Complete the proof of Property 4 of the following theorem by supplying the justification for each step. Properties of Additive Identity and Additive Inverse Let v be a vector in R", and let c be a scalar. Then the properties below are true. 1. The additive identity is unique. That is, if v + u = v, then u = 0. 2. The additive inverse of v is unique. That is, if v + u = 0, then u = -v. 3. Ov = 0 4. co = 0 5. If cv = 0, then c = 0 or v = 0. 6. -(-v) = v Use the properties of vector addition and scalar multiplication from the following theorem. Properties of Vector Addition and Scalar Multiplication in R" Let u, v, and w be vectors in R", and let c and d be scalars. 1. u + v is a vector in R". 2. u + v = v + u 3. (u + u) + w = u + (u + w) 4. u +0 = u 5. u + (-u) = 0 Closure under addition 6. cu is a vector in R". 7. c(u + v) = cu + cv 8. (c + d)u = cu + du 9. c(du) = (cd)u 10. 1(u) = u Commutative property of addition Associative property of addition Additive identity property Additive inverse property Closure under scalar multiplication Distributive property Distributive property Associative property of multiplication Multiplicative identity property Step Justification co = c(0 + 0) -Select--- c0 = c0 + co ---Select-- co + (-c0) = (c0 + c0) + (-cO) -Select--- 0 = (co + co) + (-c0) -Select--- 0 = c0 + (co + (-c0)) -Select- O = c0 + 0 Select--
Complete the proof of Property 4 of the following theorem by supplying the justification for each step. Properties of Additive Identity and Additive Inverse Let v be a vector in R", and let c be a scalar. Then the properties below are true. 1. The additive identity is unique. That is, if v + u = v, then u = 0. 2. The additive inverse of v is unique. That is, if v + u = 0, then u = -v. 3. Ov = 0 4. co = 0 5. If cv = 0, then c = 0 or v = 0. 6. -(-v) = v Use the properties of vector addition and scalar multiplication from the following theorem. Properties of Vector Addition and Scalar Multiplication in R" Let u, v, and w be vectors in R", and let c and d be scalars. 1. u + v is a vector in R". 2. u + v = v + u 3. (u + u) + w = u + (u + w) 4. u +0 = u 5. u + (-u) = 0 Closure under addition 6. cu is a vector in R". 7. c(u + v) = cu + cv 8. (c + d)u = cu + du 9. c(du) = (cd)u 10. 1(u) = u Commutative property of addition Associative property of addition Additive identity property Additive inverse property Closure under scalar multiplication Distributive property Distributive property Associative property of multiplication Multiplicative identity property Step Justification co = c(0 + 0) -Select--- c0 = c0 + co ---Select-- co + (-c0) = (c0 + c0) + (-cO) -Select--- 0 = (co + co) + (-c0) -Select--- 0 = c0 + (co + (-c0)) -Select- O = c0 + 0 Select--
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Complete the proof of Property 4 of the following theorem by supplying the justification for each step.
Properties of Additive Identity and Additive Inverse
Let v be a vector in R", and let c be a scalar. Then the properties below are true.
1. The additive identity is unique. That is, if v + u = v, then u = 0.
2. The additive inverse of v is unique. That is, if v + u =
3. Ov
0, then u = -v.
= 0
4. СО %3D 0
5. If cv = 0, then c = 0 or v = 0.
6. -(-v) = v
Use the properties of
ctor addition and scalar multiplication from the following theorem.
Properties of Vector Addition and Scalar Multiplication in R"
Let u, v, and w be vectors in R", and let c and d be scalars.
1. u + v is a vector in R".
Closure under addition
2. u + v = v + u
Commutative property of addition
3. (u + u) + w = u + (u + w)
Associative property of addition
Additive identity property
Additive inverse property
Closure under scalar multiplication
4. u + 0 = u
5. u + (-u) = 0
6. cu is a vector in R".
7. c(u + v)
8. (с + d)u
9. c(du) =
10. 1(u) = u
Distributive property
Distributive property
= cu + cV
= cu + du
(cd)u
Associative property of multiplication
Multiplicative identity property
%3D
Step
Justification
c0 =
c(0 + 0)
---Select---
c0 = c0 + c0
---Select---
c0 + (-c0) = (c0 + c0) + (-c0)
---Select---
(c0 + c0) + (-c0)
---Select---
0 =
O = c0 + (c0 + (-c0))
---Select---
0 = c0 + 0
---Select---
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