Why is the square root of three an irrational number?

Answer 1

I'm assuming that you mean 'square root'. Yes, this sum is irrational. So are each of the two numbers alone. A simple proof can be done by writing x=square root 2 + square root 3 and then "squareing away" the square roots and then use the rational roots theorem. The sum or difference of two irrational number need not be irrational! Look at sqrt(2)- sqrt(2)=0 which is rational.

Answer 2

The square root of 30 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification.

Assume the square root of 30 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the square root of 30, its square must equal 30. That is,

(a/b)2 = 30

a2/b2 = 30

a2 = 30b2.

The right side is even, so the left side must be even also, that is, a2 is even. Since a2 is even, a is also even (because the square of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now,

(2c)2 = 30b2

4c2 = 30b2

2c2 = 15b2.

The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b2 cannot be odd; it must be even. Therefore b is even as well.

Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must be false, and the square root of 30 is irrational.

Answer 3

Square root of 3 cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that is rational so that we may write square root of 3=a/b in reduced from so 3b^2=a^2 If b is odd, the so is b^2 and so is 3b^2 so a^2 is also odd which implies a is odd as well. Similarly if b is even, b^2 is even and so is 3b^2 so that means a^2 is even which implies a is even as well. Since a/b is in reduced from, any ratio of a/b where both a and b are even is not in reduced form (we could cancel 2 from the fraction) so we assume a and b are both odd. Now since both are odd, for some integers m and n we can write a=m+1 b=n+1 Now plug this into the equation 3b^2=a^2 and we have 3(4n2 + 4n + 1) = 4m2 + 4m + 1 which simplifies to 6n2 + 6n + 1 = 2(m2 + m) The right hand side of this is clearly even, but the left hand side is odd. We conclude that there is no solution and square root of 3 is therefore irrational.
Yes

Answer 4

In order to know whether or not √3 is a subset of the Irrational Numbers, all we need to show is that √3 is itself irrational.

Let's start out with the basic inequality 1 < 3 < 4.

Now, we'll take the square root of this inequality:

1 < √3 < 2.

If you subtract all numbers by 1, you get:

0 < √3 - 1 < 1.

If √3 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √3. Therefore, √3n must be an integer, and n must be the smallest multiple of √3 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √3n by (√3 - 1). This gives 3n - √3n. Well, 3n is an integer, and, as we explained above, √3n is also an integer; therefore, 3n - √3n is an integer as well. We're going to rearrange this expression to (√3n - n)√3 and then set the term (√3n - n) equal to p, for simplicity. This gives us the expression √3p, which is equal to 3n - √3n, and is an integer.

Remember, from above, that 0 < √3 - 1 < 1.

If we multiply this inequality by n, we get 0 < √3n - n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √3p < √3n. We've already determined that both √3p and √3n are integers, but recall that we said n was the smallest multiple of √3 to yield an integer value. Thus, √3p < √3n is a contradiction; therefore √3 can't be rational and so must be irrational. So, √3 is a subset of the irrational numbers.

Q.E.D.