
Very Simple Proofs
I need to prove the following:
a) Let a, b, c, d ∈ Z. Prove that if a < b and c < d then a+c<b+d. b) Suppose that ab > 0. Prove that either a > 0 and b > 0 or a < 0 and b < 0.
These are both obviously true, and yet its so simple to just look at I can't think of how to "prove" it.. Any help?
 Ricky
 Moderator
Re: Very Simple Proofs
Yea, don't you just hate that? Just start out with your assumptions and try to play around. Play around with the letters, keeping in mind what you want.
Assume a < b and c < d. Since a < b, We want something with a + c. So take a < b, and add c to both sides. a + c < b + c. But we also know that c < d. So c + b < d + b. Thus, a + c < b + c < b + d. And there you have it!
For the second one, how much do you believe about multiplication? Do you believe a * b = ab? Do you believe ab where a and b are positive is positive?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 darthradius
 Full Member
Re: Very Simple Proofs
You can do it in cases...
Let a,b∈ Z and suppose that: case 1: a<0 and b>0... Then a=(1)g ( g=abs(a) ) Then ab = (1)gb = (1g)b = (gb) [negative]
case 2: a<0, b<0 Then a=(1)g and b=(1)h ( g=abs(a) h=abs(b) ) Then ab=(1)g(1)h=(1)(1)gh=(1)gh=gh [positive]
case 3: a>0, b>0 then ab>0 [obviously positive]
Last edited by darthradius (20060331 17:49:58)
The greatest challenge to any thinker is stating the problem in a way that will allow a solution. Bertrand Russell
 Ricky
 Moderator
Re: Very Simple Proofs
Then ab = (1)gb = (1g)b = (gb)
You assume that (a)(b) = (ab). Prove it.
You assume that a negative times a negative is a positive. Prove it.
ab>0 [obviously positive]
Same. Positive times a positive is positive.
This is just what I meant by believing in multiplication. I agree with all your proofs, but I think I can prove the first two assumptions you make. Not so sure about the third one.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 Ricky
 Moderator
Re: Very Simple Proofs
Well, I'll just go ahead and get the first two out of the way.
These proofs assume two things. First, that multiplication distributes over addition. Second, that if a = b, then ac = bc and a + c = b + c.
Just a note before it gets started. a is just the symbol for "The inverse of a". It's how we choose to write inverses. The definition of an inverse is a + a = 0 and a + a = 0. Just making clear that this is not an assumption, but rather a definition.
This will be a proof of three things:
a0 = 0a = 0 (a)b = (ab) (a)(b) = ab  Show that a0 = 0a = 0
0 + 0 = 0. This is because 0 is the additive identity.
Applying a to the left side of each part: a(0 + 0) = a0 Distribute: a0 + a0 = a0 Apply a0 to each side: a0 + a0 + (a0) = a0 + (a0) Then since a0 + (a0) = 0 because (a0) is the inverse of a0, a0 = 0
As you can see, we can do the same thing by applying a to the right side, and come up with 0a = 0.
So that shows a0 = 0a = 0. This will be important in the future.  Show that (a)b = (ab)
Consider: (a)b + ab By distribution: b(a + a) a + a = 0, so: b(0) By the first proof: 0
So (a)b + ab = 0 Applying (ab) to each side: (a)b + ab + (ab) = (ab) But ab + (ab) = 0: (a)b = (ab)
And we're done.  Show that (a)(b) = ab
Consider (a)(b)  (ab) As shown above, (ab) = (a)b, so: (a)(b) + (a)b Distribution: a(b + b) b + b = 0: a(0) By the first proof: 0
So (a)(b)  (ab) = 0 Apply (ab) to both sides: (a)(b)  (ab) + (ab) = (ab) But (ab) + (ab) = 0, so (a)(b) = (ab)
And we're done. 
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 darthradius
 Full Member
Re: Very Simple Proofs
I think that by stating that a,b∈ Z, we can know for certain that (a)(b) = (ab), and (1)(1)=1 since we have that 1 is the multiplicative identity and we know that no integer squared has a negative value...
The greatest challenge to any thinker is stating the problem in a way that will allow a solution. Bertrand Russell
 Ricky
 Moderator
Re: Very Simple Proofs
I think that by stating that a,b∈ Z, we can know for certain that (a)(b) = (ab)
But why can we know for certain? What makes you think that?
(1)(1)=1 since we have that 1 is the multiplicative identity and we know that no integer squared has a negative value...
And why does no integer squared have a negative value?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 darthradius
 Full Member
Re: Very Simple Proofs
I suppose I shouldn't have said that... But since I did...Since we are going to the trouble of stating that a,b are integers and since we know that the set of integers are an integral domain, having all of the properties of an integral domain, AND since we know that by definition, sqrt(1)=i is an element of the complex numbers and NOT an integer...
I dunno...I guess I might have assumed too much... I was just thinking that since he bothered to ask the question by mentioning that a,b were integers, we ought to be able to use that fact in order to define the numbers and their properties, right?
The greatest challenge to any thinker is stating the problem in a way that will allow a solution. Bertrand Russell
 Ricky
 Moderator
Re: Very Simple Proofs
Agreed. But any property of integers we are going to use we should be able to prove, no?
Just to be clear, I'm not certain what exactly must be assumed, what must be shown, what must be proved or defined. But my basic instinct tells me: If you can prove, why the heck not?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: Very Simple Proofs
Whenever it cannot be proved, it should be defined.
I agreeee...
X'(yXβ)=0
 Ricky
 Moderator
Re: Very Simple Proofs
Hah, good one.
Whenever it can't be proved, there is a reason it can't be, and that can be proved.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
