No, -6 is rational.
No. Sqrt(2)*sqrt(18) = 6.
Irrational numbers are infinitely dense. Between any two numbers, there are infinitely many irrational numbers. So if it was claimed that some irrational, x, was the closest irrational to 6, it is possible to find an infinite number of irrationals between 6 and x. Each one of these infinite number of irrationals would be closer to 6 than x. So the search for the nearest irrational must fail.
no
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.
no it is not irrational irrational is a neverending number like2.36945856235......... that goes on and on 6 over 7 is a rational number rational numbers are whole numbers natural numbers and integers.
No -6 can be expressed as a fraction (-6/1) and therefore cannot be irrational. Irrational numbers cannot be expressed as fractions.
No, 6 is rational.
It is an irrational number.
irrational
There are infinite irrational numbers between 1 and 6.
-18
it is irrational because it's from the exponential numbers
Irrational numbers are infinitely dense. That is to say, between any two irrational (or rational) numbers there is an infinite number of irrational numbers. So, for any irrational number close to 6 it is always possible to find another that is closer; and then another that is even closer; and then another that is even closer that that, ...
No. Sqrt(2)*sqrt(18) = 6.
It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.
No, 6 over 5 is not irrational. Irrational numbers cannot be expressed as a ratio of two integers. In this case, 6 over 5 simplifies to the rational number 1.2, which can be expressed as a ratio of 6 to 5.
4.6 is rational.