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Not necessarily. The cube roots of 4, 6 and 9 are all irrational (and different). But their product is 6, not just rational, but an integer.

Q: If three different irrational numbers are multiplied is the result an irrational number?

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Any irrational number does.

You can't get an irrational number by applying a finite number of additions, subtractions, multiplications and divisions to a set of rational numbers.

Yes, unless the rational number is 0.

The product will also be irrational.

The real numbers are divided into rational numbers and irrational numbers.

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Every irrational number, when multiplied by 0.5 will produce an irrational number.

Every irrational number, when multiplied by 0.4 will produce an irrational number.

Any irrational number multiplied by 0.5 will remain irrational. Any rational number multiplied by 0.5 will remain rational.

Any number which is irrational will produce an irrational number when multiplied by 0.5

At least one of the factors has to be irrational.* An irrational number times ANY number (except zero) is irrational. * The product of two irrational numbers can be either rational or irrational.

Every irrational number, when multiplied by 0.4 will produce an irrational number.

Some irrational numbers can be multiplied by another irrational number to yield a rational number - for example the square root of 2 is irrational but if you multiply it by itself, you get 2 - which is rational. Irrational roots of numbers can yield rational numbers if they are raised to the appropriate power

Every irrational number, when multiplied by 0.4 will produce an irrational number.

Any irrational number, when multiplied by 0.5 will give an irrational number.

Any irrational number will do.

Sometimes yes, sometimes no. sqrt(2) x sqrt(8) = 4 rational sqrt(2) x sqrt(3) = sqrt(6) irrational

rational * irrational = irrational.