Can someone show me the proof by mathematical induction that 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 expression is divisible by 49:
Supose that now for expression is also divisible by 49:
From first equation we get that .
Substituting into second equation we get:
Last expression isn't divisible by 49 so isn't divisible by 49.
Here is how you do it.
You need to show there doth not exist such a such as,
Which means there are no integral solution for .
If you solve for "n",
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
is not a square then,
is also not a square.
Thus, again you need to show that.
is not a square
is not a square.
is a square.
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
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).
Don't you meanOriginally Posted by ThePerfectHacker
Need to prove either
statement true for n = k implies statement true for n = k + 1
Statememt not true for n = k + 1 implies statement not true for n = k
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