1. Prove: (x+y)+(u+v) = (x+u)+(y+v)
Proof: On the LHS of the equation, we can apply the associative property:
(x+y)+(u+v) = x+(y+(u+v))
Next, apply the commutative property - we can switch the order of any terms in a sum. This gives us:
x+(y+(u+v)) = x+((y+u)+v) (commutative property)
Apply associative property again:
x+((y+u)+v) = (x+(y+u))+v (associative property)
= (x+y+u)+v (associative property)
Thus, we have shown that:
(x+y)+(u+v) = (x+u)+(y+v)
2. Prove that the equation x+3=2 has no solution in the set of natural numbers.
Proof: Suppose our number system N consisted of:
N = {1, 2, 3, 4, 5, ...}
From x+3=2:
x+3=2 = -1+3 (cancellation property)
x+3=-1+3 (cancellation property)
x = -1 (cancellation property)
As the value of x is negative, it does not exist in the set of natural numbers.
Prove that if x+y=2, then x=y=1 is the only possible solution.
Proof: Assume x=1 and y=1
Thus, 1+1 = 1+1' (B)
(1+1) = 1' = (1+0)' (B)
Thus, this proves that when x=1 and y=1, the equation x+y=2 is true.
Prove that if x>y, z>u, then x+z>y+u.
Adding z to both sides won't change the inequality.
x+z > y+z
Adding y to both sides won't change the inequality.
y+z > y+u
Transitivity:
x+z > y+u
Thus, this proves that if x>y, z>u, then x+z>y+u.