Prove that negative of an irrational number is an irrational number?

Answer 1

Assume = a/b with positive integers a und b. Now, for some natural number n define the functions f and F as follows. Strictly speaking, f and F should each have n as an index as they depend on n but this would render things unreadable; remember that n is always the same constant throughout this proof. Let f(x) = xn(a-bx)n/n! and let F(x) = f(x) + ... + (-1)jf(2j)(x) + ... + (-1)nf(2n)(x) where f(2j) denotes the 2j-th derivative of f. Then f and F have the following properties: f is a polynomial with coefficients that are integer, except for a factor of 1/n! f(x) = f(-x) 0

Answer 2

Try it by contradiction.

Suppose that z is irrational, but -z is rational. That means that -z = p/q for some integers p and q. That yields a contradiction because z= -(-z) = ????

Answer 3

Let's start out with the basic inequality 4 < 5 < 9.

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

2 < √5 < 3.

If you subtract all numbers by 2, you get:

0 < √5 - 2 < 1.

If √5 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 √5. Therefore, √6n must be an integer, and n must be the smallest multiple of √5 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 √5n by (√5 - 2). This gives 5n - 2√5n. Well, 5n is an integer, and, as we explained above, √5n is also an integer; therefore, 5n - 2√5n is an integer as well. We're going to rearrange this expression to (√5n - 2n)√5 and then set the term (√5n - 2n) equal to p, for simplicity. This gives us the expression √5p, which is equal to 5n - 2√5n, and is an integer.

Remember, from above, that 0 < √5 - 2 < 1.

If we multiply this inequality by n, we get 0 < √5n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √5p < √5n. We've already determined that both √5p and √5n are integers, but recall that we said n was the smallest multiple of √5 to yield an integer value. Thus, √5p < √5n is a contradiction; therefore √5 can't be rational and so must be irrational.

Q.E.D.

Answer 4

The first proof of the irrationality of PI was written by Lambert in 1770 and was later published by Adrien-Marie Legendre in "Elements de Geometrie"

Answer 5

According to Wikipedia, "This was proved in 1768 by Johann Heinrich Lambert."