Proving $a_n>2$ where $a_k=sqrt{2a_{k-1}}$












0












$begingroup$



Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.




$a_1 = 3$.



Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.






discrete-mathematics







share|cite|improve this question











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  • 3




    $begingroup$
    Stop trying to delete your questions.
    $endgroup$
    – fleablood
    Nov 27 '18 at 3:12










  • $begingroup$
    The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
    $endgroup$
    – fleablood
    Nov 27 '18 at 5:26


















0












$begingroup$



Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.




$a_1 = 3$.



Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Stop trying to delete your questions.
    $endgroup$
    – fleablood
    Nov 27 '18 at 3:12










  • $begingroup$
    The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
    $endgroup$
    – fleablood
    Nov 27 '18 at 5:26
















0












0








0





$begingroup$



Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.




$a_1 = 3$.



Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.






discrete-mathematics







share|cite|improve this question











$endgroup$





Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.




$a_1 = 3$.



Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.






discrete-mathematics




discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 4:42









Blue

48k870153




48k870153










asked Nov 27 '18 at 2:20









capcap

103




103








  • 3




    $begingroup$
    Stop trying to delete your questions.
    $endgroup$
    – fleablood
    Nov 27 '18 at 3:12










  • $begingroup$
    The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
    $endgroup$
    – fleablood
    Nov 27 '18 at 5:26
















  • 3




    $begingroup$
    Stop trying to delete your questions.
    $endgroup$
    – fleablood
    Nov 27 '18 at 3:12










  • $begingroup$
    The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
    $endgroup$
    – fleablood
    Nov 27 '18 at 5:26










3




3




$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12




$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12












$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26






$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26












1 Answer
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$begingroup$

You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$



Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.



We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.



Hence by the principle of mathematical induction, we are done :)






share|cite|improve this answer









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    $begingroup$

    You first need to show your base case. For $k = 2$, we have
    $$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$



    Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.



    We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
    begin{align}
    a_{k+1} &= sqrt{2 a_k}\
    &= (sqrt{2})(sqrt{a_k})\
    end{align}
    But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.



    Hence by the principle of mathematical induction, we are done :)






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You first need to show your base case. For $k = 2$, we have
      $$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$



      Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.



      We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
      begin{align}
      a_{k+1} &= sqrt{2 a_k}\
      &= (sqrt{2})(sqrt{a_k})\
      end{align}
      But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.



      Hence by the principle of mathematical induction, we are done :)






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You first need to show your base case. For $k = 2$, we have
        $$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$



        Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.



        We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
        begin{align}
        a_{k+1} &= sqrt{2 a_k}\
        &= (sqrt{2})(sqrt{a_k})\
        end{align}
        But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.



        Hence by the principle of mathematical induction, we are done :)






        share|cite|improve this answer









        $endgroup$



        You first need to show your base case. For $k = 2$, we have
        $$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$



        Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.



        We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
        begin{align}
        a_{k+1} &= sqrt{2 a_k}\
        &= (sqrt{2})(sqrt{a_k})\
        end{align}
        But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.



        Hence by the principle of mathematical induction, we are done :)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 2:52









        AlkaKadriAlkaKadri

        1,459411




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