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Irrational Numbers
An irrational number is a number that can't be expressed by a fraction having integers in both its numerator and denominator. While their existence was once kept secret from the public for philosophical reasons, they are now well accepted, yet still surprisingly hard to prove on an individual basis. Please post all questions about irrational numbers, including the famous examples of π, e, and √2, into this category.
3,962 Questions
-1,0,or 1 the negative square root of four is negative two and the square root of four is two
Between two irrational numbers there is a rational number?
There are infinitely many rational numbers, not just one.
no, irrational numbers cannot be expressed as fractions. 16/1 is the fraction for 16.
Is irrational number can be decimal?
Any irrational number can be approximated by decimals. You can never write it exactly, since there are an infinite number of decimals, and these don't repeat.
Can an irrational number be represented on a number line?
Not normally because an irrational number can't be expressed as a fraction which can be represented on the number line.
What kind of number is the square root of -16?
Square root of -1 is an imaginary number (√-1 = i),
√-16 = √-1 * √16
√-16 = i * 4
√-16 = 4 i
Any number that you can completely write down on paper is rational.
You wrote (or typed) 67. Therefore it's rational.
Is the square root of a fraction a rational number?
Only if the square root of the numerator and the square root
of the denominator are both rational numbers.
Is it true that The difference of two rational numbers always a rational number?
Yes. The rational numbers are a closed set with respect to subtraction.
Is the square root of 14 an irrational number?
Yes, here's the proof.
Let's start out with the basic inequality 81 < 83 < 100.
Now, we'll take the square root of this inequality:
9 < √83 < 10.
If you subtract all numbers by 9, you get:
0 < √83 - 9 < 1.
If √83 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 √83. Therefore, √83n must be an integer, and n must be the smallest multiple of √83 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 √83n by (√83 - 9). This gives 83n - 9√83n. Well, 83n is an integer, and, as we explained above, √83n is also an integer, so 9√83n is an integer too; therefore, 83n - 9√83n is an integer as well. We're going to rearrange this expression to (√83n - 9n)√83 and then set the term (√83n - 9n) equal to p, for simplicity. This gives us the expression √83p, which is equal to 83n - 9√83n, and is an integer.
Remember, from above, that 0 < √83 - 9 < 1.
If we multiply this inequality by n, we get 0 < √83n - 9n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √83p < √83n. We've already determined that both √83p and √83n are integers, but recall that we said n was the smallest multiple of √83 to yield an integer value. Thus, √83p < √83n is a contradiction; therefore √83 can't be rational and so must be irrational.
Q.E.D.
Is the square route of 9 rational or irrational?
The square root of 9 is 3 which is a rational number
because it does not value
Zero is considered rational because it can still be expressed as a fraction
Why is the square root of 7 not a rational number?
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.
How are square roots and rational and irrational numbers related?
A rational number is one that can be expressed as a ratio of two integers, p/q where q > 0. In decimal form, it has a terminating or recurring representation.
An rational number is a number than cannot be expressed as a ratio of two integers. In decimal form, it has a infinitely long, non-recurring representation.
A square root can be either rational or irrational or neither.
Examples:
The square roots of 4 are -2 and +2, both rational.
The square roots of 2.25 are -1.5 and +1.5, both rational.
The square roots of 2 are -1.41421... and +1.41421, both irrational.
The square roots of -2 are -1.41421...*i and +1.41421*i, both imaginary.
How do you add rational numbers?
By finding their common denominator & adding the top numbers of-the fractions. : )
Hi, excuse my English it not my native language. Anyone can correct my text. Answer: because they internal motor and element come too hot. This kind of cushion are not ventiled and the motor are monted on bushing. Best regard! G.B. Qc Im sign in soon
No because 12 is a rational number that can be expressed as an improper fraction in the form of 12/1
Is the product of a rational and irrational number always irrational?
No, but the only exception is if the rational number is zero.
Yes, -0.7 is a rational number. It's the ratio of -7 to 10.
Any number that you can completely write down with digits is rational.
Is the square root of 200 a rational number?
First, simplify the number:
sqrt(200) = sqrt(2 x 100) = 10 x sqrt(2)
Because sqrt(2) is an irrational number, in this case, sqrt(200) is irrational.
