Proving a(b+c) ≠ ab+ac: Understanding the Non-Distributive Property

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Homework Statement


Verify using an example that a(b+c) is not equal to ab+ac. (This means that addition does not distribute over the dot product.)

Homework Equations


a(b+c)= ab+ac is ALWAYS true for whatever numbers you substitute

let a = 16
b = 3
and c = 5

The Attempt at a Solution



so LHE = a/(a + b) = 16 / (3 + 5) = 16/8 = 2
and RHE = (a/b) + (a/c) = (16/3) + (16/5) ≠ 2
so a/(a + b) ≠ (a/b) + (a/c)
LHE = a + (bc) = 16 + 3 x 5 = 16 + 15 = 31
RHE = (a + b)(a + c) = (16 + 3)(16 + 5) = 19 x 21 ≠ 31
so a + (bc) ≠ (a + b)(a + c)
[/B]
 
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a,b,c are vectors not numbers. You are using numeric division as well and that's not defined for vectors.

Your example needs to define three vectors a,b,c and show that addition does not distribute over the dot product.
 
amy098yay said:

Homework Statement


Verify using an example that a(b+c) is not equal to ab+ac. (This means that addition does not distribute over the dot product.)

Homework Equations


a(b+c)= ab+ac is ALWAYS true for whatever numbers you substitute

let a = 16
b = 3
and c = 5

The Attempt at a Solution



so LHE = a/(a + b) = 16 / (3 + 5) = 16/8 = 2
and RHE = (a/b) + (a/c) = (16/3) + (16/5) ≠ 2 [/B]
This is very confusing. In your problem statement you say that you are asked to show an example for which a(b + c) is not equal to ab + ac. But in your attempt you have a/(a + b).
Are a, b, and c numbers, vectors, or what? Is the operation multiplication (dot product) or division?

amy098yay said:
so a/(a + b) ≠ (a/b) + (a/c)
LHE = a + (bc) = 16 + 3 x 5 = 16 + 15 = 31
RHE = (a + b)(a + c) = (16 + 3)(16 + 5) = 19 x 21 ≠ 31
so a + (bc) ≠ (a + b)(a + c)
 
amy098yay said:
Verify using an example that a(b+c) is not equal to ab+ac.
It is very well known that if a, b, and c are vectors of the same dimension, then ##\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}##
amy098yay said:
(This means that addition does not distribute over the dot product.)
This sentence doesn't agree with what you wrote. The dot product distributes over vector addition.
 
If we let:

Vector A be in the y direction (Ax=0 , Ay=1 , Az = 0)
Vector B be in the x direction (Bx=1 , By=0 , Bz = 0)

so, Vector A×B components:

x = Ay * Bz - By * Az = 0
y = Az * Bx - Bz * Ax = 0
z = Ax * By - Bx * Ay = -1

so,

AxB = (0 , 0 , -1)

would this work?
 
I don't know what you're doing. See my reply to your PM to me.

What is the exact statement of the problem? It seems to be about the properties of the dot product, but the work you're doing involves the cross product. Until we know exactly what you're trying to do, we're not going to be able to help you.
 
Maybe take a picture of the problem from your book and post it here. You can use the UPLOAD A FILE button to do this.
 
amy098yay said:
If we let:

Vector A be in the y direction (Ax=0 , Ay=1 , Az = 0)
Vector B be in the x direction (Bx=1 , By=0 , Bz = 0)

so, Vector A×B components:

x = Ay * Bz - By * Az = 0
y = Az * Bx - Bz * Ax = 0
z = Ax * By - Bx * Ay = -1

so,

AxB = (0 , 0 , -1)

would this work?
I replied in a PM to @amy098yay, but I'll include the gist of what I said here. The problem is ostensibly about the dot product, but the work above is A x B for the vector given above, so seems totally unrelated to the problem given at the start of this thread.

In any case, since the dot product distributes across the sum of vectors, it won't be possible to find an example for which ##\vec{a} \cdot (\vec{b} + \vec{c}) \neq \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}##, assuming that the vectors here are of the same dimension.
 
amy098yay said:

Homework Statement


Verify using an example that a(b+c) is not equal to ab+ac. (This means that addition does not distribute over the dot product.)

Homework Equations


a(b+c)= ab+ac is ALWAYS true for whatever numbers you substitute

let a = 16
b = 3
and c = 5

The Attempt at a Solution



so LHE = a/(a + b) = 16 / (3 + 5) = 16/8 = 2
and RHE = (a/b) + (a/c) = (16/3) + (16/5) ≠ 2
so a/(a + b) ≠ (a/b) + (a/c)
LHE = a + (bc) = 16 + 3 x 5 = 16 + 15 = 31
RHE = (a + b)(a + c) = (16 + 3)(16 + 5) = 19 x 21 ≠ 31
so a + (bc) ≠ (a + b)(a + c)[/B]

What you have been asked to do is impossible; for either dot or cross products, the product DOES distributes over addition. What you tried to do in the above was to use division instead of multiplication---and it is, indeed true that usually ##a/(b+c) \neq (a/b) + (a/c)## (although it can hold in some special cases).

Who gave you this question to do? I hope it was not your instructor!
 
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