Answer to Question #313835 in Linear Algebra for Nick

Question #313835

Consider the set V = R 2 . For (x1, y1),(x2, y2) ∈ R 2 and c ∈ R, define the following operations:

I. (x1, y1) + (x2, y2) = (0, y1 + y2)

II. c(x1, y1) = (0, cy1)

Is the subset a vector of R2. If not prove the axioms that makes it false.Also prove those axioms that are true.


1
Expert's answer
2022-03-19T02:40:21-0400

V=R2I.a=(x1,y1)V,b=(x2,y2)Va+b=(x1,y1)+(x2,y2)=(0,y1+y2)VII.cR,a=(x1,y1)Vca=c(x1,y1)=(0,cy1)VV=R^2\\ I. \forall \vec{a}=(x_1,y_1)\in V, \vec{b}=(x_2,y_2)\in V\\ \vec{a}+\vec{b}=(x_1,y_1)+(x_2,y_2)=(0,y_1+y_2)\in V\\ II. \forall c\in R, \forall \vec{a}=(x_1,y_1)\in V\\ c\cdot\vec{a}=c\cdot(x_1,y_1)=(0,cy_1)\in V

Axioms

1)a+b=b+aa+b=(x1,y1)+(x2,y2)=(0,y1+y2)b+a=(x2,y2)+(x1,y1)=(0,y2+y1)==(0,y1+y2)=a+btrue1) \vec{a}+\vec{b}= \vec{b}+\vec{a}\\ \vec{a}+\vec{b}=(x_1,y_1)+(x_2,y_2)=(0,y_1+y_2)\\ \vec{b}+\vec{a}=(x_2,y_2)+(x_1,y_1)=(0, y_2+y_1)=\\ =(0,y_1+y_2)= \vec{a}+\vec{b}\\ true

2.d=(x3,y3)V(a+b)+d=a+(b+d)a+(b+d)=(x1,y1)+(0,y2+y3)==(0,y1+y2+y3)(a+b)+d=(0,y1+y2)+(x3,y3)==(0,y1+y2+y3)=a+(b+c)true2. \forall \vec{d}=(x_3,y_3)\in V\\ (\vec{a}+\vec{b})+\vec{d}=\vec{a}+(\vec{b}+\vec{d})\\ \vec{a}+(\vec{b}+\vec{d})=(x_1,y_1)+(0, y_2+y_3)=\\ =(0, y_1+y_2+y_3)\\ (\vec{a}+\vec{b})+\vec{d}=(0, y_1+y_2)+(x_3,y_3)=\\ =(0, y_1+y_2+y_3)=\vec{a}+(\vec{b}+\vec{c})\\ true

3.0=(x2,y2)V,a=(x1,y1)Va+0=aa+0=(x1,y1)+(x2,y2)=(0,y1+y2)a∄0false3. \exist\vec{0}=(x_2,y_2)\in V, \forall\vec{a}=(x_1,y_1)\in V\\ \vec{a}+\vec{0}=\vec{a}\\ \vec{a}+\vec{0}=(x_1,y_1)+(x_2,y_2)=(0, y_1+y_2)\neq\vec{a}\\ \not\exist\vec{0}\\ false

4.a=(x1,y1)Vb=(x2,y2)Va+b=0∄0false4. \forall\vec{a}=(x_1,y_1)\in V\exist\vec{b}=(x_2,y_2)\in V\\ \vec{a}+\vec{b}=\vec{0}\\ \not\exist\vec{0}\\ false

5.c,kR,a=(x1,y1)V(c+d)a=ca+da(c+d)a=(c+d)(x1,y1)=(0,(c+d)y1)ca=c(x1,y1)=(0,cy1)da=d(x1,y1)=(0,dy1)ca+da=(0,cy1)+(0,dy1)=(0,cy1+dy1)==(c+d)atrue5. \forall c, k\in R, \forall \vec{a}=(x_1,y_1)\in V\\ (c+d)\cdot\vec{a}=c\cdot\vec{a}+d\cdot\vec{a}\\ (c+d)\cdot\vec{a}=(c+d)\cdot(x_1,y_1)=(0,(c+d)y_1)\\ c\cdot\vec{a}=c\cdot(x_1,y_1)=(0,cy_1)\\ d\cdot\vec{a}=d\cdot(x_1,y_1)=(0,dy_1)\\ c\cdot\vec{a}+d\cdot\vec{a}=(0,cy_1)+(0,dy_1)=(0,cy_1+dy_1)=\\ =(c+d)\cdot\vec{a}\\true

6.cR,a=(x1,y1)V,b=(x2,y2)Vc(a+b)=ca+cbc(a+b)=c(0,y1+y2)=(0,c(y1+y2))ca=(0,cy1)cb=(0,cy2)ca+cb=(0,cy1)+(0,cy2)=(0,cy1+cy2)==c(a+b)true6. \forall c\in R, \forall \vec{a}=(x_1,y_1)\in V,\\ \vec{b}=(x_2,y_2)\in V\\ c\cdot(\vec{a}+\vec{b})=c\cdot\vec{a}+c\cdot\vec{b}\\ c\cdot(\vec{a}+\vec{b})=c\cdot(0,y_1+y_2)=(0,c(y_1+y_2))\\ c\cdot\vec{a}=(0,cy_1)\\ c\cdot\vec{b}=(0,cy_2)\\ c\cdot\vec{a}+c\cdot\vec{b}=(0,cy_1)+(0,cy_2)=(0,cy_1+cy_2)=\\ =c\cdot(\vec{a}+\vec{b})\\true

7.c,kR,a=(x1,y1)V(cd)a=c(da)(cd)a=(cd)(x1,y1)=(0,cdy1)da=d(x1,y1)=(0,dy1)c(da)=c(0,dy1)=(0,cdy1)=(cd)atrue7. \forall c, k\in R, \forall \vec{a}=(x_1,y_1)\in V\\ (c\cdot d)\cdot\vec{a}=c\cdot(d\cdot\vec{a})\\ (c\cdot d)\cdot\vec{a}=(c\cdot d)\cdot(x_1,y_1)=(0,c\cdot dy_1)\\ d\cdot\vec{a}=d\cdot(x_1,y_1)=(0,dy_1)\\ c\cdot(d\cdot\vec{a})=c\cdot(0,dy_1)=(0,cdy_1)= (c\cdot d)\cdot\vec{a}\\true

8.1R,a=(x1,y1)V1a=a1a=1(x1,y1)=(0,1y1)=(0,y1)afalse8. \exist 1\in R, \forall\vec{a}=(x_1,y_1)\in V\\ 1\cdot\vec{a}=\vec{a}\\ 1\cdot\vec{a}=1\cdot(x_1,y_1)=(0,1\cdot y_1)=(0,y_1)\neq\vec{a}\\ false

VV is not the subset a vector of R2

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment