Question

Consider the set S=R consisting of all real number . On this set S, consider the operations (+), (x) defined as:

u (+) v = max(u,v)
k (x) u = k u

a) Determine whether vector addition is commutative.
u(+) v = v(+) u for all u , v in S

b) Determine whether or not S has zero element 0 such that u(+)0= u

Answer 1

(1) addition is commutative according to the properties of real numbers. .

Answer 2

(2) yes, there exist an element/number 0 in addition, that's according to the identity property of addition-"that any number added to zero is equal to the number".

Answer 3

but (+) is not addition as defined here

Answer 4

You have \(u⊕v=\text{max}(u,v)=\text{max}(v,u)=v⊕u \text{ }\forall \text{ }u,v \in S.\) Thus ⊕ is commutative.

Answer 5

ok, that's what I thougth

Answer 6

but second one, it is no right because
if u<0 then we get 0

Answer 7

Let me make sure of one thing first. Does max(u,v) means the vector with the largest magnitude?

Answer 8

They are real numbers

Answer 9

Whoops! right. Then the answer is no, since there's no element \(a \in \mathbb{R}\) such that \(a\le u \forall u\in \mathbb{S}\).

Answer 10

a≤u∀u∈S

can you explain me what this mean?

Answer 11

By the definition we have for \(⊕\), our zero element \(\bf0\) has to satisfy \(u⊕{\bf {0}}=\text{max}(u,{\bf{0}})=u.\) This will be the case if and only if \({\bf 0}\le u\) for all \(u\in S\). But this will never happen because there is NO least real number.

Answer 12

ok, thank you very much

Answer 13

You're welcome!