={∈ | 4|^2 }, ={∈ | 2| }. Prove =.

Question

Show (a) ⊆; Show (b) ⊆.

a) Show A⊆B
Let y ∈ A, Since 4 | y^2,
y^2 = 4m for some integer m
y = 2sqrt(m), so 2 | y,
which implies y ∈ B

b) Show B⊆A Let x ∈ B,
Since 2 | x,
x = 2k for some integer k
x^2 = (2k)^2 = 4k^2 = 4(k^2), so 4 | x
which implies x ∈ A

Hence, A = B.

Is my proof correct? Any help will be appreciated. Thank you.

More generally: Show $,a^n \mid b^n\Rightarrow a \mid b,,$ i.e. $,r\in \Bbb Q,,r^n\in\Bbb Z\Rightarrow r\in\Bbb Z$ — Anne Bauval, Sep 27 '23 at 06:51

score 0 · Answer 1 · answered Sep 27 '23 at 06:41

0

No, it is not correct. You can't deduce from $y=2\sqrt m$ that $2\mid y$, since $\sqrt m$ could be a non-integer.

If $4\mid y^2$, then $y$ is an even integer, because, if $y$ was odd, then $y=2k+1$ for some integer $k$, and then $y^2=4k^2+4k+1$, in which case $4\nmid y^2$.

answered Sep 27 '23 at 06:41

José Carlos Santos

427,504

From 4 | y^2, y^2 = 4m for some integer m, can I deduce that since every number in B is even, and y is even, x ∈ A and x ∈ B? – quik214 Sep 27 '23 at 06:48
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What is $x{}{}$? – José Carlos Santos Sep 27 '23 at 07:55

={∈ | 4|^2 }, ={∈ | 2| }. Prove =.

1 Answers1