Is the relation $P$, for all real numbers $x$ and $y$ that satisfy $xPy$ iff $x^3 - y \ge y^3 - x$, a reflexive, symmetric and transitive relation?

Image of an exam question I am revising link: [1]

For (i) I have stated the relation is reflexive as $\forall x ∈ \Bbb R, xPx$ is reflexive as $x^3 \ge x$

For (ii) I have stated that the relation is not symmetric as $\forall x,y ∈ \Bbb R, xRy$ does not equal $yRx$, as for the values $x=2, y=1, x^3 - y \ge y^3 - x$ does not equal to $y^3 - x \ge x^3 - y$

For (iii) I have stated that the relation is transitive as as $xPy$ and $yPz$ imply $xPz \forall x,y,z ∈ P,$ as $\forall x,y,z$ that are elements of $R$ and that satisfy $P$, $(a,b)$ $a$ must always be greater than $b$, therefore $\forall x,y ∈ P x\ge y, \forall y,z ∈ P y\ge z,$ therefore $x\ge z$

The last one is a tad all over the place, but I just wanted to know if I am on the right track and if my answers are correct, if not I would love to know why so I can improve myself and see where I went wrong as to not make the same mistake in the future. Any and all help would be greatly appreciated.

• Alos, $x^3\ge x$ need not be true (and has nothing to do with reflexivity). – Dietrich Burde Feb 2 '16 at 16:42
• Okay everyone, it seems that OP has miscopied the question from the link, in which the relation is defined by $x^3-y\geq y^3-x$. I'm not editing the question just in case this was an intention of OP to change it to $=$ (though I doubt it is). – Wojowu Feb 2 '16 at 16:46
• is the relation not the same as the one I've copied it from the link? – roughosing Feb 2 '16 at 16:52
• I've editted your question to make it a bit more readable. Please take a look at my edits to understand how to use mathjax for following questions. – YoTengoUnLCD Feb 2 '16 at 16:56

Your resolution is unclear and incorrect in some places.

For reflexivity, that's not what you have to prove (which is also wrong! as, if $x=0.5$ then $x^3\not\ge x$), you have to show that $x^3-x\ge x^3 -x$ which is obviously true (as the expressions are equal).

Also, saying "$xRy$ does not equal $yRx$" doesn't make much sense, you should use a word like equivalent to describe that two propositions imply the same things.

Here's perhaps an easier resolution.

Note that

$$xPy\iff x^3-y\ge y^3-x\iff x^3+x\ge y^3+y.$$

From that it's clear that the relation is reflexive (explained above).

Now, Let $x=1,y=0$, we have that $2=x^3+x\ge y^3+y=0$, and so $xPy$, but $yPx$ is not true (why?), thus $P$ is not symmetric (in fact, it's asymmetric!).

Now, for transitivity, suppose $xPy$ and $yPz$, this says that $x^3+x\ge y^3+y$ and $y^3+y\ge z^3+z$, can you follow?

• thanks for this comment, it's cleared up a lot, this is the first time I've attempted a question of this format so it was quite difficult to understand, but this has helped a lot thank you, although why do you state that is symmetric then state that it is not symmetric? – roughosing Feb 2 '16 at 17:01
• @roughosing Oops I messed up the words "reflexive" and "symmetric". Fixed now. – YoTengoUnLCD Feb 2 '16 at 17:01

An easier way to deal with this relation is to define $g(x)=x^3+x$ and say that $xPy\iff g(x)\ge g(y)$. Which immediately shows that $P$ is indeed transitive but not symmetric.

• It is $\ge$, not $=$, so it is not symmetric. – Ross Millikan Feb 2 '16 at 16:59
• There was an error in the title, the relation is with "$\ge$" instead of "$=$". – YoTengoUnLCD Feb 2 '16 at 16:59
• Edited to fit the current version.... – Jack's wasted life Feb 2 '16 at 17:04