proving a function is a bijective
$begingroup$
Is $f(x)=2^x$ a bijection from $f:mathbb{Q} rightarrow mathbb{Q}^+$?
If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So would this then be bijective since the area that would make this not surjective not be accounted for?
functions
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
$endgroup$
add a comment |
$begingroup$
Is $f(x)=2^x$ a bijection from $f:mathbb{Q} rightarrow mathbb{Q}^+$?
If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So would this then be bijective since the area that would make this not surjective not be accounted for?
functions
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
$endgroup$
add a comment |
$begingroup$
Is $f(x)=2^x$ a bijection from $f:mathbb{Q} rightarrow mathbb{Q}^+$?
If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So would this then be bijective since the area that would make this not surjective not be accounted for?
functions
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
$endgroup$
Is $f(x)=2^x$ a bijection from $f:mathbb{Q} rightarrow mathbb{Q}^+$?
If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So would this then be bijective since the area that would make this not surjective not be accounted for?
functions
functions
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
asked Dec 17 '18 at 1:39
GeorgeGeorge
676
676
add a comment |
add a comment |
1 Answer
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I think in terms of being surjective, the kicker here is the choice of $mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=log_2(y)$ always rational?
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
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1 Answer
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1 Answer
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$begingroup$
I think in terms of being surjective, the kicker here is the choice of $mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=log_2(y)$ always rational?
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
$endgroup$
add a comment |
$begingroup$
I think in terms of being surjective, the kicker here is the choice of $mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=log_2(y)$ always rational?
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
$endgroup$
add a comment |
$begingroup$
I think in terms of being surjective, the kicker here is the choice of $mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=log_2(y)$ always rational?
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
$endgroup$
I think in terms of being surjective, the kicker here is the choice of $mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=log_2(y)$ always rational?
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
answered Dec 17 '18 at 1:47
MatthiasMatthias
3287
3287
add a comment |
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