Prove by induction:
For all $$ n \in N, n+3

Question

Prove by induction:
For all $$ n \in N, n+3 < 5n^2 $$

swissgirl · Answer

Mt textbook provided steps but i didnt follow

swissgirl · Answer

the first step was that it showed that this was true for 1

amistre64 · Answer

n + 3 < 5n^2

id prolly rewrite it as

0 < 5n^2 - n -3

amistre64 · Answer

induction ... never liked it ... says provide a basis step

amistre64 · Answer

its true for some value; then assume its true for some general value "k"

amistre64 · Answer

0 < 5k^2 - k -3

now prove its true for (k+1)

0 < 5(k+1)^2 - (k+1) -3

swissgirl · Answer

sorry my dad just pulled me away from the screen for a sec here

swissgirl · Answer

ill tell u the steps they gave

swissgirl · Answer

so first they showed its true for teh lowest vlaue of n=1
then they assumed that $ n \in N, n+3<5n^2 $

amistre64 · Answer

0 < 5(k+1)^2 - (k+1) -3

0 < 5(k^2 +2k +1) - k -1 - 3

0 < 5k^2 +10k +5 - k -1 - 3

rearrange

0 < (5k^2 -k -3) + (10k +5 -1)
        ^^^^^^                    ^^^  this part is always positive
this part is assumed
 true already

swissgirl · Answer

Then (n+1)+3=n+3+1
                     < $ 5n^2+1$
                     <  $ 5n^2+10n+5 $
                     <  $ 5(n+1)^2 $

swissgirl · Answer

Ur proof is correct but i am suppossed to show that it is true for K+1 or n+1 whtvr

amistre64 · Answer

right, after assuming its correct for n=k; then you readdress if for n=(k+1)

swissgirl · Answer

Like id ont get what was done by the proof i provided

amistre64 · Answer

as long as you get your original "kth" value back and a value that cant possibly contradict the setup your good

swissgirl · Answer

ya but can u explain me the proof i gave. like i didnt follow it

amistre64 · Answer

you said first step (called a basis) is to determine its value for a specific integer:  1 in this case

n+3 < 5n^2
1 + 3 < 5 , is true

swissgirl · Answer

right that i followed

amistre64 · Answer

second step is to assume its true for all n=k values

k+3 < 5k^2

swissgirl · Answer

right

amistre64 · Answer

since its true for "k"  we want to prove its also true for "k+1" so that it follows out indefinantly

swissgirl · Answer

right

amistre64 · Answer

(k+1) +3 < 5(k+1)^2

k+1 +3 < 5k^2 +10k + 5

we need to rewrite this so that we have our original "k" setup

swissgirl · Answer

ok

amistre64 · Answer

(k+3) + 1 <  (5k^2) + 10k + 5
          - 1                            - 1
------------------------------
(k+3)       <  (5k^2) + 10k + 4

amistre64 · Answer

since this is an inequality, we would have to see if 10k+4 is always postive; since increasing 5k^2 only makes it bigger

swissgirl · Answer

well it is since its all natural number

amistre64 · Answer

correct, then depending on how much your teacher wants you to reinvent the wheel, this is proof enough

swissgirl · Answer

But like y did the textbook prove like the way i showed u.
Then (n+1)+3=n+3+1
                     < 5n^ 2 +1 
                     <  5n^ 2 +10n+5 
                     <  5(n+1)^ 2

amistre64 · Answer

or you could go further and say

10k+4 > 0
      -4    -4
-----------
    10k > -4
   /10     /10
  ------------
          k > -2/5

and since k is always greater than -2/5 .... its good

amistre64 · Answer

im not sure why the author of your textbook would choose one way over another

swissgirl · Answer

whtvr ok

swissgirl · Answer

THHANKKKSSS

amistre64 · Answer

lol, youre welcome :)