Question

If X = {8 n − 7n − 1 ∶ n ∈ N} and Y= {49 (n-1) : n ∈ N }, then prove that X ⊆ Y.

Answer 1

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 2

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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