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.)
My solution: is this correct?
a(b+c)= ab+ac is ALWAYS true for whatever numbers you substitute
let a = 16
b = 3
and c = 5
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)
http://www.cliffsnotes.com/math/algebra/linear-algebra/matrix-algebra/operations-with-matrices
Your solution is not quite correct. Let me explain why.
To verify that a(b+c) is not equal to ab+ac, we need to show a counterexample, which means finding specific values for a, b, and c that make the statement false.
Let's take a = 2, b = 3, and c = 4.
Using these values, we have:
LHS = a(b+c) = 2(3+4) = 2(7) = 14.
RHS = ab + ac = (2*3) + (2*4) = 6 + 8 = 14.
In this case, the LHS is equal to the RHS, which means that a(b+c) is equal to ab+ac.
To show that a(b+c) is not always equal to ab+ac, we need to find another example where they are not equal.
Let's take a = 3, b = 2, and c = 5.
Using these values, we have:
LHS = a(b+c) = 3(2+5) = 3(7) = 21.
RHS = ab + ac = (3*2) + (3*5) = 6 + 15 = 21.
In this case, the LHS is equal to the RHS again, which means that a(b+c) is equal to ab+ac.
Based on these counterexamples, it appears that a(b+c) is indeed equal to ab+ac in all cases, indicating that addition does distribute over the dot product.