If X = {8 n − 7n − 1 ∶ n ∈ N} and Y= {49 (n-1) : n ∈ N }, then prove that X ⊆ Y.
Answers
Step-by-step explanation:
n is a natural number,
Lets prove by induction that 8n−7n−1 is always divisble by 49.
let p(n) be the statement: 8n−7n−1is divisible by 49.
p(1) is true. Let p(n) be true
⟹8n−7n−1=49k
check p(n+1)
8n+1−7(n+1)−1=8n+1−7n−8
=8(8n−1)−7n=8(49k+7n)−7n=49(8k+n)
thus p(n)⟹p(n+1) and p(1) is true. Therefore it is true for all n>0.
thus the set X contains numbers which are divisible by 49.
and set Y is clearly the set of “ALL” the non negative multiples of 49.
Before we conclude X is a subset of Y lets also prove X cannot contain all the multiples of 49. It is sufficient to show if Y can take a value(s) which X cannot, we are done.
let f(n)=8n−7n−1.
f(1)=0
f(2)=49
f(3)=490.
note that for all n>0
8n+1−7(n+1)−1>8n−7n−1
⟺8n>1 which is true.
thus f(n) is monotonic. Thus no multiple of 49 between 49 and 490 i.e between f(2) and f(3)is obtainable in X which are all obtainable in Y.
Thus, X⊂Y
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Answer:
Step-by-step explanation:
We know that n is a natural number,
By induction We know that 8n−7n−1 is always divisble by 49.
let p(n) = 8n−7n−1is divisible by 49.
p(1) is true. Let p(n) be true
so, 8n−7n−1=49k
lets then check p(n+1)
8n+1−7(n+1)−1=8n+1−7n−8
=8(8n−1)−7n=8(49k+7n)−7n=49(8k+n)
thus p(n)⟹p(n+1) and p(1) is true. Therefore it is true for all n>0.
thus the set X contains numbers which are divisible by 49.
and set Y is clearly the set of “ALL” the non negative multiples of 49.
Before we conclude X is a subset of Y lets also prove X cannot contain all the multiples of 49. It is sufficient to show if Y can take a value(s) which X cannot, we are done.
let f(n)=8n−7n−1.
f(1)=0
f(2)=49
f(3)=490.
note that for all n>0
8n+1−7(n+1)−1>8n−7n−1
⟺8n>1 which is true.
thus f(n) is monotonic. Thus no multiple of 49 between 49 and 490 i.e between f(2) and f(3)is obtainable in X which are all obtainable in Y.
Thus, X⊂Y
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