Answer:
1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that 
where 
and 
are non negative integers, and the fraction 
is irreducible, i.e., the numbers and have no common factors.
Now, squaring the equality at the beginning we get that
(1)
which is equivalent to 
. From this we can deduce that 3 divides the number 
, and necessarily 3 must divide . Thus, 
, where 
is a non negative integer.
Substituting into (1), we get
which is equivalent to
.
Thus, 3 divides 
and necessarily 3 must divide . Hence, 
where 
is a non negative integer.
Notice that
.
The above equality means that the fraction is reducible, what contradicts our initial assumption. So, 
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 
where and are non negative integers. Also, assume that 
. So, we want to prove that 
. Recall that an integer 
can be written as
.
Then,
.
Notice that the product 
is an integer. Thus, the fraction 
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 
.
Write 
and let us suppose 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, 
is irrational.
