mathematical induction (1 Viewer)

[namburger]

inequality mathematical induction
just say ASSUME a > b
then when i go to the k+1 step, i get an expression: LHS > b
Can i replace the b with an a since a > b

and therefore, try and prove LHS > a

 

[vds700]

inequality mathematical induction
just say ASSUME a > b
then when i go to the k+1 step, i get an expression: LHS > b
Can i replace the b with an a since a > b

and therefore, try and prove LHS > a

yes u can. It would be something like:

To Prove LHS > c for n = k+1

LHS = .... Replace a wit b and u can say that
LHS > ... because b is smaller than a

 

[YannY]

yes u can. It would be something like:

To Prove LHS > c for n = k+1

LHS = .... Replace a wit b and u can say that
LHS > ... because b is smaller than a

Ahaha you made that up in your head

LHS>b a>b

say LHS=5 b=3 a=6

5>3 true!
6>3 true!
now you want to replace b with a so that LHS>a
lets see
5>6 hmmmmm in what world does that work?

Okay you can argue that a could be 5 and lhs could be 6 but theres no point
cause if one example dosnt work then it should not work at all

 

[vds700]

Ahaha you made that up in your head

Let me demonstarte with an example.

2005 ext1 HSC Q4d)

Prove by induction that 4^n -1 - 7n >0 for n >or= 2

Step 1: for n = 2, 16-1-14=1 >0

Step 2: Assume that
4^k-1-7k>0


Step 3: Prove that 4^(k+1) - 1 -7(k+1) >0

LHS =4.4^k - 8 - 7k
now from the assumption, 4^k>1+7k, so if we replace 4^k with (1+7k) which is smaller, then
LHS > 4(1+7k) -8-7k
>4+28k-8-7k
>21k -4 >0 as k>or=0