Question

Define A = {a | a = n^2 for some n ∈ Z}
and B = {b | b = m^6 for some m ∈ Z}.
Prove that B ⊆ A.

Answer 1

@hartnn

Answer 2

I get that for n^2 and m^6 since m is in a higher power it will have the same even values that n will eventually have, but overall less values

Answer 3

@Preetha

Answer 4

@UnkleRhaukus

Answer 5

do you just have to show that
a^3 = b
n^3 = a
?

Answer 6

I guess that makes sense

Answer 7

why n^3=a ?

Answer 8

maybe better, is to say that any b can be expressed in terms of condition for a. I mean
b=m^6=(k^3)^2
where k^3 is some n∈ Z

Answer 9

an I think that's it

Answer 10

*ah typo,

Answer 11

and*

Answer 12

Thanks Myko

Answer 13

to prove that \[B\subset A\]
show that every \(b\in B\) implies \(b\in A\)

let \(b\in B\)

\(b=m^6\) for some \(m\in Z\)
\[b=m^6=(m^3)^2\]
put \(n=m^3\)

\(b=m^6=n^2\in A\)