For every \(n\in\mathbb{N}\), \[\sum_{k=1}^{n} \frac{1}{k(k+1)}=1-\frac{1}{n+1}\]

Base case: \(n=1\) \[\frac{1}{1\cdot2}=\frac{1}{2}\] \[1-\frac{1}{1+1}=1-\frac{1}{2}=\frac{1}{2}\] So \(\sum_{k=1}^{n} \frac{1}{k(k+1)}=1-\frac{1}{n+1}\) for \(n=1\)

Induction step: \[\begin{align} \sum_{k=1}^{n+1} \frac{1}{k(k+1)} &= \sum_{k=1}^{n} \frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\ &= 1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}\\ &= 1-\frac{n+2-1}{(n+1)(n+2)}\\ &= 1-\frac{1}{n+2} \end{align}\]

|x+y| is less than or equal to |x|+|y| for all real numbers x and y

 Prf. :

Let x and y be real numbers. We can consider the following cases:

x and y are of the same 

If x and y are both positive, then the statement is automatically true, similar to the case when both x and y are negative

 one of x and y is negative

WLOG we let y = -n:

x-n <= x+n.  If x-n is negative then it will be less than x+n, showing it to be true. If it were positive, then we would have to be greater than x+n, which tells us that x is greater than 0. Thus completing the proof.

For all real numbers x and y, |x+y| is less than or equal to |x|+|y|.

If p and q are primes > 2, then pq+1 is not prime.

Prove \(p*q+1 \) is not prime

prime number p is defined as \(1|p\) &  \(p|p\) where \(p \in \mathbb {Z} \)

This also implies that a prime number is odd therefore any prime number can be written in the form:

\(2k+1 \) where \(k \in \mathbb {Z} \)

let \(p= 2a+1,    q = 2b+1 \) where   \(a,b \in \mathbb {Z} \)

\(p*q+1 = (2a+1)*(2b+1) +1 \) 

 \( (2a+1)*(2b+1) +1 = (4ab+2a+2b+1) +1 \)

 \( (4ab+2a+2b+1) +1 = (2(2ab+a+b)+1) +1  \)

\(2ab+a+b \in \mathbb{Z}\) so let \( m = 2ab+a+b \)

 \((2m+1) +1\)

this is an odd number + 1 which is an even number.

by deffiniton a prime number is not an even number therefore  \(pq+1\) will not be prime.

If x is irrational and r is rational, then: a) x+r is irrational b) If r ≠ 0, then xr is irrational

 a) \(x + r\) is irrational

    Proof:

    Let \(x\) be irrational and \(r \in Q\).

    By was of contradiction, prove  \(x + r\) is rational.

    Assume \(x + r = k\), where \(k ∈ Q\).

    Since \(r\) and \(k\) are rational numbers, \(r = a/b\), \(k = c/d\) where \(∃ a, b, c, d ∈ Z, b, d ≠ 0\).

    We can rewrite as \(x + a/b= c/d\) and simplify to \(x = c/d - a/b\) and further to \(x = (cb - ad)/db\). 

    This is a contradiction, \(x ≠ (cb - ad)/db\) since \(x\) is irrational and \((cb - ad)/db\) is a rational number (fraction).

    So \(x + r\) must be irrational

b) If \(r \ne 0, xr \) is irrational

    Proof.

    Let \(x \) be irrational and \(r \in Q\).

    By way of contradiction, \(xr\) is rational.

    So we can assume \(xr = a/b\), where \(a, b \in \mathbb{Z}, b ≠ 0\).

    Since \(r\) is rational, \(r = c/d\), where \(c, d \in Z, d \ne 0\).

    We can rewrite as \(xc/d = a/b\) and simplify to \(x = a/b / cd \) and further to \(x = ad/cd\).

    This is a contradiction, \(x \ne ad/cb\) since \(x\) is irrational and \(ad/cb\) is a rational number (fraction),

    So \(xr\) must be irrational.