If a is rational then there exist integers p and q such that a = p/q where q>0.
Similarly, b = r/s for some integers r and s (s>0)
Then a*b = p/q * r/s = (p*r)/(q*s)
Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0
Thus a*b can be expressed as x/y where
p and r are integers implies that x = p*r is an integer
q and s are positive integers implies that y = q*s is a positive integer.
That is, a*b is rational.
No, and I can prove it: -- The product of two rational numbers is always a rational number. -- If the two numbers happen to be the same number, then it's the square root of their product. -- Remember ... the product of two rational numbers is always a rational number. -- So the square of a rational number is always a rational number. -- So the square root of an irrational number can't be a rational number (because its square would be rational etc.).
1) Adding an irrational number and a rational number will always give you an irrational number. 2) Multiplying an irrational number by a non-zero rational number will always give you an irrational number.
9.546 as an improper fraction in its simplest form is 4773/500 which proves that it is a rational number
You cannot, because the statements are false! (The second is rational only if r = 0).
A rational number is a number that can be written in the form a/b with a and b relatively prime integers - a and b are whole numbers with no common factors (eg if a=3 then b can't be 3,6,9,12,etc). Rational numbers have decimal representations that either terminate (like 3/4=0.75) or are infinitely recurring (like 1/9=0.1111111111... or 5/7=0.714285|714285|714285...). Irrational numbers (numbers that aren't rational) have infinite decimals that never repeat (like pi=3.1415926535..., e=2.7182818284590...). It is possible to prove that unless n is a square number, the square root of n is irrational - if n can't be written as m^2 then n^0.5 is irrational. Since you can't find a and b such that (a/b)^2=7 the square root of 7 is irrational. It should be noted that you can get as close as you like to 7^0.5 with rational numbers but you can never reach it exactly. See related question.
It must be a generalised rational number. Otherwise, if you select a rational number to multiply, then you will only prove it for that number.
No, they are not. 1/2 is a ratio of two integers and so it is rational. But it is not a whole number.
No, and I can prove it: -- The product of two rational numbers is always a rational number. -- If the two numbers happen to be the same number, then it's the square root of their product. -- Remember ... the product of two rational numbers is always a rational number. -- So the square of a rational number is always a rational number. -- So the square root of an irrational number can't be a rational number (because its square would be rational etc.).
If the number can be expressed as a ratio of two integer (the second not zero) then the number is rational. However, it is not always a simple matter to prove that if you cannot find such a representation, then the number is not rational: it is possible that you have not looked hard enough!
They are not. They are countably infinite. That is, there is a one-to-one mapping between the set of rational numbers and the set of counting numbers.
1+sqrt(2) and 1-sqrt(2) are both irrational but their sum, 2, is rational.
Yes. Not only that, but there are an infinite number of rationals between any two distinct rationals - however close. We can prove it like this: Take any three rational numbers, call them A, B and C, where B is larger than A, and C is any rational number greater than 1: D = A + (B - A) / C That gives us another rational number, D, no matter what the values of the original numbers are.
2 and 1/2 are rational numbers, but 2^(1/2) is the square root of 2. It is well known that the square root of 2 is not rational.
No. If we let x be irrational, then 0/x = 0 is a counterexample. However, if we consider nonzero rational numbers, then our conjecture is true. We shall prove this by contradiction. Suppose we have nonzero rational numbers x and y, and an irrational number z, such that x/z = y. Since z is not equal to 0, x = yz. Since y is not equal to 0, x/y = z. Since x/y is a quotient of rational numbers, x/y is rational. Therefore, z is rational, contradicting our assumption that z was irrational. QED.
1) Adding an irrational number and a rational number will always give you an irrational number. 2) Multiplying an irrational number by a non-zero rational number will always give you an irrational number.
Johann Heinrich Lambert
It's the ratio of 121 to 1, so it's rational. I think the square of any rational number is a rational number. In fact, I'm sure of it, because I know how I could prove it.