Chapter 3.1, Problem 1PS

To determine

Which of the rule is not satisfied with the given condition of vector addition and scalar multiplication?

Answer to Problem 1PS

   $x+y\ne y+x$

   $(x+y)+z\ne x+(y+z)$

   $(c_1+c_2)x\ne c_{1}x+c_{2}x$

Explanation of Solution

Given information:

   $\begin{array}{l}\left(1\right)x+y=y+x\\ \left(2\right)x+\left(y+z\right)=\left(x+y\right)+z\\ \left(3\right)\text{There is a unique "zero vector" such that }x+0=x\forall x\\ \left(4\right)\text{For each }x\text{there is a unique vector }-x\text{such that }x+\left(-x\right)=0\forall x\\ \left(5\right)\text{1 times }x\text{equals }x\text{, i}\text{.e}\text{., }1\cdot x=x\\ \left(6\right)\left(c_1{c}_2\right)x=c_1\left(c_{2}x\right)\\ \left(7\right)c\left(x+y\right)=cx+cy\\ \left(8\right)\left(c_1+c_2\right)x=c_{1}x+c_{2}x\end{array}$

(1) to (4) about $x+y$

(5) to (6) about $cx$

(7) to (8) connects them

We suppose

   $\begin{array}{l}x=(x_1,x_2)\\ y=(y_1,y_2)\\ z=(z_1,z_2)\end{array}$

Then,

   $1)$ First condition

   $x+y=(x_1,x_2)+(y_1,y_2)$

$=(x_1+y_2,x_2+y_1)$

and

   $y+x=(y_1,y_2)+(x_1,x_2)$

   $=(y_1+x_2,y_2+x_1)$

Clearly, we can see that,

   $x+y\ne y+x$

Therefore, the first condition is not satisfied.

   $2)$ We check the second condition

   $x+y=(x_1,x_2)+(y_1,y_2)$

   $=(x_1+y_2,x_2+y_1)$

   $(x+y)+z=(x_1+y_2,x_2+y_1)+(z_2+z_1)$

   $=(x_1+y_2+z_2,x_2+y_1+z_1)$

and

   $y+z=(y_1,y_2)+(z_1,z_2)$

   $=(y_1+z_2,y_2+z_1)$

   $x+(y+z)=(x_1+y_2+z_1,x_2+y_1+z_1)$

Clearly, we can see that

   $(x+y)+z\ne x+(y+z)$

Therefore, the second condition is not satisfied.

   $3)$ We check the third condition

Since,

   $x+0=(x_1,x_2)+(0,0)$

   $=(x_1+0,x_2+0)$

   $=(x_1,x_2)$

   $=x$

Therefore, the third condition is satisfied.

   $4)$ We check the fourth condition.

   $-x=(-x_2,-x_1)$

Then,

   $x+(-x)=(x_1,x_2)+(-x_2,-x_1)$

   $=(x_1-x_1,x_2-x_2)$

   $=(0,0)$

   $=0$

   $5)$ This condition is automatically satisfied.

   $6)$ We suppose there are two scalars

   $c_1,c_2$

   $\begin{array}{l}(c_1{c}_2)x=(c_1{c}_2)(x_1,x_2)\\ \Rightarrow (c_1{c}_2)x=(c_1{c}_2{x}_1,c_1{c}_2{x}_2)\\ \Rightarrow (c_1{c}_2)x=(c_1(c_2{x}_1),c_1(c_2{x}_2))\\ \Rightarrow (c_1{c}_2)x=c_1(c_2{x}_2)\end{array}$

Therefore, this condition is satisfied.

$7)$ We check the condition for scalar $c$

   $x+y=(x_1+x_2)+(y_1+y_2)$

   $=(x_1+y_2,x_2+y_1)$

   $c(x+y)=(c(x_1+y_2),c(x_2+y_1))$

   $=(c{x}_1+c{y}_2,c{x}_2+c{y}_1)$

and

   $cx+cy=c(x_1,x_2),c(y_2+y_2)$

   $=(c{x}_1+c{x}_2)+(c{y}_2+c{y}_2)$

   $=(c{x}_1+c{y}_2,c{x}_2+c{y}_1)$

Therefore,

   $c(x+y)=cx+cy$

Therefore, his condition is satisfied.

   $8)$ We suppose there are two scalars

   $c_1,c_2$

Therefore,

   $(c_1+c_2)x=(c_1+c_2)(x_1,x_2)$

   $=((c_1+c_2)x_1,(c_1+c_2)x_2)$

   $=(c_1{x}_1+c_2{x}_1,c_1{x}_2+c_2{x}_2)$

and

   $c_{1}x+c_{2}x=c_1(x_1,x_2)+c_2(x_1,x_2)$

   $=(c_1{x}_1,c_2{x}_2)+(c_2{x}_1,c_2{x}_2)$

   $=(c_1{x}_1+c_2{x}_1,c_1{x}_2+c_2{x}_2)$

Therefore

   $(c_1+c_2)x\ne c_{1}x+c_{2}x$

Therefore, this condition is not satisfied.

Conclusion:

Therefore

   $x+y\ne y+x$

   $(x+y)+z\ne x+(y+z)$

   $(c_1+c_2)x\ne c_{1}x+c_{2}x$