# How do I prove something that is blatantly obvious?

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## 11 Replies - 4805 Views - Last Post: 31 March 2010 - 01:05 PM

### #1 Midwest Product

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# How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 02:19 PM

So, this might not be in the right section, but it has nothing to do with programming, and everything to do with computer/conceptual thinking.

I've been looking at some rules for the Cartesian Product, and one is of the distributive property, which is pretty obvious and straightforward. It states (I'm using "n" as intersect, because I cannot find the actual symbol): A x (B n C) = (A x B ) n (A x C)

This is pretty obvious...it is just like basic algebra. You just distribute the "A x" to B and C. So, if it is this easy, how do I prove it is true using a logic sequence? I honestly have no idea.

This post has been edited by Midwest Product: 05 March 2010 - 02:20 PM

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## Replies To: How do I prove something that is blatantly obvious?

### #2 BenignDesign

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## Re: How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 02:23 PM

It's almost quitting time on a Friday, my brain is not equipped to handle logic and/or conceptual thinking at this exact point in time.

### #3 Midwest Product

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## Re: How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 03:08 PM

well, my brain is never in a mood to handle logic. I really don't know how to show the steps...I know that A x B and A x C can each have a "A x" factored out, revealing the left hand side of the equation, but it seems to easy to be correct. Usually proofs have more steps...

### #4 Gloin

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## Re: How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 03:27 PM

You have to consider this,

Let s be an element from the set A and t be an element from the set B∩C.
What you need to prove is that if the pairing (s, t) is an element in Ax(B∩C) then that pairing is an element in both AxB and AxC (or else it cannot be in the intersection (AxB)∩(AxC))
Best way to prove it is to assume the opposite.
If the pairing is not in (AxB)∩(AxC) then that means that t is either not in B or not in C and therefore cannot be in B∩C.

This is not the full proof but hopefully will get you a bit on your way.

### #5 KYA

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## Re: How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 03:28 PM

*Moved to Student Campus so it gets the right kind of attention*

### #6 NeoTifa

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## Re: How do I prove something that is blatantly obvious?

Posted 05 March 2010 - 03:50 PM

Depends on if the x is a x or a multiplication symbol. If it's a multiplication symbol,

A U (B ∩ C)<-> (A U ∩ (A U C)

### #7 evinrows

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## Re: How do I prove something that is blatantly obvious?

Posted 06 March 2010 - 07:50 AM

You just need to prove the distributive property, which is so simple that it's based off of rudimentary math principles, so proving it isn't exactly.. interesting.

### #8 Gloin

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## Re: How do I prove something that is blatantly obvious?

Posted 06 March 2010 - 02:10 PM

evinrows, on 06 March 2010 - 02:50 PM, said:

You just need to prove the distributive property, which is so simple that it's based off of rudimentary math principles, so proving it isn't exactly.. interesting.

If A is a constant, this is rather evident but if A is another set it may not be entirely obvious and definitely not trivial. Either way a proof is reuired and you can't just say that it follows from this or that without backing it up.

### #9 Meister_Chicho

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## Re: How do I prove something that is blatantly obvious?

Posted 31 March 2010 - 10:07 AM

evinrows, on 06 March 2010 - 02:50 PM, said:

You just need to prove the distributive property, which is so simple that it's based off of rudimentary math principles, so proving it isn't exactly.. interesting.

Actually I think it is interesting.

I assume you are starting to study "hard-maths" so in my opinion that's the best way to learn. you will be able to open your mind to maths if you work with demonstrations.

Anyway, the best way to demonstrate something that is "blatantly obvious" in many cases (90%) is "reductio ad absurdum" (i expect that the same term is used in english). "Reductio ad absurdum" means (or what you are supposed to do if you want to demonstrate something this way) that you must consider that what you are demonstrating is false, and so you hypothesize with the opposite idea. If you start calculating things and making steps with that "opposite" hypothesis, you will get into an irrational conclusion (like 1=0), so you demonstrated that way was worng, and so IF THERE WERE ONLY TWO POSSIBILITIES (your hypothesis and the opposite hypothesis) then you have demonstrated too that the first hypothesis is correct.

This post has been edited by Meister_Chicho: 31 March 2010 - 10:09 AM

### #10 Galois

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## Re: How do I prove something that is blatantly obvious?

Posted 31 March 2010 - 12:02 PM

Wait, just because distribution of Cartesian product over the intersection operator looks like a grade school distribution you call it "blatantly obvious" and "easy"? If you were correct in your statements you would be able to prove it at that instant. Would you say that it is obvious that multiplying any number by a zero equals to 0? See how quickly you can come up with a proof. Anyway...

So you have to prove that 2 sets are equal. One way to do it is to show that these sets are subsets of each other. So, lets do the first part.

1. Let element y belong to A x (B n C). That means that y = (s, t) where s belongs to A and t belongs to B n C. That means that t belongs to both B and C. Therefore, (s, t) belongs to both A x B and A x C. This shows that the set A x (B n C) is a subset of [A x B] n (A x C).

Now prove the other way.

2. Let element y belong to [A x B] n (A x C). That means that y belongs to both A x B and A x C. So y = (s, t) where s belongs to A and t belongs to both B and C. In other words, t belongs to B n C. And that means that y belongs to A x (B n C). This proves the second half.

Since both sets are subsets of each other, they must be equal.

This post has been edited by Galois: 31 March 2010 - 12:04 PM

### #11 mojo666

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## Re: How do I prove something that is blatantly obvious?

Posted 31 March 2010 - 12:37 PM

Galois, on 31 March 2010 - 11:02 AM, said:

Wait, just because distribution of Cartesian product over the intersection operator looks like a grade school distribution you call it "blatantly obvious" and "easy"? If you were correct in your statements you would be able to prove it at that instant. Would you say that it is obvious that multiplying any number by a zero equals to 0? See how quickly you can come up with a proof. Anyway...

Not being mathmatically formal or anything, x*y is defined as adding x a total number of y times or
```x_1 + x_2 + ... + x_y
```
. Replace x with 0 and you end up with
```0_1 + 0_2 + ... 0_y
```
which is 0. Replace y with 0 and you get
```
```
which is 0. This is obvious. Also, if A is in Both B and C then it is obvious that A is in B and A is in C.

This post has been edited by mojo666: 31 March 2010 - 12:40 PM

### #12 Galois

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## Re: How do I prove something that is blatantly obvious?

Posted 31 March 2010 - 01:05 PM

mojo666, on 31 March 2010 - 11:37 AM, said:

Not being mathmatically formal or anything, x*y is defined as adding x a total number of y times or
```x_1 + x_2 + ... + x_y
```
. Replace x with 0 and you end up with
```0_1 + 0_2 + ... 0_y
```
which is 0. Replace y with 0 and you get
```
```
which is 0. This is obvious.

Then I guess it's not obvious to you either. What if the numbers are negative? What if they are real? Then your whole argument, which is already based on some informal elementary school definition, falls apart. If this was simple and obvious to you, you would have no problem proving it.

But it's ok though. I guess my other point is that many of us tend to take a lot of math for granted. Some results seem "obvious" until we actually try to understand and prove why they are true. However, they are intuitive, which is what I would describe them.

Quote

Also, if A is in Both B and C then it is obvious that A is in B and A is in C.

That's the definition of an intersection of a set. It's a given. But that's not what OP is trying to prove. He is proving the distributive property of Cartesian products over intersection operator.

This post has been edited by Galois: 31 March 2010 - 01:18 PM