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Let $\mathbb{R}$ be the set of all real numbers. Define scalar multiplication by

$\alpha x = \alpha \cdot x$ (the usual multiplication of real numbers)

and define addition by

$x \oplus y = \max(x, y)$ (the maximum of two numbers)

Prove that $\mathbb{R}$ with these operations is NOT a vector space. Which of the eight axioms fail to hold?

Please help I know I need to prove it by using the 8 axioms but I get confused with the addition part.

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I believe your addition does not have inverse element, so the following is not true: $$\forall v \in V, \exists w(\text{also called }-v) : v \oplus w = 0$$ because $max(v,w)$ can be only $v$ or $w$ but not $0$.

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In any vector space, $x \oplus y = x$ if and only if $y$ is the additive identity (usually $0$).

If we suppose this is a vector space, $1 \oplus 0 = \max(1,0) = 1$ implying that $0$ is indeed the additive identity. But then $1 \oplus 1 = 1$ so $1 = 0$, a contradiction. Hence $\mathbb{R}$ is not a vector space under these operations.

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The problem is that you can't find an element $u \in\mathbb{R}$ such that $x \oplus u= u \oplus x=x$ for all $x\in \mathbb{R}$. The reason is because the real numbers have no lower limit, so you can always find a number less than any other given, so the value of $\max(x,u)$ can't be fixed for $x$. The axiom that does not hold is the existence of the neutral element for the operation $\oplus$.

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I think a really intuitive way to show that this isn't a vector space is to show that there's no zero vector (additive identity element.)

There's no $\mathbf 0$ such that $\mathbf 0 + \mathbf a = \mathbf a$ for all $\mathbf a \in \Bbb R$ when a + b is defined to be max{a, b}. You can see this because when addition is defined as max{x, y} the number that gets returned is always the larger number, and because the zero vectors job is to always return the other number it gets added with (again that means $\mathbf 0 + \mathbf a = \mathbf a$ when a is any number) then your zero vector would have to be less than every other number that exists and there's no such number. Because of this, this set and operations can't be a vector space.

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Note that $(-1)(1 \oplus 2)=(-1) \cdot 2=-2$, but $(-1)(1) \oplus (-1)(2)=-1 \oplus -2=-1$. That is, scalar multiplication doesn't distribute over vector addition.

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