Mathematical induction

[lime aurora]

the task number 5 i need to prove by mathematical induction that 10^n - 4 is divisible by 6

[patent lagoon]

!status

[verbal idol]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[lime aurora]

1

[patent lagoon]

Do you know what mathematical induction is?

[lime aurora]

i know that you use it to prove a statment

in math

[patent lagoon]

But do you know how?

[lime aurora]

well it's 50/50 for me

because i can solve some problems for example i would easly do 13^n - 7 divisible by 6 but i cant and do not know how to do it with those numbers

[patent lagoon]

you can do 13^n-7 but not 10^n-4?
I just want to be sure I understand you correctly

[lime aurora]

yes because i use a technique that leads me to in first one but with the second numbers i get to 3(3 * 10^n +k)

k = an (10^n-4)

and by getting to point where 6(2 * 13^n +k) it is a proof for my equasion to be divisible by 6

and i cant get to that point with the other one

[patent lagoon]

I'm struggling to read this, sorry

[lime aurora]

'an+1 = 13^n+1 - 7 = 13 * 13^n -7 = '12*13+13^n -7 = 6**2 *13^n + 6k = 6(2 ** 13^n+k)'

[patent lagoon]

Surround your math in ` backticks to prevent the asterisks from being formatted

[patent lagoon]

Those are single quotes, not back ticks. But I see your work now

[lime aurora]

🙂

so can you help me solve it or show me how to solve it

i will go take a shower now

[patent lagoon]

So you are stating that $a_n=13^n-7$

[sick owl]

SWR

[patent lagoon]

So you've calculated $a_{n+1}=13^{n+1}-7=13\cdot13^n-7$

[sick owl]

SWR

[patent lagoon]

I do not see how you arrived at $13\cdot13^n-7=12\cdot13+13^n-7$

[sick owl]

SWR

[lime aurora]

well i helped myself with a youtube video and it is 12 * 13 ^n

[lime aurora]

and can we get to the main problem which is that task because i have to wake up at 5,20 am and rn it is 9.30 pm

[patent lagoon]

I'm disagreeing. But sure we'll get to the main problem

From what I am seeing, there's a total lack of understanding on mathematical induction, so we should start there

[lime aurora]

okay i am all ears

[patent lagoon]

Mathematical induction is a strategy for proving that something is true for output of some particular sequence.

[lime aurora]

understood

[patent lagoon]

Let $s_n$ be some arbitrary sequence for $n=0, 1, 2, ...$

[sick owl]

SWR

[patent lagoon]

Let's say there's some property of $s_n$. The property could be anything: $s_n$ is even. $s_n$ is divisible by 6. $s_n$ is a perfect square. It doesn't matter. It's just whatever property. We'll algebraically write this property as $P(s_n)$.

[sick owl]

SWR

[patent lagoon]

Say we want to prove that $P(s_n)$ is true for every $n=0, 1, 2, ...$

[sick owl]

SWR

[patent lagoon]

We can do this precisely through mathematical induction.

Here's the process to prove this through induction.

1. Prove the base case. That is, prove that $P(s_0)$ is true.

[sick owl]

SWR

[patent lagoon]

2. For any $k=0, 1, 2, ...$, you ASSUME that $P(s_k)$ is true. With this assumption, you PROVE that $P(s_{k+1})$ is also true.

[sick owl]

SWR

[patent lagoon]

That's it.

If you prove those two things, you are done.

Here's one example. Let $s_n=(n+2)^2$. Let $P(s_n)$ be the expression "$(n+2)^2>n+2$".

[sick owl]

SWR

[patent lagoon]

That is, for every $n=0, 1, 2, ...$, we will prove $(n+2)^2>n+2$

[sick owl]

SWR

[patent lagoon]

First, we prove the base case: $s_0=(0+2)^2=4$. And clearly $4>0+2$. That prove the base case.

[sick owl]

SWR

[patent lagoon]

Next. We are going to ASSUME that $(k+2)^2>(k+2)$ for any $k=0, 1, 2, ...$. With this assumption, we will PROVE that $((k+1)+2)^2>((k+1)+2)$

[sick owl]

SWR

[patent lagoon]

We rewrite $((k+1)+2)^2=((k+2)+1)^2=(k+2)^2+2(k+2)+1$

[sick owl]

SWR

[patent lagoon]

We are assuming that $(k+2)^2>(k+2)$, so, we can infer that $(k+2)^2+2(k+2)+1>(k+2)+2(k+2)+1$

[sick owl]

SWR

[patent lagoon]

And we simplify $(k+2)+2(k+2)+1$ to $3k+7$. But $k=0, 1, 2, ...$, so $k\ge 0$ which means $3k+7>0$ is always true.

[sick owl]

SWR

[patent lagoon]

Thus, (with the assumption $(k+2)^2>k+2$), we have shown that $((k+1)+2)>(k+1)+2$ simplifies to $3k+7>0$, which is always true, which means that $((k+1)+2)^2>(k+1)+2$ is also true.

[sick owl]

SWR

[patent lagoon]

That ends the proof.

@lime aurora

[lime aurora]

that is a lot

[patent lagoon]

At first, yes. Perhaps inequalities are hard to start with.

Here's another example:

Let $s_0=0$ and $s_{n+1}=s_n+1$ for $n=0, 1, 2, ...$

[sick owl]

SWR

[patent lagoon]

We will prove that $s_n=\frac{n(n+1)}{2}$

[sick owl]

SWR

[patent lagoon]

Base case: we must prove that $s_0=\frac{0(0+1)}{2}$

[sick owl]

SWR

[patent lagoon]

But $\frac{0(0+1)}{2}=0$, so the base case is proven.

[sick owl]

SWR

[patent lagoon]

With me so far?

[patent lagoon]

Correction: $s_{n+1}=s_n+(n+1)$

[sick owl]

SWR

[lime aurora]

ee

a bit

[patent lagoon]

any questions?

[lime aurora]

no

i am just kind of awaiting for doing the 10 ^n - 4 one

[patent lagoon]

But you're understanding what I'm saying up to now?

[lime aurora]

yes but i must say that my way to understand something is very specified and i would prolly understand it much more on the equasion i 10^n - 4

[patent lagoon]

Okay, we'll try it that way. What do you think is the first step should you do for your 10^n-4 problem?

[lime aurora]

do
an = 10^n-4
a1 = 6 so 6 is divisible by itself so it is good

[patent lagoon]

yup

perfect

now what

[lime aurora]

an+1 = 10^n+1 - 4 so i think we can go 10^n * 10^1 - 4

[patent lagoon]

and what next?

[lime aurora]

well can i create a variable "k" and write k * 6 = 10^n-4

[patent lagoon]

works for me

[lime aurora]

so we can go an+1 = 10^n * k6

[patent lagoon]

How did you get this?

[lime aurora]

made a mistake it should be more like 10^1 * k6

[patent lagoon]

Not quite. No

[lime aurora]

why

was the first one good?

[patent lagoon]

I get why you are writing $10^n-4=6k$. You are using the assumption that $10^n-4$ is divisible by 6, and yes, you did that correctly.

[sick owl]

SWR

[patent lagoon]

And you also correctly wrote that $a_{n+1}=10\cdot10^n-4$

[sick owl]

SWR

[patent lagoon]

But $10\cdot10^n-4\ne10\cdot6k$. You are confusing that with $10(10^n-4)$

[sick owl]

SWR

[lime aurora]

i see

@patent lagoon y ther mate?

[patent lagoon]

yes

[patent lagoon]

Do you know how to correct your mistake?

[lime aurora]

i dont because tbh i am too tired to think well

so please correct me

[patent lagoon]

Solve for 10^n in this equation

And you can put it in $10\cdot 10^n-4$

[sick owl]

SWR

[lime aurora]

10^n =k6+4

and then we go 10 * k6+4 - 4
we have k60 and 40 - 4
we drag 6 outside the (10k +6) so we proved our point?

[patent lagoon]

You need to use parentheses

[lime aurora]

i know i just didnt do it rn because i am to tired

but anything by it the induction is finished at this point?

[patent lagoon]

Yeah you got it

[lime aurora]

it took 2 hrs

thanks

mate