Proving an if-then statement with a contrapositive example [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving statements by its contrapositiveSimple question about proof by contrapositivity.Discrete Mathematics - Understanding Proof by ContrapositiveContrapositive verificationsProof by Contrapositive?Prove by Contrapositive/Contradiction/DirectProve the following statement by proving its contrapositive: if $r$ is irrational, then $ r ^ frac 1 5 $ is irrationalProof by contrapositive, what should I be assuming?Prove that for every integer $x$, if $x^2 - 2x + 7$ is even, then $x$ is odd. (Contrapositive)Proving modular arithmetic using its contrapositive

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Proving an if-then statement with a contrapositive example [closed]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving statements by its contrapositiveSimple question about proof by contrapositivity.Discrete Mathematics - Understanding Proof by ContrapositiveContrapositive verificationsProof by Contrapositive?Prove by Contrapositive/Contradiction/DirectProve the following statement by proving its contrapositive: if $r$ is irrational, then $ r ^ frac 1 5 $ is irrationalProof by contrapositive, what should I be assuming?Prove that for every integer $x$, if $x^2 - 2x + 7$ is even, then $x$ is odd. (Contrapositive)Proving modular arithmetic using its contrapositive










-1












$begingroup$


Show that:



if $k$ is an odd integer greater than 1, then $log_2(k)$ is an irrational number.



I am not quite sure how to prove this statement, would proving it using a contrapositive example be enough?










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ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



closed as unclear what you're asking by Austin Mohr, Leucippus, Cesareo, YiFan, uniquesolution Apr 9 at 7:29


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    What does "contrapositive example" mean?
    $endgroup$
    – Austin Mohr
    Apr 8 at 19:55










  • $begingroup$
    You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
    $endgroup$
    – J. W. Tanner
    Apr 8 at 20:45















-1












$begingroup$


Show that:



if $k$ is an odd integer greater than 1, then $log_2(k)$ is an irrational number.



I am not quite sure how to prove this statement, would proving it using a contrapositive example be enough?










share|cite|improve this question







New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



closed as unclear what you're asking by Austin Mohr, Leucippus, Cesareo, YiFan, uniquesolution Apr 9 at 7:29


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    What does "contrapositive example" mean?
    $endgroup$
    – Austin Mohr
    Apr 8 at 19:55










  • $begingroup$
    You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
    $endgroup$
    – J. W. Tanner
    Apr 8 at 20:45













-1












-1








-1





$begingroup$


Show that:



if $k$ is an odd integer greater than 1, then $log_2(k)$ is an irrational number.



I am not quite sure how to prove this statement, would proving it using a contrapositive example be enough?










share|cite|improve this question







New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Show that:



if $k$ is an odd integer greater than 1, then $log_2(k)$ is an irrational number.



I am not quite sure how to prove this statement, would proving it using a contrapositive example be enough?







discrete-mathematics proof-verification






share|cite|improve this question







New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Apr 8 at 19:34









ph-quiettph-quiett

213




213




New contributor




ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






ph-quiett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




closed as unclear what you're asking by Austin Mohr, Leucippus, Cesareo, YiFan, uniquesolution Apr 9 at 7:29


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Austin Mohr, Leucippus, Cesareo, YiFan, uniquesolution Apr 9 at 7:29


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    What does "contrapositive example" mean?
    $endgroup$
    – Austin Mohr
    Apr 8 at 19:55










  • $begingroup$
    You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
    $endgroup$
    – J. W. Tanner
    Apr 8 at 20:45
















  • $begingroup$
    What does "contrapositive example" mean?
    $endgroup$
    – Austin Mohr
    Apr 8 at 19:55










  • $begingroup$
    You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
    $endgroup$
    – J. W. Tanner
    Apr 8 at 20:45















$begingroup$
What does "contrapositive example" mean?
$endgroup$
– Austin Mohr
Apr 8 at 19:55




$begingroup$
What does "contrapositive example" mean?
$endgroup$
– Austin Mohr
Apr 8 at 19:55












$begingroup$
You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
$endgroup$
– J. W. Tanner
Apr 8 at 20:45




$begingroup$
You mean like if $log_2(k)$ were rational, say $p/q$ with $p,qin mathbb N$, then $2^p=k^q$, which is a contradiction if $k$ is an odd integer greater than $1$?
$endgroup$
– J. W. Tanner
Apr 8 at 20:45










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