University:
Gonzaga University
Course:
Math 413 | Real Analysis I
Academic year:

2021
Views:

316

Pages:

2
Author:

Audrey F.

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.

Next, Apply the Commutative Property We Can

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