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# DM - Grand Test -Q15

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+1 vote

Suppose S(n) is a predicate on natural numbers (Positive integers) n and suppose

$\dpi{100} \forall k \in N(S(k)\rightarrow S(k+2))$.

Now if above assertion holds then Consider the following statements:

$\dpi{100} S_1:(\forall n \leq 100(S(n)))\land (\forall n > 100 (\lnot S(n)))$

$\dpi{100} S_2:S(1)\rightarrow \forall n S(2n + 1)$

Which of the following is correct for above statements :

(A) Both Assertions $\dpi{100} S_1, S_2$ Always holds.

(B) $\dpi{100} S_1$ Always holds but $\dpi{100} S_2$ Never holds.

(C) $\dpi{100} S_1$ Never holds but $\dpi{100} S_2$ Always holds

(D) Both $\dpi{100} S_1, S_2$ can hold but Not always.

asked Aug 30, 2020
reshown Aug 31, 2020

Suppose S(n) is a predicate on natural numbers (Positive Integers), n and suppose

$\dpi{100} \forall k\in N(S(k)\rightarrow S(k+2)).$   \\ This says that If S(k) is true then S(k+2) is also true.

Now, if above assertion holds then Consider the following statements:

$\dpi{100} S_1:(\forall n \leq 100(S(n)))\land (\forall n > 100 (\lnot S(n)))$

S1 is NEVER true. S1 becomeing true means S(n) is true for all n <= 100 and S(n) is false for all n > 100. But this is impossible because  If S(99) is true then S(101) will have to be true. So, S1 never holds.

$\dpi{100} S_2:S(1)\rightarrow \forall n S(2n + 1)$

S2 is always true. If S(1) is true then S(3), S(5), S(7)... etc will be true so if S(1) is true then S(m) is true for all odd m.

So, S2 always holds.

answered Sep 9, 2020 by (226,240 points)