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Relation R symmetry for x= +- y

  1. Oct 10, 2013 #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!
     
  2. jcsd
  3. Oct 10, 2013 #2

    CompuChip

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    Homework Helper

    (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!
     
  4. Oct 10, 2013 #3

    HallsofIvy

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    Staff Emeritus
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    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.

    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?

    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."

    WHAT??? You had better rethink this! y- x= -(x- y).

    Again, you should show this: x- z= (x- y)+ (y- z) and the set of rational numbers is closed under addition.

    Not for all. It is true for x= 0. But x= 1 is a sufficient counter example.

    Again that is not always true. However, x= 4, y= 2 is a counter-example and that is sufficient.

    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/

    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.
     
  5. Oct 10, 2013 #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!
     
  6. Oct 10, 2013 #5

    CompuChip

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    Science Advisor
    Homework Helper

    You're welcome. I wish everyone I replied to was this enthusiastic :-)

    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.
     
  7. Oct 13, 2013 #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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