Answer:
1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that  where
 where   and
 and  are non negative integers, and the fraction
 are non negative integers, and the fraction  is irreducible, i.e., the numbers
 is irreducible, i.e., the numbers  and
 and  have no common factors.
 have no common factors.
Now, squaring the equality at the beginning we get that
 (1)
 (1)
which is equivalent to  . From this we can deduce that 3 divides the number
. From this we can deduce that 3 divides the number  , and necessarily 3 must divide
, and necessarily 3 must divide  . Thus,
. Thus,  , where
, where  is a non negative integer.
 is a non negative integer.
Substituting  into (1), we get
 into (1), we get

which is equivalent to
 .
.
Thus, 3 divides  and necessarily 3 must divide
 and necessarily 3 must divide  . Hence,
. Hence,  where
 where  is a non negative integer.
 is a non negative integer. 
Notice that
 .
.
The above equality means that the fraction  is reducible, what contradicts our initial assumption. So,
 is reducible, what contradicts our initial assumption. So,  is irrational.
 is irrational.
2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So,  , which is equivalent to say that
, which is equivalent to say that  where
 where   and
 and  are non negative integers. Also, assume that
 are non negative integers. Also, assume that  . So, we want to prove that
. So, we want to prove that  . Recall that an integer
. Recall that an integer  can be written as
 can be written as 
 .
.
Then, 
 .
.
Notice that the product  is an integer. Thus, the fraction
 is an integer. Thus, the fraction  is a rational number. Therefore,
 is a rational number. Therefore,  .
.
3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have  is irrational and
 is irrational and  .
. 
Write  and let us suppose that
 and let us suppose that  is a rational number. So, we get that
 is a rational number. So, we get that 
 .
.
But the subtraction or addition of two rational numbers is rational too. Then, the number  must be rational too, which is a clear contradiction with our hypothesis. Therefore,
 must be rational too, which is a clear contradiction with our hypothesis. Therefore,  is irrational.
 is irrational.