# Thread: prove by induction help

1. ## prove by induction help

I'm doing homework and getting stuck at this problem

I'm confused with the concept of induction and I don't know how to do this

a) Prove by induction that
5^n + 2.3^(n+1) + 1 is divisible by 4 for all n N.

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

any idea guys ?

Thanks,

2. Originally Posted by zisi
I'm doing homework and getting stuck at this problem

I'm confused with the concept of induction and I don't know how to do this

a) Prove by induction that
5^n + 2.3^(n+1) + 1 is divisible by 4 for all n N.

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

any idea guys ?

Thanks,
The first one is just plain old induction. What have you tried? Can you use non-induction methods? Namely, $5^n+2\cdot3^{n+1}+1\equiv 1+2\cdot\left(-1\right)^{n+1}+1=\begin{cases}0 & \mbox{if}\quad n\text{ is even}\\ 4 & \mbox{if} \quad n\text{ is odd }\end{cases}\equiv 0\text{ mod }4$

For the second one, what is the context of the problem? What are you doing in class right now?

3. Originally Posted by Drexel28
The first one is just plain old induction. What have you tried? Can you use non-induction methods? Namely, $5^n+2\cdot3^{n+1}+1\equiv 1+2\cdot\left(-1\right)^{n+1}+1=\begin{cases}0 & \mbox{if}\quad n\text{ is even}\\ 4 & \mbox{if} \quad n\text{ is odd }\end{cases}\equiv 0\text{ mod }4$

For the second one, what is the context of the problem? What are you doing in class right now?
for the second one , I guess we have to prove it by contradiction

4. Here is how I attempted the question

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

5. Prove by induction that $
5^n+2\cdot3^{n+1}+1
$
is divisble by 4 for all integer $n\in \mathbb{N}$.

Proof:

Base case:

For n=1, $
5^n+2\cdot3^{n+1}+1= 5^1+2\cdot3^2+1=5+18+1= 24= 6 \times 4$
.

For induction hypothesis:

Suppose for ingeter $k$ that $5^k+2\cdot3^{k+1}+1$ is divisible by 4. Then

$5^{k+1}+2\cdot3^{(k+1)+1}+1= 5\cdot5^{k}+2\cdot3\cdot3^{k+1}+1$
$=(4+1)5^k+2(2+1)3^{k+1}+1$
$
= 4\cdot5^k+5^k+ 4\cdot3^{k+1} +2\cdot3^{k+1}+1
$

$= 4(5^k+ 3^{k+1}) +(5^k+2\cdot3^{k+1}+1)
$

By induction hypothesis $
(5^n+2\cdot3^{n+1}+1)$
is divisble by 4 for all integer $n\in \mathbb{N}$