Need help coming up with a proof for the following?

Question

I need help writing proofs for the following using common Axioms etc.

1) If a < b and c < d, then a + c < b + d

2) If a2= b2 and a, b >= 0, then a = b

some Axioms given to me are:
AC - Commutativity of Addition
AA - Associativity of Addition
AID- Existence of Additive Identity
AIV- Existence of Additive Inverse
MC- Commutativity of Multiplication
MA- Associativity of Multiplication
MID- Existence of Multiplicative Identity
MIV- Existence of Multiplicative Inverse
D- Distributivity
OTC - Trichotomy
OTR - Transitivty
OA - Compatibility with Addition
OM - Compatibility with Multiplication

Answer 1

I forget the technical details of the axioms, but here's how I'd approach writing the proofs. I'm afraid you'll have to figure out which axioms apply. Hopefully this is enough to get you started.


1.
Since a is less than b, (b-a) > 0.
Since c is less than d, (d-c) > 0.

Since (b-a) and (d-c) are both greater than zero, (b-a) + (d-c) > 0.
Then you just add a and c to both sides of the inequality.

2.
Do a2 and b2 mean 2*a and 2*b? If not, then this problem doesn't make any sense to me. If they do mean that, I'd approach the problem this way:

Suppose a is not equal to b.
Then a*2 is not equal to b*2
Therefore, if a*2 = b*2, a = b.

Answer 2

For your first question:

a a-b<0
c d-c>0

So we can consider:
a-b < d-c

And by adding b+c to both sides we have:
a+c < b+d

And about your second question:

Not that a2=b2 means in two different forms:
1) a*2= b*2; Which my last one answered perfectly.
2) a^2=b^2 (a,b>0)
In this case you can have this:

a = |b|

In this you can see b=a or b=-a and because of b is positive only we have b=a.

Be Succeed.

Answer 3

1)
a < b => a + c < b + c
b+c < b+d
thus a+c < b+d

Answer 4

um, yeah. Good luck with that.

Answer 5

one...two...three....fourfivesixseveneightnineten....BYE!