Reasonings that explain why the following are true.?

Question

Using the six basic laws of arithmetic (associative, commutative, distributive) one at a time, explain why each real number of the following is true:

(a) for all real numbers x, y, z, and w, (x+y) + (z+w) = (w+x) + (z+y)

(b) for all real numbers x and y, (x+1) 6 + y * (x+1) = (y+6) * ( x+1)

(c) (x+4) * 6 + y * x + (x+4) + 4 * y = (x+4) * (y+7) for all x and y

(d) (x*y + v*x) + (v*y + x^2) = (x+y) * (x+y) for all x, y, and v

(e) for all real numbers x, y, a, & b, (x+y)(a+b) = xa + xb + ya + yb


Where the *, is times (multiplication)
If someone could write these out and explain why they are true, it would be greatly appreciated!

Answer 1

I wrote them out, but you didn't come to my house to look at them. I'll type them out.

(a) for all real numbers x, y, z, and w, (x+y) + (z+w) = (w+x) + (z+y)

(b) for all real numbers x and y, (x+1) 6 + y * (x+1) = (y+6) * ( x+1)

(c) (x+4) * 6 + y * x + (x+4) + 4 * y = (x+4) * (y+7) for all x and y

(d) (x*y + v*x) + (v*y + x^2) = (x+y) * (x+y) for all x, y, and v

(e) for all real numbers x, y, a, & b, (x+y)(a+b) = xa + xb + ya + yb


They are true because it is the law.

Answer 2

Lets take the first one:
(x+y)+(z+w) dropping parenthesis gives
x+y+z+w by commutativity becomes
x+y+w+z by commutativity again becomes
x+w+y+z by commutativity twice more becomes
w+x+z+y and grouping gives
(w+x)+(z+y)

The others are done in a similar fashion. It's tedious, but it's good practice ☺


Doug