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Let $\mathbb R_0^+$ is the set of all non-negative real numbers. Prove that $(\mathbb R_0^+,+,\cdot,\mathbb R)$ is a real vector space if operations are defined as: $$(\forall x,y\in\mathbb R_0^+)x+y=x\cdot y$$ $$(\forall \alpha\in\mathbb R)(\forall x\in\mathbb R_0^+)\alpha\cdot x=x^{\alpha}$$

Note: $x,y$ are vectors, and $\alpha$ is a scalar.

Could someone check statements $3,8,9$, because it seems from these statements that this is not a vector space?

$1.$ Closure under addition:

$$V=\mathbb R_0^+$$

$$ x+y\in V; \forall x,y\in V$$

We can intuitively conclude that $x+y=x\cdot y\in V$ from the definitions of operators.

$2.$ Addition associativity:

$$x+(y+z)=(x+y)+z ; \forall x,y,z\in V$$

$$LHS=x+(y+z)=x+(y\cdot z)=x\cdot y\cdot z$$ $$RHS=(x+y)+z=(x\cdot y)+z=x\cdot y\cdot z$$

$3.$ Existence of neutral element with respect to addition operator:

$$\exists 0\in V: x+0=x,\forall x\in V$$

Doesn't this contradicts the statement that $(\mathbb R_0^+,+,\cdot,\mathbb R)$ is a vector space because $$x+0=x\cdot 0\neq x$$

From here, this is not a vector space. Is this right?

$4.$ Existence of inverse element with respect to addition operator:

$$\forall x\in V, \exists (-x)\in V : x+(-x)=0$$

$$x+(-x)=x\cdot (-x)$$

If $x=0$, the statement is true.

$5.$ Addition commutativity:

$$x+y=y+x;\forall x,y\in V$$ $$LHS=x+y=x\cdot y$$ $$RHS=y+x=y\cdot x$$

$6.$ $\alpha x\in V;\forall\alpha \in\mathbb F=\mathbb R,x\in V$

Intuitively, $\alpha x=x^{\alpha}\in V$.

$7.$ $(\alpha\beta)x=\alpha(\beta x), \forall \alpha,\beta\in\mathbb F,x\in V$

$$LHS=(\alpha\beta)x=x^{\alpha\beta}$$ $$RHS=\alpha(\beta x)=\alpha(x^{\beta})=(x^{\beta})^{\alpha}=x^{\alpha\beta}$$

$8.$ $\alpha(x+y)=\alpha x+\alpha y,\forall \alpha\in \mathbb F,x,y\in V$

$$LHS=\alpha(x+y)=(x+y)^{\alpha}=(xy)^{\alpha}=\alpha xy$$ $$RHS=\alpha x+\alpha y=x^{\alpha}+y^{\alpha}$$

This statement is not true. I s this right?

$9.$ $(\alpha+\beta)x=\alpha x+\beta y,\forall \alpha,\beta\in\mathbb F,x\in V$ $$LHS=(\alpha+\beta)x=x^{\alpha+\beta}$$ $$RHS=\alpha x+\beta y=x^{\alpha}+y^{\beta}$$

How is this true?

$10.$ $1\cdot x=x, \forall x\in V$ $$1\cdot x=x^{1}=x$$

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    $\begingroup$ What is confusing you is that you are using $+$ and $.$ in two different ways: both the normal operations and these perverse ones. For (3) the zero element of the vector space is $1$, because $x.1=x$, or in the vector space language $x+0=x$. Personally I'd give these "new" operations different names from $+$ and $.$ ... $\endgroup$ – ancientmathematician Apr 18 '17 at 15:39
  • $\begingroup$ @ancientmathematician, I don't understand how the zero element in statement $3)$ is $1$? What about statements $8)$ and $9)$? $\endgroup$ – user300045 Apr 18 '17 at 15:49
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    $\begingroup$ Are you sure the vector space is $\mathbb{R}^+_0$? I think it must be $\mathbb{R}^+$. $\endgroup$ – amrsa Apr 18 '17 at 15:49
  • $\begingroup$ $0$ doesn't have an inverse for the multiplication... $\endgroup$ – amrsa Apr 18 '17 at 15:50
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    $\begingroup$ But as ancientmathematician wrote, the 'zero' element of this vector space is the real number $1$; the real number $0$ shoudn't even belong to the space. $\endgroup$ – amrsa Apr 18 '17 at 15:53
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For (8) we want:

$$\alpha.(x+y)=\alpha. x+\alpha .y,\forall \alpha\in \mathbb F,x,y\in V$$

where these operations are the ones DEFINED in the question.

So $$\alpha.(x+y)=\left(xy\right)^\alpha$$ and $$\alpha. x+\alpha .y=x^\alpha y^\alpha=\left(xy\right)^\alpha$$ where the operations on the right hand side of these equalities are the usual ones in $\mathbb{R}$.

The question is a Joke: enjoy it.

It would spoil the joke, but we could re-write the question in a more friendly way:

Let $\mathbb{R}^{+}$ be the set of all positive real numbers. Prove that $(\mathbb{R}^{+},\oplus,\odot,\mathbb{R})$ is a real vector space if operations are defined as:

$$(\forall x,y\in \mathbb{R}^{+})\ \ x\oplus y:=xy$$ and $$(\forall \alpha\in\mathbb{R})(\forall x \in \mathbb{R}^{+})\ \ \alpha\odot x:=x^\alpha.$$

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  • $\begingroup$ How to know when to jump from real operators in $\mathbb R$ to those that are randomly defined? $\endgroup$ – user300045 Apr 18 '17 at 16:30
  • $\begingroup$ I have expanded the answer reluctantly. $\endgroup$ – ancientmathematician Apr 18 '17 at 17:17

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