Turn on thread page Beta
 You are Here: Home >< Maths

# I need a little help watch

1. Hi

here is my solution and need a little help with part a a) Prove by induction that 5^n + 2.3^(n+1) + 1 is divisible by 4 for all n ∈ N.
solution:

Let p(n) be the statement
then we have to prove p(n) is true for every natural number

p(0) = 5^0 + 2 . 3^1 + = 8 which is devisible by 4

p (1) = 5^1 + 2.3^2 + 1 = 24 which is again devisible by 4

so p(n) is true for every natural number

now I have to do the inductive hypothesis step which the most difficult bit in this question

what I have done is :

let K= N

p(K) = p (N) AND that implies p ( k+1) is true

5 ^ (k+1) + 2 . 3^(k+2) +1

I don't know what to do after this step

---------------------------

b) Prove that if 2^x = 5 then x is irrational.

i used two methods for this one

the first one is by calculating

2^x = 5
ln 2^x = ln 5
x ln 2 = ln 5
x = ln 5 / ln 2
x= 2.3219

so x is irrational

OR

suppose that x is rational
a,b are positive intergers

2^x = 5
2 ^ a/b = 5 raise both sides by the power (b)

2^ (a/b) ^b = 5 ^b

2 ^a = 5^b

and from that we can say x is irrational

cheers,
2. (Original post by EffieFlowers)
what
3. (Original post by A.braham)
Hi

b) Prove that if 2^x = 5 then x is irrational.

i used two methods for this one

the first one is by calculating

2^x = 5
ln 2^x = ln 5
x ln 2 = ln 5
x = ln 5 / ln 2
x= 2.3219

so x is irrational
Woah! How does this prove that x is irrational? I don't see it; you've just truncated a decimal to 4 dp and claimed the whole thing is irrational!

OR

suppose that x is rational
a,b are positive intergers

2^x = 5
2 ^ a/b = 5 raise both sides by the power (b)

2^ (a/b) ^b = 5 ^b

2 ^a = 5^b

and from that we can say x is irrational
You're pretty much there, but some explanation is needed to say that there are no integers solutions to 2^a = 5^b. (Why isn't there? Note that there would be if the equation was 2^a = 4^b.)
4. (Original post by A.braham)
Hi

here is my solution and need a little help with part a a) Prove by induction that 5^n + 2.3^(n+1) + 1 is divisible by 4 for all n ∈ N.
solution:

Let p(n) be the statement
then we have to prove p(n) is true for every natural number

p(0) = 5^0 + 2 . 3^1 + = 8 which is devisible by 4

p (1) = 5^1 + 2.3^2 + 1 = 24 which is again devisible by 4

so p(n) is true for every natural number

,
All you have proved is that p(n) is divisible four in the limited cases that n=0 and n=1 How does this prove the general case?
5. lool you start confsuing me now guys

ok for part a

we have to do p(0) or p (1) and after that we have to do n+1

so what is wrong about that ?
6. Prove by induction that is divisible by 4 for all .

You rightly showed that:
which is again divisible by 4

And now, assume that is divisible by 4.

Then investigate k+1, and show that if p(k) is divisible by 4, then so p(k+1) must be.
The first step you might like to consider:

After messing around with this, I think that taking out a factor of p(k) - which you know is a multiple of 4 - might help you.

Once you've shown that it's true for the number 1 and, {if it's true for k, then it's true for k+1}, you can then say it's true for all of the natural numbers!

Hope this helps
7. (Original post by A.braham)
lool you start confsuing me now guys

ok for part a

we have to do p(0) or p (1) and after that we have to do n+1

so what is wrong about that ?
That sounds about right

Just make sure that you're showing that n+1 works when any n works.
8. (Original post by placenta medicae talpae)
Prove by induction that is divisible by 4 for all .

You rightly showed that:
which is again divisible by 4

And now, assume that is divisible by 4.

Then investigate k+1, and show that if p(k) is divisible by 4, then so p(k+1) must be.
The first step you might like to consider:

After messing around with this, I think that taking out a factor of p(k) - which you know is a multiple of 4 - might help you.

Once you've shown that it's true for the number 1 and, {if it's true for k, then it's true for k+1}, you can then say it's true for all of the natural numbers!

Hope this helps

you see this

that is where I'm stuck at

I couldn't do any factorisation

I ended up with
5^k . 5 + 18.3^k + 1 = 4a

which is incorrect
9. (Original post by A.braham)
you see this

that is where I'm stuck at

I couldn't do any factorisation

I ended up with
5^k . 5 + 18.3^k + 1 = 4a

which is incorrect
Oh right, okay, no worries!

Is the same as:

Maybe some factorisation might work now?
10. (Original post by placenta medicae talpae)
Oh right, okay, no worries!

Is the same as:

Maybe some factorisation might work now?

i got it now cheeers
11. (Original post by A.braham)
i got it now cheeers
No probs!

Turn on thread page Beta

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 2, 2010
The home of Results and Clearing

### 2,594

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
General Open Day Undergraduate
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University
Tue, 21 Aug '18
Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE