proof that i^n = i^(n+4)

Can someone explain: for any positive integer n, i^n = i^(n+4)?

I know it's pretty obvious, but how can it be proved for all positive integers n?

Thank you :smile:

(edited 10 years ago)

Reply 1

Noble.

17

What's

i^4

? - think of it as

(i^2)^2

Reply 2

PerArduaAdAstra

9

Induction's probably the way forward I would have thought.

Reply 3

username1258398

19

Induction is completely unnecessary; just simplify i^(n+4) using your basic exponentiation rules:

x^{a+b} = x^ax^b

(edited 10 years ago)

Reply 4

alow

19

I'd agree with proving by induction, seems far easier than any other option.

Reply 5

Noble.

17

Original post by StrangeBanana

Induction is unnecessary; just simplify i^(n+4) using your basic exponentiation rules:

x^{a+b} = x^a + x^b

You mean

x^{a+b} = x^ax^b

Reply 6

alow

19

Original post by StrangeBanana

Induction is unnecessary; just simplify i^(n+4) using your basic exponentiation rules:

x^{a+b} = x^a + x^b

Do you mean

x^{a+b}=x^{a}x^{b}

Reply 7

username1258398

19

Original post by Noble.

You mean

x^{a+b} = x^ax^b

Ah, yeah; silly me! Fixed it.

Method still works though. :tongue:

(edited 10 years ago)

Reply 8

Fryschia

OP

9

Original post by StrangeBanana

Induction is completely unnecessary; just simplify i^(n+4) using your basic exponentiation rules:

x^{a+b} = x^a+x^b

Thank you. This was helpful. So you know, it's x^{a+b} = x^a * x^b.

Reply 9

brianeverit

9

Original post by Fryschia

Can someone explain: for any positive integer n, i^n = i^(n+4)?

I know it's pretty obvious, but how can it be proved for all positive integers n?

Thank you :smile:

i^(n+4)=i^n x i^4 and i^4 =?

Reply 10

username1258398

19

Original post by Fryschia

Thank you. This was helpful. So you know, it's x^{a+b} = x^a * x^b.

No problem, and yes sorry for the typo.

Reply 11

interstitial

18

You don't need induction, use the basic rules of indices and the value of i^4 and you're sorted

Posted from TSR Mobile

Reply 12

Solivagant

18

Original post by brianeverit

i^(n+4)=i^n x i^4 and i^4 =?

i^4=(i^2)^2

Reply 13

Noble.

17

Original post by Fryschia

Can someone explain: for any positive integer n, i^n = i^(n+4)?

I know it's pretty obvious, but how can it be proved for all positive integers n?

Thank you :smile:

Was this asked on a sheet where you're supposed to use induction? Given the question asks you to prove it for the natural numbers, and this in fact holds for all integers, I'm thinking it was? If so, it's worth solving this by induction just to get to grips with induction.

Reply 14

user2020user

16

Induction?

\mathbb{Z^{+}} = \left\{ 1, 2, 3, 4, \cdots \right\}

so the base case is

n=1

Base case

Assumption step

I'll leave the inductive step to you.

(edited 10 years ago)

Reply 15

Kvothe the Arcane

20

Anyone who uses induction unnecessarily is just :puke:

.
As

i=\sqrt{-1}

and

i^4=(i^2)^2

,

i^4

should be easy enough to evaluate OP.

Reply 16

Fryschia

OP

9

Original post by Noble.

Was this asked on a sheet where you're supposed to use induction? Given the question asks you to prove it for the natural numbers, and this in fact holds for all integers, I'm thinking it was? If so, it's worth solving this by induction just to get to grips with induction.