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I'm trying to do the following problem in my book, but I don't understand how the book got their answer.

The problem:
Determine whether the following relations are equivalence relations:

The relation R on R given by xRy if and only if |x-y| <= 1

The answer only says it isn't transitive and gives this example:
1R2 / 2R3, butn 1 R 3. Where did they get those numbers from?

As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:

Reflexive: For any x such that xRx ---> x <= 1

Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|

• Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.


• Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.

Now transitivity is not verified.

• transitivity means $xRytext{ AND } yRzimplies xRz$
  

So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?

A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$

Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.

Let $xRy$ if and only if $|x-y|le 1$.

Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.

Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.

The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:

$|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.

This is contradicts the requirement of transitivity, which would imply (in particular) that
$$1R2;wedge; 2R3quadRightarrowquad 1R3.$$

Reflexive: For any x such that xRx ---> x <= 1

No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$

Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|

No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$

Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.

Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.

$$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$

So what real numbers could possible make that so?

Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$

An equivalence relation $sim$ satisfies three axioms.

1. 
   Reflexivity. $x sim x$ for all $x$.


2. 
   Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


3. 
   Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.

If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.

To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.