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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.

Q: Prove that if a and b are rational numbers then a multiplied by b is a rational number?

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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).

4.56 = 456/100 = 114/25 or 414/25

Related questions

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.