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 May 16th, 2015, 10:21 AM #1 Newbie   Joined: May 2015 From: Utah Posts: 3 Thanks: 0 Need help with Closure under scalar multiplication I am having difficulty proving this axiom. My last response from my instructor was this: Support for law 6 is not sufficient. The requirements are to solve and explain each step of the axiom. This is what I have so far, I hope someone can help me fix my errors. Closure under scalar multiplication If X is any vector in V and r is any real scalar, then rX ∈ V. r(a, b) ∈ V substituting, coordinate form of vectors. (r(a), r(b)) Scalar multiplication r(a) by the closure property (r) and (a) are real number. Therefore (r) and (a) are closed under real number multiplication. r(b) by the closure property (r) and (b) are real number. Therefore (r) and (a) are closed under real number multiplication. ((r)(a), (r)(b) we know is in R^2 because (r) is in R^2, and X is in R^2 Coordinate X are real numbers, and is in R^2 Therefore r(X) is in R^2 May 16th, 2015, 12:54 PM   #2
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 Originally Posted by bgi11in I am having difficulty proving this axiom. My last response from my instructor was this: Support for law 6 is not sufficient. The requirements are to solve and explain each step of the axiom. This is what I have so far, I hope someone can help me fix my errors. Closure under scalar multiplication If X is any vector in V and r is any real scalar, then rX ∈ V. r(a, b) ∈ V substituting, coordinate form of vectors.
• "Coordinate for of vectors"? Are told that V is two dimensional?

Quote:
 (r(a), r(b)) Scalar multiplication r(a) by the closure property (r) and (a) are real number. Therefore (r) and (a) are closed under real number multiplication.
?? I don't know what you are saying here. (r) and (a) are specific numbers. It makes no sense to say that they are "closed under real number multiplication". That would be like saying that "4" and "18" are "closed under real number multiplication".

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 r(b) by the closure property (r) and (b) are real number. Therefore (r) and (a) are closed under real number multiplication.
Same thing.

[quote]
• ((r)(a), (r)(b) we know is in R^2 because (r) is in R^2, and X is in R^2[quote]
No, in this problem, r is in R, not R^2. Once again, were you specifically told that X was in R^2? I see that nowhere in your statement of the problem which was just to "show closure under scalar multiplication".

Quote:
 Coordinate X are real numbers, and is in R^2 Therefore r(X) is in R^2
What was the exact statement of the problem? Was it specifically to show that R^2, with the usual definition of "scalar multiplication", is "closed under scalar multiplication"?

Last edited by Country Boy; May 16th, 2015 at 12:57 PM. May 16th, 2015, 03:15 PM #3 Newbie   Joined: May 2015 From: Utah Posts: 3 Thanks: 0 That you for the replay. The assignment is to prove "Closure under scalar multiplication If X is any vector in V and r is any real scalar, then rX ∈ V." I have to break each step down and use the following to explain what I did. Substitution of coordinate form for vector form Substitution of vector form for coordinate form Vector addition of coordinate form Scalar multiplication of coordinate form Real number associativity of addition or multiplication Real number commutativity of addition or multiplication Closure of real number addition or multiplication Distributive property of real number multiplication over real number addition Identity property of real number addition or multiplication Inverse property of real number addition or multiplication May 16th, 2015, 04:02 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra So $\vec v \in V$ is an $n$-tuple $\langle a_1, a_2, \ldots, a_n\rangle$ of real numbers. (Substitution of vector form for coordinate form. Thus for $r \in \mathbb R$ we have $r\vec v = \langle ra_1, ra_2, \ldots, ra_n\rangle$ (scalar multiple of coordinate form). Since $r$ and all the $a_i$ are real, the $ra_i$ are also real (closure of the reals under multiplication). Thus $\vec w = \langle ra_1, ra_2, \ldots, ra_n\rangle$ is an $n$-tuple of reals and $\vec w = r\vec v$ is thus in $V$ (substitution of coordinate form for vector form). Thanks from bgi11in May 16th, 2015, 04:12 PM #5 Newbie   Joined: May 2015 From: Utah Posts: 3 Thanks: 0 Thanks V8archie. Yeah the instructor said my steps are correct however my explanation is incorrect. Tags closure, multiplication, scalar Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jjpbq6 Real Analysis 3 March 6th, 2013 05:26 PM 450081592 Real Analysis 4 November 2nd, 2011 07:25 AM waytogo Real Analysis 1 October 16th, 2011 12:30 AM Teddy Linear Algebra 3 December 11th, 2009 03:08 PM ElMarsh Algebra 3 October 3rd, 2009 07:01 AM

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