prove that root3 is irrational?

Answer 1

Proof by contradiction:

Assume that √(3) is rational. Then √(3) can be expressed as a quotient of integers m and n, such that

√(3) = m/n.

Let us also assume that m and n have no common factors and is an irreducible fraction. {If m and n did have a common factor, we can reduce it such that it does, so it is perfectly valid to make this assumption.}

Squaring both sides, we get

[√(3]² = (m²)/(n²)

But, squaring a square root eliminates the radical, so

3 = (m²)/(n²)

Multiply both sides by n²,

3n² = m²

This implies m² is a multiple of 3, which means m is a multiple of 3.

If m is a multiple of 3, then m can be expressed in the form
m = 3k (for some integer k). If that's the case, then, since

3n² = m²,
3n² = [3k]²
3n² = 9k². Divide both sides by 3,

n² = 3k²

This now implies n² is a multiple of 3, which means
n is a multiple of 3. As a result, both m and n are a multiple of 3. This is a contradiction because n and m have no common factors.

Therefore, √(3) is irrational.

Answer 2

We shall prove this by contradiction. Contradiction is a method used to prove statements by assuming the opposite of the given statement, and proving the opposite of the statement as false, thereby proving that the given statement is true.


Proof:
First, we shall define a rational number. It is any number of the form a/b where b is not 0, a and b are integers and a and b are coprime. Coprime means that their HCF is 1.

Then we need the information that is the square of any number a is divisible by any prime number p, the a is divisible by p. Take 2 as an example of a prime number. The statement means that if a^2 is divisible by 2, a is divisible by 2.

Assume sqrt 3 is rational.
sqrt 3 = a/b where a and b satisfy the conditions given above.

Square both sides

3 = a^2/b^2
a^2 = 3b^2.....(1)

3b^2 is obviously divisible by 3 as it is a multiple of 3.
And a^2 = 3b^2 = multiple of 3

Since a^2 is divisible by 3, a is divisible by 3.
a can be written as 3c for some integer c.

Substitute a = 3c in (1)
(3c)^2 = 3b^2
9c^2 = 3b^2
3c^2 = b^2

Since b^2 is divisible by 3, b is divisible by 3.

So, both a and b have 3 as a common factor.
But this contradicts the fact that a and b are coprime.

We have bumped into this contradiction since what we assumed is wrong. (sqrt 3 is not rational)

So, sqrt 3 is irrational

Answer 3

We'll prove this by contradiction. This means we'll assume that √3 is rational and come to a contradiction, which will prove that √3 is in fact irrational.

So let's assume √3 is rational. Then, by definition of rational numbers, we can write √3 as a fraction in its lowest terms:
√3 = a/b
and gcd(a,b) = 1.

Now square both sides:
3 = a²/b²
gcd(a², b²) = 1 because gcd(a,b) = 1 (if a and b have no prime factors in common, then a² and b² cannot have any common prime factors).

But 3 = a²/b² means a² = 3b² so
a²/b² = 3b²/b²
Now this fraction is obviously not in its lowest terms - the numerator and the denominator have b² in common!

This is a contradiction!

Hence √3 is irrational.

Hope this helps.

Answer 4

Consider the polynomial f(x)=x^2-3. The only possible rational roots (see my source) are -3,-1,1, and 3. But the values of f at these numbers is 6,-2,-2, and 6. Therefore f has no *rational* roots, which says that any root must be irrational.

Answer 5

In addition to these answers, the square root of any number that is not a perfect square in itself is irrational.

Answer 6

See dear any number containing (square root )is called irrational like 3,5,6,8,7,etc while numbers like 4,9 are rational as they do not have sq. root.