Prove by mathematical induction that $n^2 > n$ for all $n geq 2$

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Prove by mathematical induction that $n^2 > n$ for all $n geq 2$.




My attempt:



When $n=2$



LHS , $2^2 =4$



RHS , $2$
When $n=2$, LHS $>$ RHS



Assume true for $n=k$



$k^2>k$



RTP true for $n=k+1$



$(k+1)^2 > k + 1$



Proof



By assumption
$k^2>k$

Thus, $k^2 +2k +1 > k + 2k +1 $

$(k+1)^2 > (k+1) + 2k $

Since $k > 1$

$2k>2 $

Hence $(k+1) + 2k > k+1 $

Therefore $(k+1)^2 > k+1$



Hence by induction the statement is true




inequality induction self-learning formal-proofs




share|cite|improve this question

















  • 1




    Could you please write down the whole proof so that we can tell you whether it is ok or not?
    – Suzet
    Jul 31 at 3:42










  • Okay np I'll hand write it and submit it since it's a bit long
    – user122343
    Jul 31 at 3:43










  • Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
    – Suzet
    Jul 31 at 5:17















up vote
0
down vote

favorite













Prove by mathematical induction that $n^2 > n$ for all $n geq 2$.




My attempt:



When $n=2$



LHS , $2^2 =4$



RHS , $2$
When $n=2$, LHS $>$ RHS



Assume true for $n=k$



$k^2>k$



RTP true for $n=k+1$



$(k+1)^2 > k + 1$



Proof



By assumption
$k^2>k$

Thus, $k^2 +2k +1 > k + 2k +1 $

$(k+1)^2 > (k+1) + 2k $

Since $k > 1$

$2k>2 $

Hence $(k+1) + 2k > k+1 $

Therefore $(k+1)^2 > k+1$



Hence by induction the statement is true




inequality induction self-learning formal-proofs




share|cite|improve this question

















  • 1




    Could you please write down the whole proof so that we can tell you whether it is ok or not?
    – Suzet
    Jul 31 at 3:42










  • Okay np I'll hand write it and submit it since it's a bit long
    – user122343
    Jul 31 at 3:43










  • Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
    – Suzet
    Jul 31 at 5:17













up vote
0
down vote

favorite









up vote
0
down vote

favorite












Prove by mathematical induction that $n^2 > n$ for all $n geq 2$.




My attempt:



When $n=2$



LHS , $2^2 =4$



RHS , $2$
When $n=2$, LHS $>$ RHS



Assume true for $n=k$



$k^2>k$



RTP true for $n=k+1$



$(k+1)^2 > k + 1$



Proof



By assumption
$k^2>k$

Thus, $k^2 +2k +1 > k + 2k +1 $

$(k+1)^2 > (k+1) + 2k $

Since $k > 1$

$2k>2 $

Hence $(k+1) + 2k > k+1 $

Therefore $(k+1)^2 > k+1$



Hence by induction the statement is true




inequality induction self-learning formal-proofs




share|cite|improve this question














Prove by mathematical induction that $n^2 > n$ for all $n geq 2$.




My attempt:



When $n=2$



LHS , $2^2 =4$



RHS , $2$
When $n=2$, LHS $>$ RHS



Assume true for $n=k$



$k^2>k$



RTP true for $n=k+1$



$(k+1)^2 > k + 1$



Proof



By assumption
$k^2>k$

Thus, $k^2 +2k +1 > k + 2k +1 $

$(k+1)^2 > (k+1) + 2k $

Since $k > 1$

$2k>2 $

Hence $(k+1) + 2k > k+1 $

Therefore $(k+1)^2 > k+1$



Hence by induction the statement is true





inequality induction self-learning formal-proofs





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 6:39









Rodrigo de Azevedo

12.6k41751




12.6k41751









asked Jul 31 at 3:40









user122343

445




445







  • 1




    Could you please write down the whole proof so that we can tell you whether it is ok or not?
    – Suzet
    Jul 31 at 3:42










  • Okay np I'll hand write it and submit it since it's a bit long
    – user122343
    Jul 31 at 3:43










  • Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
    – Suzet
    Jul 31 at 5:17













  • 1




    Could you please write down the whole proof so that we can tell you whether it is ok or not?
    – Suzet
    Jul 31 at 3:42










  • Okay np I'll hand write it and submit it since it's a bit long
    – user122343
    Jul 31 at 3:43










  • Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
    – Suzet
    Jul 31 at 5:17








1




1




Could you please write down the whole proof so that we can tell you whether it is ok or not?
– Suzet
Jul 31 at 3:42




Could you please write down the whole proof so that we can tell you whether it is ok or not?
– Suzet
Jul 31 at 3:42












Okay np I'll hand write it and submit it since it's a bit long
– user122343
Jul 31 at 3:43




Okay np I'll hand write it and submit it since it's a bit long
– user122343
Jul 31 at 3:43












Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
– Suzet
Jul 31 at 5:17





Your proof is perfectly correct, there is no problem with it. A more straightforward way to prove the statement directly without using induction simply is by multiplying the inequality $n>1$ by $n$.
– Suzet
Jul 31 at 5:17
















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