use mathematical induction to show that (7^n)-1 is divisible by 6
Let P(n) be the statement 7^n - 1 is divisible by 6. Then first of all P(1) is clearly true as 7^1 - 1 = 6. Now suppose that P(k) is true for some k >= 1, we want to show that we can deduce that P(k+1) is true.
Well, P(k) being true means that 7^k - 1 is divisible by 6 What does that imply for 7^(k+1) - 1 The 'trick' here is finding a way to express this last expression in terms of 7^k - 1. Think about it for a few minutes.
so for n=1, it is obiviously true since 6|6 now assume for some k that \[6|(7^k-1)=>7^k-1 =6a, a \in \mathbb{Z} \] so now we to need to show the expression is true for k+1 \[7^{k+1}-1=7^k7-1=7(7^k-\frac{1}{7})=7(7^k-1+\frac{6}{7})=7(7^k-1)+7(\frac{6}{7})\] \[7(6a)+6=6(7a+1)\]
but since a is an integer then 7a+1 is an integer so we have that \[6|(7^k-1)\] for all integer k>=1
so james u mean the thesis will b 7^(k+1)-1
I'm not exactly sure how you're using the word thesis here. We are trying to show that P(k) ==> P(k+1). I.e., we can deduce the statement P(k+1) from P(k).
myininaya, where does "a" come from? We started with variable n
a is integer that i pulled out of the skies
ok i will tell you again if \[6|(7^k-1) => 6a=7^k-1, a \in \mathbb{Z}\]
We can also do this in an "a"-less way. Note that \[7^{k+1} = 7.7^k - 1 = 7(7^k -1) + 6\] Now we already know something about that expression in brackets, 7^k - 1 ....
6|something means there is some integer k such that 6k=something
i usually say k but something else was called k so i used a
James, for less confusion, write: \[7 \dot\ 7^k -1\]
i have actually seen 7.7^k written in a paper and it was defined as multiplication
well not that exact expression
but . was defined as multiplication
we can make any symbol to define any operation we want
Myininaya, stop defending James
lol
lol
i like james
ok james its that all
I know you do
You're a biological computer. Biological computers are not allowed to have "favorites" or take sides...: :P
thanks ... @remainder: do you have your solution?
thanks every1
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