Can someone show me the proof by mathematical induction that $\displaystyle n^2+n+2$ isn't divisible by 49 for every n (natural number).
Firstly observe that n^2+n+2 is divisible by 7 if and only if n is congruent to 3 modulo 7: that is, if n is of the form 7x+3. See this by writing it as n^2 - 6n + 9 + 7(n-1), so that it is (n-3)^2 modulo 7.
Now substitute n=7x+3 into n^2 + n + 2 to get 49x^2 + 49x + 14, which cannot be divisible by 49.
Is this proof valid?
Supose that for $\displaystyle n=k$ expression is divisible by 49:
$\displaystyle k(k + 1) + 2 = 49a$
Supose that now for $\displaystyle n=k+1$ expression is also divisible by 49:
$\displaystyle (k + 1)(k + 2) + 2 = 49b$
From first equation we get that $\displaystyle k + 1 = \frac{{49a - 2}}{k}$.
Substituting into second equation we get:
$\displaystyle \begin{array}{l}
(\frac{{49a - 2}}{k})(k + 2) + 2 = 49b \\
\frac{{49ak + 98a - 2k - 4}}{k} = 49b - 2 \\
49ak + 98a - 2k - 4 = 49bk - 2k \\
49ak + 98a - 2k - 4 - 49bk + 2k = 0 \\
49ak + 98a - 49bk - 4 = 0 \\
\end{array}
$
Last expression isn't divisible by 49 so $\displaystyle n^2+n+2$ isn't divisible by 49.
You need to show that,
If,
$\displaystyle n^2+n+2\not = k$
then,
$\displaystyle n^2+n+2\not = k+1$
Equivalently (contropositive),
If,
$\displaystyle n^2+n+2=k+1$
then,
$\displaystyle n^2+n+2=k$
You shown that:
if,
$\displaystyle n^2+n+2=k$
then,
$\displaystyle n^2+n+2\not = k+1$
This is not equivalent to the above two statements.
Here is how you do it.
---
You need to show there doth not exist such a $\displaystyle 49k$ such as,
$\displaystyle n^2+n+(2-49k)=0$
Which means there are no integral solution for $\displaystyle n$.
If you solve for "n",
$\displaystyle n=\frac{-1\pm \sqrt{1-4(2-49k)}}{2}$.
Note the following, if the expression in the radical is a square then "n" is an integral number because the square root of an odd square is an odd whole number and when added of subtracted to 1 produces an even number and when finally divided by two produces an integral number, thus, if
$\displaystyle 1-4(2-49k)=196k-7$ is not a square then,
$\displaystyle 1-4(2-49(k+1))=1-4(-47-49k)=196k+189$ is also not a square.
Thus, again you need to show that.
IF,
$\displaystyle 196k-7$ is not a square
THEN,
$\displaystyle 196k+188$ is not a square.
~~Contropositive Time~~
IF,
$\displaystyle 196k+188$ is a square.
THEN,
$\displaystyle 196k-7$ is a square.
For some reason I cannot continue. I am hoping someone will see this and tell me the next step.
From his silence on this and departure off on another tangent I thinkOriginally Posted by Quick
ImPerfectHacker agrees with you and he doesn't understand his
argument either
It is either so subtle that we can't see the point, or its wrong. I know
where I'm putting my money (no not in that box under the bed).
RonL
Don't you meanOriginally Posted by ThePerfectHacker
Need to prove either
statement true for n = k implies statement true for n = k + 1
or
Contropositive
Statememt not true for n = k + 1 implies statement not true for n = k
OReilly showed
Statement true for n = k implies statement not true for n = k + 1
I assume you are unfamiliar with the principle of mathematical inductionOriginally Posted by Quick
Usually you are trying to prove something for all natural numbers n
You show that it is true when n = 1
Then you assume it is true for some n = k
It is okay to assume this as we have already proved it for n = 1 so we already know there is at least one possible value for k.
We then try to prove that it also holds for n = k + 1 assuming that it holds for n= k.
Once we have done this we have proved it for all natural numbers n as
True for n = 1 means true for n = 2 means true for n = 3 and so on ...
This method can be adapted to suit several different purposes