Proving an equivalence relation

[shelly89]

[A relation is defined on the set of all rational numbers as follows,

a~b means that there is a rational number k >0 , such that \(\displaystyle a = b^{k} \)

prove that ~ is an equivalence relation. on the set of all natural numbers N.

Okay so I have to show it is reflective, symmetric and transitive.

1) if a~a than [ tex] a = a^{k} [/tex] \(\displaystyle a = a^{1} \) thus a is reflective,

2) not sure about this one but

a~b than \(\displaystyle a=b^{k} \) than b~a [ tex] b = a^{1/k} [/tex]

1/k is in the set of positive rationals so a~b is symmetric.


3) a~b than \(\displaystyle a = b^{k} \)

b~c \(\displaystyle b = c^{k} \)

how do i prove this is transitive? And I also are the aboce correct?

thank you very much.

 

[Romsek]

[A relation is defined on the set of all rational numbers as follows,

a~b means that there is a rational number k >0 , such that \(\displaystyle a = b^{k} \)

prove that ~ is an equivalence relation. on the set of all natural numbers N.

Okay so I have to show it is reflective, symmetric and transitive.

1) if a~a than [ tex] a = a^{k} [/tex] \(\displaystyle a = a^{1} \) thus a is reflective,

2) not sure about this one but

a~b than \(\displaystyle a=b^{k} \) than b~a [ tex] b = a^{1/k} [/tex]

1/k is in the set of positive rationals so a~b is symmetric.


3) a~b than \(\displaystyle a = b^{k} \)

b~c \(\displaystyle b = c^{k} \)

how do i prove this is transitive? And I also are the aboce correct?

thank you very much.

a) is fine, but the term is reflexive, not reflective
b) is fine
c) is pretty simple, I'll post it below so you can try it again w/o looking. It's not any harder than a or b, scroll down if you still need help.




























a~b, b~c, show a~c. a=b^k, b=c^m, a=c^km, km is the product of two positive rationals and is thus a positive rational, q.e.d.

 

[pka]

[A relation is defined on the set of all rational numbers as follows,
a~b means that there is a rational number k >0 , such that \(\displaystyle a = b^{k} \)
prove that ~ is an equivalence relation. on the set of all natural numbers N.
Okay so I have to show it is reflective, symmetric and transitive.
1) if a~a than [ tex] a = a^{k} [/tex] \(\displaystyle a = a^{1} \) thus a is reflective,
2) not sure about this one but
a~b than \(\displaystyle a=b^{k} \) than b~a [ tex] b = a^{1/k} [/tex]
1/k is in the set of positive rationals so a~b is symmetric.


3) a~b than \(\displaystyle a = b^{k} \)

b~c \(\displaystyle b = c^{k} \).

For 3) you must use different variables.

\(\displaystyle a=b^k~\&~b=c^j\)