Irrational numbers
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Transcript of Irrational numbers
Made by Karan
An irrational number is any real number that cannot be
expressed as a ratio a/b, where a and b are integers, with
b non-zero, and therefore not a rational number.
We have studied many properties of irrational numbers,
we have also located irrational numbers on the number
line and have seen that every irrational number
corresponds to a definite point on the number line. But
we have not prove that they are irrational. Here we shall
prove that if ‘p’ is a prime number then √p is irrational.
For this we take help of the Fundamental Theorem of
Arithmetic.
• Let ‘p’ be a prime number. If p divides a², then p divides a
where a is a positive integer.
• Proof:- Given a is a positive integer. So, let p1 , p2 ,p3, ..
pn be prime factor (not necessarily all distinct) of a.
• Then a=p1.p2.p3 ….pn
• Therefore, a2 =(p1 p1.p2.p3 ….pn) (p1 p1.p2.p3….pn)
• =p12.p2
2.p32 ….pn
2
• It is given that p divides a².
• So, p is a prime factor of a², from the fundamental
theorem of arithmetic.
• But the only prime factors of a² are p1 , p2 ,p3, .. Pn.
Therefore p is one factor of p1 , p2 ,p3, .. Pn. Hence p
divides a.
• In this method we assume a contradictory statement of what is
to be prove and come to a conclusion that contradicts the
hypothesis. And so the theorem is proved. For example:
• By the method of contradiction prove that √2 is irrational.
• Solution: We assume that √2 is rational. So, in its simplest
form, let √2= p/q, where p and q are integers having no
common factor other than 1 and q≠0.
• Then p=q √2 → p² = 2q²
• From (1) p² is divisible by 2, so p is divisible by 2.
• Let p= 2m→ p²=(2m)² = 2q²
• 4m²= 2q² →2m² = q²
• Hence, q² is divisible by 2 and consequently q is divisible
by 2. So , p is divisible by 2 as well as q is divisible by 2.
• Thus, 2 is a common factor of p and q which contradicts
the assumption, that p and q have no common factor other
than 1.
• This contradiction is due to the incorrect assumption that
√2 is rational. Hence √2 is irrational.
• Show that 5-√3 is irrational.
• Solution: Assume that 5-√3 is rational.
• That is, we can find co prime a and b (b≠0) such that 5-
√3 =a/b.
• Therefore, 5-a = √3
b
• Rearranging this equation, we get √3= 5-a = 5b-a
b b
• Since a and b are integers we get 5-a is rational and so
• b
• √3 is rational. But this contradict the fact that √3 is
rational.
• This contradiction has arisen due to incorrect assumption
that 5- √3 is irrational.
• So 5-√3 is irrational.
• So we conclude that:-
• The sum or difference of a rational and irrational number
is irrational.