Solution 1: To prove for an integer.

Solution 1: To prove for an integer ,
Consider left hand side
Simplify further we get,
Hence, . This completes the proof.
Now, let’s simplify .
We know that,
Differentiating with respect to x we get,
Substitute we get,
Solution a: Suppose that is an integer such that . This implies
that
And
For some integers
Subtract equation (1) from equation (2) we get,
Suppose we get,
Since, and are the two consecutive integers and we know that product of two
consecutive integer is divisible by 2. That is must be multiple of 2, and hence
.
Solution b: To prove that for an even integer n.

By Euclid algorithm, . Since 2 is second smallest prime
number and n is even this implies that is odd. We know that the gcd of a prime
number and any number is 1.
Hence
Therefore,
Solution: Given that A relation is defined on R by is and only if for some
integer
To prove that is partial order.
We know that is partial order if the relation is reflexive, anti-symmetric and transitive.
Reflexive: For , is true for some integer k. hence relation R is reflexive.
Anti-symmetric: Suppose and this implies that
Hence, relation is Anti-symmetric
Transitive: Suppose . This implies that
Hence, the relation is Transitive.
Since, the given relation is reflexive, anti-symmetric and transitive. Therefore the relation
is partial order. This completes the proof.