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Let V=(2,∞). For u,v∈V and a∈R define vector addition by u⊞v:=uv−2(u+v)+6 and scalar multiplication by a⊡u:=(u−2)^a+2. It can be shown that (V,⊞,⊡) is a vector space over the scalar field R. Find the following:

the sum:

6⊞10=34

the scalar multiple:

−3⊡6=2.015625

the additive inverse of 6:

⊟6=

the zero vector:

0V=3

the additive inverse of x:

⊟x=

As you can see I found the sum and the scalar multiple as well as the zero vector.However, I tried doing the additive inverse of 6 and the additif inverse of x, but I have no idea on how to find the answers.

needhelp.com
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    Hint: the “sum” is really multiplication: first shift $u$ and $v$ “back” by $2$, then shift the answer forward by $2$: $(u-2)(v-2)+2$. The zero vector is the identity of multiplication, $1$, shifted forward by $2$. So the inverse of $6$ would be... – Arturo Magidin Apr 25 '21 at 03:19
  • I still don't get it .... – needhelp.com Apr 25 '21 at 04:19
  • Shift everything back $2$ and multiply; then shift everything forward $2$. That’s all you are doing. The inverse of $4$ if you were not shifting would be $\frac{1}{4}$, so the inverse of $6$ will be the inverse of $6-2$, shifted forward for $2$. – Arturo Magidin Apr 25 '21 at 04:21
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    Remark: specific examples a proof does not make. – Sean Roberson Apr 25 '21 at 05:07
  • Please use MathJax. – Toby Mak Apr 25 '21 at 08:23
  • Have you checked the text of your homework ? The fact that $V=(2,+\infty)$ is not stable for external multiplication is more than a big problem... – Jean Marie Apr 25 '21 at 16:27
  • The text is erroneous and the OP hasn't rectified it :$a⊡u:=(u−2)a+2$ should be $a⊡u:=(u−2)^a+2$... – Jean Marie Apr 27 '21 at 09:19

2 Answers2

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Initial remark: You have a problem with external multiplication

$$a⊡u:=(u−2)a+2$$

If I take $u=3$ and $a=-1$, I get $a⊡u = 1 \notin V$...

$V$ isn't stable for external multiplication...

In fact, here is a very analogous question where you will see that the external multiplication must use an exponent

Besides, writing the definition of your operation under the form:

$$u \oplus v -2 = (u-2) (v-2)$$

and setting

$$f(x):=x-2,\tag{1}$$

you see that you have a "transfer rule" :

$$f(u \oplus v)=f(u)f(v)\tag{2}$$

Same type of operation with scalar multiplication:

$$a⊡u:=(u−2)a+2 \ \iff \ f(a⊡u):=af(u) \tag{3}$$

Therefore you can solve your questions mecanically with relationships (1), (2) and (3).

Remark:

It would be natural in fact to replace (3) by:

$$\ln f(u \oplus v)=\ln f(u)+ \ln f(v)\tag{4}$$

in order to find a bijective correspondence with an additive rule.

Jean Marie
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Before you can find an additive inverse you must find the additive identity. That is $e\square x = x$ for all $x$.

so $ex -2(e+x) +6 = x$. Solver for $e$

so $e(x-2)=3x -6$ so $e=\frac {3x-6}{x-2} = 3$

The additive identity is $3$.

Verify: $k\square 3= 3k -2(k+3) + 6 = k -6 + 6 = k$ so that is indeed the indentity.

Now to find the additive inverse of $6$ we must solve: $x\square 6 = 3$.

So $6x - 2(x+6) + 6 = 3$

$4x =9$

$x = \frac 94$.

SO $x=\frac 94$ is the additive inverse of $6$.

Verify: $\frac 94 \square 6 = \frac 94\cdot 6 -2(6 +\frac 94) +6= \frac {27}2 - 12-\frac 93 + 6 = 9-12 + 6 = 3$. It checks.

fleablood
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  • Do I have to use the same formula to find the additive inverse of x ? – needhelp.com Apr 26 '21 at 15:03
  • To find the additive inverse of $x$ solve $y\square x = 3$ for $y$. $xy-2(x+y) + 6 =3$ so solve for $y$. $xy-2y =-3+2x; y=\frac{-3+2x}{x-2}$. Verfy: $\frac{-3+2x}{x-2}\square x = \frac {-3+2x}{x-2}x - 2(\frac {-3+2x}{x-2} + x) + 6 = \frac {x(-3+2x)-2(-3+2x)}{x-2} -2x + 6 =(-3+2x) -2x +6=-3+6=3$. – fleablood Apr 26 '21 at 16:27
  • The external multiplication should be $(u-2)^{a}+2$ with $a$ an exponent. See the analogous question here. – Jean Marie Apr 26 '21 at 23:09