Mixing
Consider the function $θ:{0,1}×{Bbb N}→{Bbb Z}$ defined as $θ(a,b)=(-1)^{a}b$
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How do I prove the function $theta$ is injective, surjective, and bijective, if any of these?
functions proof-writing proof-explanation
asked Jan 1 at 8:36
Fifi12Fifi12
62
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add a comment |
$begingroup$
How do I prove the function $theta$ is injective, surjective, and bijective, if any of these?
functions proof-writing proof-explanation
asked Jan 1 at 8:36
Fifi12Fifi12
62
$endgroup$
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where have you been stuck?
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– dmtri
Jan 1 at 8:42
add a comment |
$begingroup$
How do I prove the function $theta$ is injective, surjective, and bijective, if any of these?
functions proof-writing proof-explanation
asked Jan 1 at 8:36
Fifi12Fifi12
62
$endgroup$
How do I prove the function $theta$ is injective, surjective, and bijective, if any of these?
functions proof-writing proof-explanation
functions proof-writing proof-explanation
asked Jan 1 at 8:36
Fifi12Fifi12
62
asked Jan 1 at 8:36
Fifi12Fifi12
62
asked Jan 1 at 8:36
Fifi12Fifi12
62
asked Jan 1 at 8:36
Fifi12Fifi12
62
asked Jan 1 at 8:36
Fifi12Fifi12
62
62
$begingroup$
where have you been stuck?
$endgroup$
– dmtri
Jan 1 at 8:42
add a comment |
$begingroup$
where have you been stuck?
$endgroup$
– dmtri
Jan 1 at 8:42
$begingroup$
where have you been stuck?
$endgroup$
– dmtri
Jan 1 at 8:42
$begingroup$
where have you been stuck?
$endgroup$
– dmtri
Jan 1 at 8:42
add a comment |
1 Answer
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$0$ is not in the range so it is not surjective. Can you see that it is injective? Hint: if $theta (a,b)=theta(a',b')$ take absolute value on both sides to prove that $b=b'$ and then prove that $a=a'$.
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
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1 Answer
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1 Answer
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$begingroup$
$0$ is not in the range so it is not surjective. Can you see that it is injective? Hint: if $theta (a,b)=theta(a',b')$ take absolute value on both sides to prove that $b=b'$ and then prove that $a=a'$.
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
add a comment |
$begingroup$
$0$ is not in the range so it is not surjective. Can you see that it is injective? Hint: if $theta (a,b)=theta(a',b')$ take absolute value on both sides to prove that $b=b'$ and then prove that $a=a'$.
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
add a comment |
$begingroup$
$0$ is not in the range so it is not surjective. Can you see that it is injective? Hint: if $theta (a,b)=theta(a',b')$ take absolute value on both sides to prove that $b=b'$ and then prove that $a=a'$.
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
$endgroup$
$0$ is not in the range so it is not surjective. Can you see that it is injective? Hint: if $theta (a,b)=theta(a',b')$ take absolute value on both sides to prove that $b=b'$ and then prove that $a=a'$.
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
answered Jan 1 at 8:48
Kavi Rama MurthyKavi Rama Murthy
52.5k32055
52.5k32055
add a comment |
add a comment |
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where have you been stuck?
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– dmtri
Jan 1 at 8:42