No, you've proven it for a = 2 but not for all a.
I have to prove that is dividable with 30 for every integer value of .
My proof is:
and .
So if we put instead of 2 (or 2 instead of a) the proof is proven.
I have found in book much complicated solution but this seems ok to me.
Am I right?
P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.
It would help if you told us what level course thisOriginally Posted by DenMac21
is related to, otherwise you risk getting a totally
incomprehensible explanation of how to prove the
result.
What you want is the LaTeX guide, its in the followingP.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.
thread:
http://www.mathhelpforum.com/math-he...read.php?t=266
RonL
[QUOTE=DenMac21]I have to prove that is dividable with 30 for every integer value of .
/[QUOTE]
I'm not sure if this qualifies as high school maths but here goes:
(read as divides so means divides .
if then and , also if
and then because
and have no common factor.
So we need only show that for any positive integer
that and .
Now any positive integer may be written
,
for some positive integer and remainder .
Now:
,
for some positive integer . so if:
then:
where , and these five cases can be
checked by hand.
Now repeat the same type of argument for in place
of to complete the proof.
RonL
Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for then it must work for . Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
Thus, let be divisible by 30.
Then show that is divisible by 30. Use the binomial theorem
Thus,
But, is divisible by 30,
thus,
if divisible by 30 then the whole expression is divisible by 30.
But
is divisible by 5
. Factor,
.
Notice if this is true for thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.
In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
This is the proof from the book which is (at the moment) most understable to me:
are five successive integer numbers and they are divisible with 30.
have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.
So, both expressions are divisible with 30 and their sum is then also divisible with 30.
You have lost me around here, presumably you are trying to show thatOriginally Posted by ThePerfectHacker
.
is divisible by 6 (or 2 and 3), but I can't follow how you are doing so
RonLNotice if this is true for thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.
In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
Seems to me that you might have to prove this last assertion (not tooOriginally Posted by DenMac21
difficult) unless its a result you are expected to know. Same comment
applies to similar assertion below.
RonLhave as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.
So, both expressions are divisible with 30 and their sum is then also divisible with 30.