Answer:
Step-by-step explanation:
The Strong Induction Principle establishes that if a a subset S of the positive integers satisfies:
- S is a non-empty set.
- If m+1, m+2, ..., m+k ∈ S then m+k+1 ∈ S.
Then, we have that n ∈ S for all n ≥ k.
- <u>Base case</u>: Now, in our problem let S be the <em>set of positive numbers than can be written as a sum of distinct powers of 2</em>. Note that S is non-empty because, for example, 1, 2, 3 and 4 belongs to S:
This is the so called <em>base case</em>, and in the definition above we set k = 1. - <u>Inductive step</u>: Now suppose that 1, 2, 3, .., k ∈ S. This is the <em>inductive hypothesis.</em> We are going to show that k+1 ∈ S. By hypothesis, since k ∈ S, it can be written as a sum of distinct powers of two, namely,
where 
, i.e., every power of 2 occurs only once or not appear. Using the hint, we consider two cases:
- k+1 is odd: In this case, k must be even. Note that
. If not were the case, then 
and we can factor 2 in the representation of k: 
This will lead us to the contradiction that k is even. Then, adding 1 to k we obtain:
- k+1 is even: Then
is an integer and is smaller than k, which means by the inductive hypothesis that belongs to S, that is, 
where 
, for all 
. Therefore, multiplying both sides by 2, we obtain 
This is a sum of distinct powers of 2, which implies that k+1 ∈ S.
Then we can conclude that n ∈ S , for all n ≥ 1, that is, every positive integer n can be written as a sum of distinct powers of 2.
18=2(5)+b
b=8 y intercept is 8
0=2(x)+8
-8=2x
x=-4
X intercept is -4
X - the number of hats on Monday
So, he had 40 on Monday.
The sum is the gcf, if 60 plus 84 is 148 then wouldn't it be 148 that's what they all have in common, you can use a calculator to figure this out like say 148 times 2 equals 296, divide 296 by 60. Does it go in?
Answer:
study :)
Step-by-step explanation:
