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Relation R symmetry for x= +- y

  • #1
Hi. I'm working through one of my first problems on sets and relations, and I need some help understanding if I'm getting this right. Any help/suggestions on my through process is greatly appreciated.

The question is:
Determine whether the relation R on the set of all real numbers is reflexive, symmetric, anti-symmetric, and/or transitive, where (x,y) ∈ R if and only if
(a) x+y=0
(b) x = ± y
(c) x – y is a rational number (d) x=2y
(e) xy≥0

Here are my answers, and my reasons:
a)
reflexive: true (if x = 0, then x+x = 0)
symmetric: true (if xRy = 0 then yRx = 0)
antisymmetric: false (it is symmetric, so it cannot be antisymmetric)
transitive: true (if xRy = 0 and yRz = 0 then z = x)

b) x = +- y
reflexive: true (xRx can equal positive or negative x)
symmetric: true (if x is positive or negative y, then y can be positive or negative x)
antisymmetric: false (even though xRy and yRx can be true, it doesn't mean x=y)
transitive: true (if x is positive or negative y, and y is positive or negative z, then x is positive or negative z)

c) x-y is rational
reflexive: true (x-x=0, and 0 is rational)
symmetric: false ( just because x-y is rational doesn't mean that y-x will be )
antisymmetric: false ( if x-y is rational, and y-x is rational, y != x in all cases )
transitive: true (if x-y is rational, and y-z is rational, then x-z will be rational )

d) x = 2y
reflexive: false ( x does not equal 2x )
symmetric: false ( if x = 2y, y != 2x )
antisymmetric: true ( if x = 2y and y = 2x then y = x )
transitive: false ( if x = 2y and y = 2z then x != 2z )

e) xy >= 0
reflexive: true ( some number times itself will be >= 0 )
symmetric: true ( if x * y >= 0 then y * x >= 0 )
antisymmetric: false ( it is symmetric )
transitive: true ( if x * y >= 0 and y * z >= 0 then x*z >= 0 )

Thanks for any guidance!
 

Answers and Replies

  • #2
CompuChip
Science Advisor
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(a) If x = 0 then xRx. But you want to check that xRx for all x.

For (c), I don't agree with your argument that if x - y is rational, y - x is not necessarily. As far as I know, if q is rational then so is -q. Maybe you can also explain the transitive reasoning more: why does x-z being rational follow from x-y and y-z being rational?

(d) Note that x = 2y and y = 2x actually implies x = y = 0.

(e) Again, I think the transitivity could use some more explanation. It is in fact true, but not trivial - why can't x be positive and z be negative, for example?

Apart from the two small mistakes pointed out above, you got them all right. Well done!
 
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  • #3
HallsofIvy
Science Advisor
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Hi. I'm working through one of my first problems on sets and relations, and I need some help understanding if I'm getting this right. Any help/suggestions on my through process is greatly appreciated.

The question is:
Determine whether the relation R on the set of all real numbers is reflexive, symmetric, anti-symmetric, and/or transitive, where (x,y) ∈ R if and only if
(a) x+y=0
(b) x = ± y
(c) x – y is a rational number (d) x=2y
(e) xy≥0

Here are my answers, and my reasons:
a)
reflexive: true (if x = 0, then x+x = 0)
No. "Reflexive" does NOT say "for some x, xRx. It says for all x, xRx. And it is certainly NOT true that 1+ 1= 0.

symmetric: true (if xRy = 0 then yRx = 0)
You should show this: if x+y= 0 then y+ x= 0. Do you see why I have used the "commutive property" of addition to swap x and y?

antisymmetric: false (it is symmetric, so it cannot be antisymmetric)
transitive: true (if xRy = 0 and yRz = 0 then z = x)
You seem to be very confused about what "xRz" means and what you want to show. If xRy (NOT "xRy= 0") then x+y= 0. If yRz then y+ z= 0. Yes, it follows that "z= x" but that does NOT directly show that "xRz" is not true. Better is "1R(-1) because 1+(-1)= 0. (-1)R1 because -1+ 1= 0. But "1R1" is NOT true because 1+ 1= 2, not 0."

b) x = +- y
reflexive: true (xRx can equal positive or negative x)
symmetric: true (if x is positive or negative y, then y can be positive or negative x)
antisymmetric: false (even though xRy and yRx can be true, it doesn't mean x=y)
transitive: true (if x is positive or negative y, and y is positive or negative z, then x is positive or negative z)

c) x-y is rational
reflexive: true (x-x=0, and 0 is rational)
symmetric: false ( just because x-y is rational doesn't mean that y-x will be )
WHAT??? You had better rethink this! y- x= -(x- y).

antisymmetric: false ( if x-y is rational, and y-x is rational, y != x in all cases )
transitive: true (if x-y is rational, and y-z is rational, then x-z will be rational )
Again, you should show this: x- z= (x- y)+ (y- z) and the set of rational numbers is closed under addition.

d) x = 2y
reflexive: false ( x does not equal 2x )
Not for all. It is true for x= 0. But x= 1 is a sufficient counter example.

symmetric: false ( if x = 2y, y != 2x )
Again that is not always true. However, x= 4, y= 2 is a counter-example and that is sufficient.

antisymmetric: true ( if x = 2y and y = 2x then y = x )
transitive: false ( if x = 2y and y = 2z then x != 2z )
You keep making statements that mean "this is never true" when you mean to say "there are cases when this is not true". It is sufficient to point out x= 8, y= 4, z= 2 satisfy xRy and yRz but not xRz/

e) xy >= 0
reflexive: true ( some number times itself will be >= 0 )
symmetric: true ( if x * y >= 0 then y * x >= 0 )
antisymmetric: false ( it is symmetric )
transitive: true ( if x * y >= 0 and y * z >= 0 then x*z >= 0 )

Thanks for any guidance!
Notice that (e) is the same as "x and y have the same sign", (c) is the same as saying "either x and y are both rational or they are both irrational". Every "equivalence relation", xRy, is saying that x and y "are the same" in some way.
 
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  • #4
You guys are all awesome. This site is a huge help. Thanks everybody!

Here's what I've learned.

a,reflexive
Yes, for all x got it!

a,symmetric
So just showing x+y=0 then y+x=0 is "showing" the symmetry?

a,transitive
I did sort of take a shortcut by saying the relation was equal to something. I see that the transitive property is false. I should probably slow down, wait a few hours, then come back and review my work rather than just throwing the pencil down and shouting "done"!

c
I completely forgot that a negative number is rational. I haven't picked this stuff up in years. Realizing that this is the case, c is completely wrong.
reflexive: true (x-x=0, and 0 is rational)
symmetric: TRUE!
transitive: true

** Saying that the set of rational numbers is closed under addition (or subtraction,multiplication,and division), does this mean that either of the 4 operations, when applied to x and y, will result in a rational number?

d,reflexive
So, false is correct, buy my reasoning is unsound because I did not show the counter example with x=1 (or some other counter example). Got it.

d,transitive
The sensitivity of using English to describe mathematics is not my strong suite. I will work on this.

Thanks a lot for all the help. I'm going to do these from scratch again, and work on my reasoning skills!
 
  • #5
CompuChip
Science Advisor
Homework Helper
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You guys are all awesome. This site is a huge help. Thanks everybody!
You're welcome. I wish everyone I replied to was this enthusiastic :-)

The sensitivity of using English to describe mathematics is not my strong suite. I will work on this.

Thanks a lot for all the help. I'm going to do these from scratch again, and work on my reasoning skills!
Unfortunately that is a big part of the job. Just scribbling down some symbols is not very useful, it is the text in between that makes it a coherent story. Explaining why something is true - either formally or on an intuitional level - or how you get from one formula to another, is really the bread and butter of mathematics. I don't know about you, but English is not my native language, which makes it all the more important that you express yourself unambiguously.
 
  • #6
It looks like the transitivity of e is in fact false. If x=1, y=0, z=-1 then

1*0 <= 0
0*(-1) <= 0
but
1*(-1) < 0
 

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