- #1

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Ive a question in my textbook and I dont really know what to do!

The question is:

I know I need to prove that the relation is

1)Symmetric

2)Reflexive

3)Transitive

But how do I prove this

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- Thread starter BMath
- Start date

- #1

- 8

- 0

Ive a question in my textbook and I dont really know what to do!

The question is:

I know I need to prove that the relation is

1)Symmetric

2)Reflexive

3)Transitive

But how do I prove this

- #2

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- 25

Do you know what reflexivity, symmetry, and transitivity mean?

- #3

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Reflexive: x+x= 2.x and not 3.x so its not reflexive

Symmetric: x+y = 3k so y+x = 3k

Transitive: xRy and yRz

x+y=3k for keZ

x+y=3n for neZ

So x=3k-y and y=3n-y so x+z=3k-y+3n-y= 3(k+n)-2y We see that its not transitive

Because it hasnt al three properties it isnt an equivalence relationship

- #4

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I dont understand it really I read my textbook over and over but I still cant imagine how it works

- #5

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Even better would be to give a specific counterexample. Notice that if ##x = 3##, then ##x + x = 6 = 3\cdot 2## so the equation holds for some values of ##x##. Find one for which it does not hold.Yes and this is my proof:

Reflexive: x+x= 2.x and not 3.x so its not reflexive

OK. Writing in more detail, you might say something like "if ##xRy## then ##x + y = 3k##. Since addition of integers is commutative, we have ##x + y = y + x##. Therefore ##y + x = 3k##, which means that ##yRx##."Symmetric: x+y = 3k so y+x = 3k

I think you mean ##y + z = 3n## for that last equation. What can you conclude about ##x + z##?Transitive: xRy and yRz

x+y=3k for keZ

x+y=3n for neZ

- #6

Mark44

Mentor

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Some values of x belong to the relation, but others don't. As you note, x + x = 2x, so the only way a number x could belong to the relation is if x itself is a multiple of 3.Yes and this is my proof:

Reflexive: x+x= 2.x and not 3.x so its not reflexive

0R0 = 0 + 0 = 3 * 0

3R3 = 6 = 3 * 2

but 1 doesn't work, nor does 2.

I think you mean y + z = 3n for n ##\in## Z.Symmetric: x+y = 3k so y+x = 3k

Transitive: xRy and yRz

x+y=3k for keZ

x+y=3n for neZ

If y is a multiple of 3, then we do have transitivity. For example, 0R3 because 0 + 3 = 3, and 3R6 because 3 + 6 = 9. From this, we see that 0R9 because 0 + 9 = 9 = 3*3.So x=3k-y and y=3n-y so x+z=3k-y+3n-y= 3(k+n)-2y We see that its not transitive

Because it hasnt al three properties it isnt an equivalence relationship

- #7

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Even better would be to give a specific counterexample. Notice that if ##x = 3##, then ##x + x = 6 = 3\cdot 2## so the equation holds for some values of ##x##. Find one for which it does not hold.

OK. Writing in more detail, you might say something like "if ##xRy## then ##x + y = 3k##. Since addition of integers is commutative, we have ##x + y = y + x##. Therefore ##y + x = 3k##, which means that ##yRx##."

I think you mean ##y + z = 3n## for that last equation. What can you conclude about ##x + z##?

That x+z is not an integer?

- #8

Mark44

Mentor

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No. The context of what both jbunniii and I said was the part where you were investigating transitivity or the relation. You're given that xRy and yRz, so do x and z also belong to the relation?That x+z is not an integer?

- #9

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0R2 and 2R5

0+2=2=3k and 2+5=7=3k we see that there doesnt exist an integer k such that 3k=2 and 3k=7 so

xRz doesnt exists.

- #10

Mark44

Mentor

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But 0 and 2 don't belong to the relation R, so you can't say 0R2. And 2 and 5 don't belong to the relation either, so you can't say 2R5.x,y,z are integers. xRy and yRz

0R2 and 2R5

Whatever numbers x, y, and z you work with, x and y have to belong to the relation, and y and z have to belong as well.

0+2=2=3k and 2+5=7=3k we see that there doesnt exist an integer k such that 3k=2 and 3k=7 so

xRz doesnt exists.

- #11

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but then its always transitive cause i cant find any false relations 3R6 and 6R9

x+z=24=3*8

x+z=24=3*8

- #12

Mark44

Mentor

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That's not a good reason. Possibly you didn't look hard enough.but then its always transitive cause i cant find any false relations

How did you get 24? x = 3 and z = 9, so x + z = 12.3R6 and 6R9

x+z=24=3*8

What about 1R2 and 2R4?

BTW, it's probably better to work with this algebraically rather than to try to guess combinations that work or don't work.

For the record, you have already shown enough to convince me or your instructor that R as defined in this thread is NOT a relation. It is not in general reflexive (e.g. 1 and 1 aren't in the relation, nor are 2 and 2, but 0 and 0 are, as are 3 and 3).

Also, the relation is transitive for some values of x, y, and z, but not for all such values. Most of what we've been doing here is trying to help you get a handle on what this is about.

- #13

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How did you get 24? If x = 3, y = 6, and z = 9, then x + z = 12. [EDIT: as Mark44 already pointed out before I submitted my reply.]but then its always transitive cause i cant find any false relations 3R6 and 6R9

x+z=24=3*8

Also, to prove transitivity, you must show that it is true in general, not for a specific example. Start by assuming that x + y and y + z are both multiples of 3. Then prove that this implies that x + z is also a multiple of 3. If you cannot show this in general, then try to find a counterexample.

- #14

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So if I have x+y 1R2 1+2=3=3.1

and y+z 2R7 2+7=9=3.3

then x+z 1R7 1+7=8=3.k we see that theres no integer k so that 3k=8 thus the relation is not transitive

So if its wrong I only have to give a counterexample? And if its true you need to prove it

- #15

HallsofIvy

Science Advisor

Homework Helper

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Yes, to prove that a general statement is NOT true you only have to give one example in which is does not work- a "counter-example".

2 is NOT related to itself in this way because 2+ 2= 4 is not a multiple of 3. Therefore, it is not "reflexive".

It

4+ 2= 6= 2(3) so 4 and 2 are related. 2+ 1= 3 so 2 and 1 are related. But 4+ 1= 5 which is NOT a multiple of 3 so this is NOT transitive.

- #16

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sumis a multiple of 3.

Yes, to prove that a general statement is NOT true you only have to give one example in which is does not work- a "counter-example".

2 is NOT related to itself in this way because 2+ 2= 4 is not a multiple of 3. Therefore, it is not "reflexive".

Itisobviously symmetric.

4+ 2= 6= 2(3) so 4 and 2 are related. 2+ 1= 3 so 2 and 1 are related. But 4+ 1= 5 which is NOT a multiple of 3 so this is NOT transitive.

Yes thank you very much for the help!

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