Find the factors of 582.
1 answer:
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Solution :
We have to prove that (De-Morgan's law)
Let then
and
and so and
.
Thus, and so
Hence, .........(1)
Now we will show that
Let ⇒
Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.
and
Therefore, and we have
.............(2)
From (1) and (2),
Hence proved.
Depending on the first term, the sequences
would work.
If the sequence began at 6,
If the sequence began at 1,
If the sequence began at 3,
would work.
If the sequence began at 6,
If the sequence began at 1,
If the sequence began at 3,