3,(k+m)u2,(k+m)u1)?

That way, does it not equal = ((k+m)u1,(k+m)u2,(k+m)u3))?

or is it not equal because the order of the triple is different?

I'm quite confused and any help would be much appreciated!

Have a great day all