- #1

moocav

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**2 Induction Questions**

## Homework Statement

I have quite a few questions and so i just made it an image. Also attached.

http://img411.imageshack.us/img411/1002/inductionforlife.jpg" [Broken]

Only need help with questions

**2**and

**5**now

Oh and so far my lecturer has taught well-ordering, strong induction and simple induction. But I could only follow simple induction... the other two I'm quite clueless about >< Though tell me which method is best for each question.

## Homework Equations

Refer to image

**3. The Attempt at a Solution s**

__Question 2__I have no clue to how to start it..

All i have done is

24 = 7 + 7 + 5 + 5

25 = 5 + 5 + 5 + 5 + 5

26 = 7 + 7 + 7 + 5

27 = 7 + 5 + 5 + 5 + 5

28 = 7 + 7 + 7 + 7

29 = 7 + 7 + 5 + 5 + 5

no idea what to do now

[STRIKE]

**Question 3**I'm not sure if my method is correct but I've proved that when x = 0 and n = 1, x = 1 n = 2 are true. But I get stuck whilst proving n = k + 1

Let S

_{n}be (1 + x)

^{n}>= 1 + nx

For n = 1 and x = 0, S

_{1}=

LHS = (1+0)

^{1}= 1

RHS = 1 + (1)(0) = 1

Therfore LHS >= RHS Hence n = 1 is true.

Assume n = k is true

S

_{k}--> (1 + x)

^{k}>= 1 + kx

For n = k + 1, S

_{k+1}=

I know that I need to get to

(1 + x)

^{k+1}>= 1 + (k+1)x

(1 + x)

^{k}>= 1 + kx

(1 + x)

^{k}(1 + x)

^{1}>= (1 + kx)(1 + x)

^{1}(multiplied both sides by (x + 1)

(1 + x)

^{k+1}>= 1 + x + kx + kx

^{2}

I can see that on the RHS there is 1 + x + kx I'm not sure what to do with it... hints/help?

__(Just needs checking)__

*Question 4*Let S

_{n}be ƒ

_{1}+ ƒ

_{2}+ ... + ƒ

_{n}= ƒ

_{n+2}-1

For n = 1, S

_{1}

LHS = ƒ

_{1}= 1

RHS = ƒ

_{1+2}- 1 = 2 - 1 = 1

LHS = RHS

Therefore n = 1 is true

Assume true for n = k

S

_{k}--> ƒ

_{1}+ ƒ

_{2}+ ... + ƒ

_{k}= ƒ

_{k+2}-1

For n = k + 1, S

_{k+1}=

RHS = ƒ

_{k+3}- 1

LHS = ƒ

_{1}+ ƒ

_{2}+ ... + ƒ

_{k}+ ƒ

_{k+1}

= ƒ

_{k+2}-1 + ƒ

_{k+1}

= ƒ

_{k+2}+ ƒ

_{k+1}- 1

= ƒ

_{k+3}- 1 (should I write any reason here? if yes..what should i write?)

= RHS

Hence n = k + 1 is true

By mathematical induction S

_{n}is true for all positive integers n.

[/STRIKE]

*Question 5*Show that n/t - 1/(q+1) is positive and numerator is less than n

where t = nq + r with 0 < r < n

(get common denominator then expand and simplify)

n/t - 1/(q+1)

= n(q + 1)/[t(q+1)] - t/[t(q+1)]

= [n(q+1) - t] / [t(q+1)]

= [nq - t + n] / [t(q+1)]

t = nq + r

nq - t = -r

hence n/t - 1/(q+1)

= [n-r] / [t(q+1)]

from 0 < r < n

n > r therefore n - r > 0 (proved that numerator is positive)

and since r > 0 then n - r < n (proved that numerator is < n)

I'm not sure where to go from here

Please someone help me however you can..

Thank you in advance!!

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