If you divide a rational number by an irrational number, or vice versa, you will ALMOST ALWAYS get an irrational result. The sole exception is if you divide zero (which is rational) by any irrational number.

Irrational numbers are non-repeating, non-terminating decimals. If your number repeats, it's rational. If it doesn't, it's irrational.

Rational

rational

It is rational.

Any number that has a digit, or group of digits, that repeat forever is rational.

NO !!!

However, the square root of '5' is irrational

5^(1/2) = 2.236067978...

Casually an IRRATIONAL NUMBER is one where the decimals go to infinity and there is no regular order in the decimal numbers.

pi = 3.141592.... It the most well known irrational number.

However, 3.3333.... Is NOT irrational because there is a regular order in the decimals.

Here is a definitive statement of irrational numbers. Irrational numbers are real numbers that cannot be represented as simple fractions. An irrational number cannot be expressed as a ratio, such as p/q, where p and q are integers, q≠0. It is a contradiction of rational numbers. Irrational numbers are usually expressed as R\Q, where the backward slash symbol denotes ‘set minus’. It can also be expressed as R – Q, which states the difference between a set of real numbers and a set of rational numbers.

Irrational

0.33333333333 is irrational, with the numeral 3 going on and on in that decimal.

rational

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

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.

rational

is it

There is no prefix.

'Quarante' is the french word for forty.

Negative pi

peenutbudr

No because integers are whole numbers

The square roots of 36 are +6 & -6

Remember +6 x + 6 = (+)36

Also

-6 x -6 = (+)36 Double minus becomes positive, so square roots must show both plus and minus answers.

= +36

It is rational because it can be expressed as a fraction which is 21/5

They never repeat in any pattern.

If they never repeated, you could have at most 10 digits after the decimal point and therefore the decimal representation would be terminating.

Yes, and here's the proof:

Let's start out with the basic inequality 4 < 7 < 9.

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

2 < √7 < 3.

If you subtract all numbers by 2, you get:

0 < √7 - 2 < 1.

If √7 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 √7. Therefore, √7n must be an integer, and n must be the smallest multiple of √7 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 √7n by (√7 - 2). This gives 7n - 2√7n. Well, 7n is an integer, and, as we explained above, √7n is also an integer; therefore, 7n - 2√7n is an integer as well. We're going to rearrange this expression to (√7n - 2n)√7, and then set the term (√7n - 2n) equal to p, for simplicity. This gives us the expression √7p, which is equal to 7n - 2√7n, and is an integer.

Remember, from above, that 0 < √7 - 2 < 1.

If we multiply this inequality by n, we get 0 < √7n - 2n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √7p < √7n. We've already determined that both √7p and √7n are integers, but recall that we said n was the smallest multiple of √7 to yield an integer value. Thus, √7p < √7n is a contradiction; therefore √7 can't be rational, and so must be irrational.

Q.E.D.