Weak Induction (1) Using weak induction, prove that 3" < n! for all integers n > 6.
(2) Prove that log(n!) < n log(n) for all integers n > 1.
Reminder 1: log(1) = 0.
Reminder 2: log(a*b) = log(a) + log (6).
Reminder 3: If a < b then log(a) < log(6).
Note: The base of the logarithms doesn't matter for any of the above.
(3) Prove for all integers n > 1 that if A1, A2, ..., An and B are sets, then:.
(A1∩A2∩...∩An)UB = (A1UB)∩(A2UB)∩...∩(AnUB)
Hint: Use the fact that (XNY) UZ = (XUZ)n(Y UZ) where X, Y, and Z are sets.

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

when n=k for the true condition:

when n=k+1 it tests the value:

since k>6  hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

when n=k+1

In option 3:

when n=1:

when n=k

hence n=k+1 is true.

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