Inductive reasoning in Math

How in the HELL do you do this???

This is an example of what I am told to do;

Making and Proving Conjectures

Consider this procedure. Select a number. Multiply the number by 9. Add 6 to the product and divide the sum by 3. Subtract 2 from the quotient.

(a) Follow the procedure for four different numbers and make a conjecture about the relationship between the original number and the final result.

(b) Represent the original number by n and prove the conjecture in part (a).

Solution
Let us select four different numbers--say, 5, 10, 21, and 100--and apply the procedure given.

WHAT THE SHIT, MAN!?!

OP HERE:

so here is what they say (a) is supposed to be;

(a)
Select a number                 (5) | (10) | (21) | (100)
Multiply the number by 9.(45)| (90) | (189)| (900)
Add 6 to the product.    (51)| (96) | (195)| (906)
Divide the sum by 3.     (17)| (32) | (65) | (302)
Subtract 2 from the quotient. (15)| (30) | (63) | (300)

Since we have to make a conjecture relating the original number and the final result, let us look at the original numbers and the final results.

5|10|21|100
15|30|63|300

Do you see the pattern? Using inductive reasoning, our cnjecture is that the final result is three times the original number. But can we prove it? Let us go to part (b) and repeat the process using n as the original number.

(b) Select a number. n
    Multiply n by 9  9・n = 9n
    Add 6 to the product. 9n+6
    Divide the sum by 3. (9n+6)/3 = 3n+2
{It gets a little tricky here. You have to divide 9n by 3 and get 3n and 6 by 3 and get 2.}

The final result 3n is indeed three times the original number n; this proves our original conjecture.


Okay, now I just accidentally 12 years of math. The fuck man!?!
Anyone wanna help out here?

woops, forgot a line;

just after Dividing the sum by 3;

Subtract 2 from the quotient. 3n+2-2=3n

It's so bizarre cause it appears that this method is not only creating a conjecture, but proving it using the same method as creating it....it may be my own misgivings, but isn't that always going to be the case?

Your subject is not correct.  Inductive reasoning is not the same as mathematical induction, which is what the instructions appear to outline; the latter is actually a form of deductive reasoning.

That is all I can say.  I am not terribly good at working with either.

This is what my book says;

B. Inductive Reasoning

As we mentioned, one of the strategies used in problem solving is to find a pattern. Reasoning based on examning a variety of cases, discovering patterns, and forming conclusions is called inductive reasoning. For example, if you want to make sure that you have a good instructor in your next mathematics course. You ask several of your friends about different instructors. They all say that Professor X is a good instructor. You conclude that Professor X is indeed a good instructor. This is an example of inductive reasoning, but it may be a hasty generalization. Now consider this example: Suppose you become a doctor and one of your patients comes to you and says,
     It hurts when I press here (pressing his side).
     And when I press here (pressing his other side).
     And here (pressing his leg).
     And here, here, and here (pressing his other leg, and both arms).

You examine the patient all over, trying to invoke your inductive powers. What can you conclude?

The patient has a broken finger.

Inductive reasoning is the process of arriving at a general conclusion on the basis of repeated observations of specific examples.

Please, let's go from what is inductive reasoning and inductive math to how to use induction in math? Thanks in advance!

The process of proof by mathematical induction has three steps.

Say P(n) is a statement involing the set of  all natural numbers N.

Show that a trivial solution satisfies P

i.e that P(1) is true

show that for any number k in N, if P(k) is true, then P(k+1) is true. I.E in the statement on P(k) replace the k's with k+1.

then you will have shown that P is satisfied for any natural number k.

Example

show that 1+2+3+4+5+6+...+n=(n/2)*(n+1)

First we observe P(1) to be true, this should be trivial.

Then the inductive step (from whence the nomenclature of induction) suppose that P(k) is true for any k in N

1+2+3+...+k=(k/2)*(k+1)

then check P(k+1);

1+2+3+...+k+(k+1)=((k/2)*(k+1))+(k+1)

RHS =(k+1)((k/2)+1

RHS =((k+1)/2)*(k+2)
RHS =((k+1)/2)*((k+1)+1)

showing that P(k+1) is true, and thus, that

for all k in N, P(k) is true implies P(k+1) is true.

Hope I have been of help, and sorry I didn't use Latex, my I havent yet developed the skills to pay the bills.

Okay, thought I'd try my hand at induction on primes.
BIG MISTAKE!!!
Between 1 and 37 in primes;

On the third tier there are zeros and negative numbers.
The final tier was 13 and the final move was -1452.

One thing I did notice was a distinct sense of chaos. The moves increased even greater with each tier both into the negative and positive ranges. At this point, I'm almost certain that I've stepped over the tier where the pattern is supposed to emerge.

Later on I'm going to try my hand at other arithmetic to see if there are any other means to make the moves...we'll see what happens. God, I thought maybe I'd spot something, but then again someone probably already thought of this which is probably why we have primes standing out amongst number sets.

Wish me luck!!!

I decided to hold off on using induction on primes until I learn more methods of analysis in mathematics. Well, back to basics, amirite?

Hello sir,

Let us example the procedure.
"Select a number. Multiply the number by 9. Add 6 to the product and divide the sum by 3. Subtract 2 from the quotient."

5*9 = 45 ----> 45+6 = 51  ----> 51/3 = 17  ----> 17-2 = 15
10*9 = 90 ---> 90+6 = 96 -----> 96 /3 = 32 ----> 32 -2 = 30
21*9 = 189 ---> 189 + 6 = 195 ----> 195/3 = 65 ---> 65 -2 = 63
It seems like the relation is you multiply it by 3.
WHYYYYYY?? Lets do it in the general case.
n*9 = 9n ----> 9n+6 ---->(9n+6)/3 ---->  (9n+6)/3  -2
Now lets simplify
(9n+6)/3 -2 = 9n/3 + 6/3 -2 = 3n +2 -2 = 3n

>>9
here's the thing. The control isn't the random set of numbers {5, 10, 21, 100}...it's the {x9, +6, ÷3, -2}

Any numbers you put into this process equates to 3n. Try it.