Is $k \geq 3$ enough to conclude $(k - 1)^2 > 2$ in mathematical induction?

[ineedhelpnow]

Did i do this right?

$2^n>n^2$, $n \ge 5$

$S_{k+1}: 2^{k+1}>k^2+2k+1$

$2^k>k^2$

$2(2^k)>2k^2$

$2^{k+1}>k^2+k^2$

And $k^2+k^2>k^2+2k+1$

RHS: $k^2+2k+1$

So $2^{k+1}>(k+1)^2$

 

[kaliprasad]

Did i do this right?

$2^n>n^2$, $n \ge 5$

$S_{k+1}: 2^{k+1}>k^2+2k+1$

$2^k>k^2$

$2(2^k)>2k^2$

$2^{k+1}>k^2+k^2$

And $k^2+k^2>k^2+2k+1$

RHS: $k^2+2k+1$

So $2^{k+1}>(k+1)^2$

the steps are right but you need to show that
$k^2\gt 2k+ 1$
which is true for k >4(as you need for k > 4)
as $k^2\gt 4k \gt 2k +1$

then the proof becomes compelte

 

[Nono713]

And you need to show the base case, i.e. show it holds for $n = 5$ (easy, but you must not forget it, otherwise the proof is useless).

 

[ineedhelpnow]

the steps are right but you need to show that
$k^2\gt 2k+ 1$
which is true for k >4(as you need for k > 4)
as $k^2\gt 4k \gt 2k +1$

then the proof becomes compelte

how do i show that though? do i just state it or do i have to prove it?

 

[Deveno]

You have to prove it, of course, like kaliprasad did.

 

[ineedhelpnow]

how would i prove that though? do because i mentioned in my notes that $k^2>2k+1$ but i don't know how to prove it.

 

[MarkFL]

how would i prove that though? do because i mentioned in my notes that $k^2>2k+1$ but i don't know how to prove it.

Try adding $1-2k$ to both sides...then you can find a lower bound for the positive real numbers.

 

[ineedhelpnow]

if you add that to both sides you end up getting $k^2-2k+1>0$

 

[MarkFL]

if you add that to both sides you end up getting $k^2-2k+1>0$

Well, you actually get:

\(\displaystyle k^2-2k+1>2\)

Now, factor the left side, and concern yourself only with \(\displaystyle 0<k\)...

 

[ineedhelpnow]

oh yes your right.

$(k-1)^2>2$

 

[MarkFL]

oh yes your right.

$(k-1)^2>2$

So, what is the lower bound for positive $k$?

 

[ineedhelpnow]

2?

 

[MarkFL]

2?

I really should have said concern yourself with \(\displaystyle 1\le k\) where \(\displaystyle k\in\mathbb{N}\)

Take the positive root of both sides...what do you find?

 

[Deveno]

Suppose we want to be able to CONCLUDE:

$(k - 1)^2 > 2$

What conditions does $k$ have to satisfy, first?

Note that if $a,b > 0$ and $a^2 \geq b^2$, then it must be that $a \geq b$, for if we had:

$a < b$, then $a^2 < ab < b^2$, contradiction.

If we could show $(k - 1)^2 \geq 4$, it would surely be greater than 2, since $4 > 2$.

Now 4 has the advantage of being a perfect square, so from the argument above, it would be sufficient if we have:

$k - 1 \geq 2$.

Almost there...can you finish?