How the hell does one go about proving x <= x ???

[Hoeboy]

How can one prove x<=x (x is less than or equal to x)???? I can see how to prove x to be equal to x because if x-x=0, then they're equal. But how can something be less than itself?

 

[Cerebus451]

If x is equal to x, then x is also less than or equal to x. <= implies an or condition, not an and condition.

 

[gopunk]

what he said. the compound proposition of (T OR p) where p is any proposition, will *always* be true

 

[Mr.IncrediblyBored]

Cerebus is correct. You can break out the single equation into 2 separate equations:

x <= x

Becomes:

x < x OR x = x

Since we know:

x - x = 0

We can assume:

x <= x

 

[Chaotic42]

Originally posted by: gopunk
what he said. the compound proposition of (T OR p) where p is any proposition, will *always* be true

Someone has been spending too much time in advanced math classes...

 

[gopunk]

Originally posted by: Chaotic42

Originally posted by: gopunk
what he said. the compound proposition of (T OR p) where p is any proposition, will *always* be true

Someone has been spending too much time in advanced math classes...

*rubs eyes*

yup, spent all night doing hw. ok, well spent up to midnight out with friends, but the rest of it was doing hw

 

[Hoeboy]

What I don't get is that for every real # x, it will ALWAYS be equal to itself. I would understand if the equation was something like y<=x and there are two different variables. In that case, y is either less than x, OR equal to x. But how can a number be EITHER less than, OR equal to itself? I would assume It would ALWAYS have to be equal to itself.

 

[gopunk]

Originally posted by: Hoeboy
What I don't get is that for every real # x, it will ALWAYS be equal to itself. I would understand if the equation was something like y<=x and there are two different variables. In that case, y is either less than x, OR equal to x. But how can a number be EITHER less than, OR equal to itself? I would assume It would ALWAYS have to be equal to itself.

yes, but it is an OR statement! you only need one of the two propositions to always be true in order for it to be a tautology. so if you know that one of them is always true (x=x), then it doesn't matter what they put on the other side!

http://www.cs.washington.edu/education/courses/cse321/CurrentQtr/slides/logic.pdf

look at the 8th slide, under "Domination Laws"

 

[Hoeboy]

So I would basically just have to say:


x-x=0

Since x-x=0, then x=x. We can then assume that since x=x, then x<=x by definition.



Looks like a pretty short proof

 

[gopunk]

this is how i would do it... i guess it depends on what your instructor wants

x<=x
x < x \/ x = x
x < x \/ T
T (by domination law)



but yea, it is pretty short. x=x... what more can we say?

 

[Hoeboy]

well i don't think i can start out w/ x<=x since that is what we're trying to prove. before i started out with x=x which i don't think i can do either. so this is what i got:


x+ (-x) = 0
x - x = 0

if x-x = 0, then x=x. we can then safely assume x <= x by definition.

 

[gopunk]

Originally posted by: Hoeboy
well i don't think i can start out w/ x<=x since that is what we're trying to prove. before i started out with x=x which i don't think i can do either. so this is what i got:


x+ (-x) = 0
x - x = 0

if x-x = 0, then x=x. we can then safely assume x <= x by definition.

no, you can start out with x<=x because you are making no assumption about whether or not it is true. you just start out with that statement, and keep on finding logically equivalent statements until you conclude that x<=x is logically equivalent to TRUE.

well this is my belief anyways... i could be wrong...

 

[CSMOOTH]

I am going to assume that you are in a foundations of math class or some sort of real analysis. The question is what propositions have you proven already? Here is a proof by contradiction with very little assumed.

x <= x is proved as follows:
assume the opposite is true and show there is a contradiction
x not <= x
if x is not <= x this means that x > x ( this is because of the exclusive nature of the <= operation)
x>x implies that x + n = x for some n
add the inverse of x to both sides:
-x + x + n = -x + x
-x + x = 0 so you have 0 + n = 0
which yields n = 0
this is clearly a contradiction so x must be less than or equal to x.
From here you would prove the opposite arguement (x >= x) and the union of the two statements yields that x = x.
Christopher Clark

 

[lilFajita]

Originally posted by: CSMOOTH
I am going to assume that you are in a foundations of math class or some sort of real analysis. The question is what propositions have you proven already? Here is a proof by contradiction with very little assumed.

x <= x is proved as follows:
assume the opposite is true and show there is a contradiction
x not <= x
if x is not <= x this means that x > x ( this is because of the exclusive nature of the <= operation)
x>x implies that x + n = x for some n
add the inverse of x to both sides:
-x + x + n = -x + x
-x + x = 0 so you have 0 + n = 0
which yields n = 0
this is clearly a contradiction so x must be less than or equal to x.
From here you would prove the opposite arguement (x >= x) and the union of the two statements yields that x = x.
Christopher Clark

I don't know if it is neccessary, but I would change
x>x implies that x + n = x for some n
to say
x>x implies that x + n = x, where |n| > 0 (absolute value of n is bigger than zero)

 

[JayHu]

Originally posted by: LilFajita

Originally posted by: CSMOOTH
I am going to assume that you are in a foundations of math class or some sort of real analysis. The question is what propositions have you proven already? Here is a proof by contradiction with very little assumed.

x <= x is proved as follows:
assume the opposite is true and show there is a contradiction
x not <= x
if x is not <= x this means that x > x ( this is because of the exclusive nature of the <= operation)
x>x implies that x + n = x for some n
add the inverse of x to both sides:
-x + x + n = -x + x
-x + x = 0 so you have 0 + n = 0
which yields n = 0
this is clearly a contradiction so x must be less than or equal to x.
From here you would prove the opposite arguement (x >= x) and the union of the two statements yields that x = x.
Christopher Clark

I don't know if it is neccessary, but I would change
x>x implies that x + n = x for some n
to say
x>x implies that x + n = x, where |n| > 0 (absolute value of n is bigger than zero)

|n|>0 is definitely something that would be required.
And if it's length you wanted from a proof, there you go!

 

[CSMOOTH]

yes...yes... I know.... it was an outline not the complete proof and I doubt that I included the syntax to be totally rigerous..... But that is how you do it....

 

[ndee]

That's why I hate math. How can 1cm be smaller than 1cm? Math is not logical. Period.

 

[CSMOOTH]

No no you are reading too much into it. This the first part of the proof that x = x in real analysis. You need to use your foundation principles prove x <= x and then x>=x and the 2 together mean that x = x. All that x <= x says that x is less than or *equal* to x. Though this is true for our real number system there are a number of cases you can construct in other systems that don't hold to x = x.