Whether the cube root of 36 is an irrational number

Be. I'll prove it:

If the cube root of 36 is a rational number, let it be equal to a b (a and b are both natural numbers and coprime), then a 3 = 36 * b 3, and it is easy to know that a is an even number.

Let a=2c, then 2*c 3=9*b 3, and it is easy to know that a is also an even number.

Contradicts "A and B are co-prime".

Be. ives10 You are talking nonsense, irrational numbers refer to infinite non-cyclic decimals, and irrational numbers are not meaningless! Have you ever studied math?

Not. Irrational numbers, meaningless numbers are called irrational numbers, such as the number of negative numbers under the root number, the cube root of 36 is obviously meaningful, do not consider the number after the decimal point to be irrational if it is not infinitely looping.

To make a mistake, the above definition is the definition of an imaginary number. This number is a real number, and the real number is divided into rational and irrational numbers, and the cubic root of 36 is an irrational number.

Be. Irrational numbers refer to infinite non-cyclic decimals, not imaginary numbers. Irrational numbers are a part of real numbers, which are divided into rational numbers and irrational numbers. The cube root of 36 is an infinite non-cyclic decimal number, such as pai, and the number ...... that cannot be squared

Be. Because 36 is not a perfect cubic number, the cubic root of 36 is an irrational number.

Although I was only in the sixth grade of elementary school, my sixth sense told me.

Yes, yes.

Although I was only in the second year of junior high school, my calculations told me yes. Be.

Be. Be. Be. Be.

Be. Be. Be.

Yes, of course! It can be proved by counter-evidence Oh......

The cube root of 36 is.

is an irrational number.

1) Any number has a cubic root, the cubic root of 0 is 0, and what is said on the high ceiling is wrong.

2) Irrational numbers are infinite non-cyclic decimals, as long as they are full of Qi Yin, this condition is sufficient, and they are all irrational numbers.

3) Any real number to the power of 0 is equal to 1What was said upstairs was also wrong.

The cube root of 16 is an irrational number. Irrational numbers, also known as infinite non-cyclic decimals, cannot be written as a ratio of two integers. If the state writes it as a decimal place, there are an infinite number of numbers after the decimal point, and it does not circulate.

Common irrational numbers include the square root of a non-perfect square number, and e (where the latter two are transcendent numbers), etc.

If the cube of a number is equal to a, then the number is called the cubic root of a, also known as the cubic root. That is, if x = a, then x is called the cube root of a. Note:

The root index 2 in the square root can be omitted and not written early, but the root index 3 in the cube root cannot be omitted.

Counter-evidence. Suppose there is an irrational number x whose square root is the rational number y, then y 2=y*y=x

Because the family shouts that the product of 2 rational numbers is still a rational number, then x is the number of rational families, which contradicts the hypothesis.

So the square root of an irrational number must be an irrational number.

The cube root of 26 is not a rational number, but an irrational number. This is because the cube root collapse of 26 cannot be expressed as the ratio of two integers, that is, it cannot be expressed as a rational number.

Counterproofs can be used to prove that the cube root of 26 is not a rational number. Suppose that the cube root of 26 is a rational number, i.e., it can be expressed as the ratio of two integers, i.e. 26 = p q, where p and q are integers, and p and q have no common factor. Then, we can get 26 = p q) 3 = p 3 q 3, i.e. p 3 = 26q 3.

This means that p 3 is a multiple of 26 and therefore p is also a factor of 26. However, 26 has only two factors, 1 and 26, and this p can only be 1 or 26. If p=1, then there is 1 3=26q 3, that is, the cube of q must be a factor of 1, but the cube of 1 has only 1, so q=1, which contradicts the assumption that p and q have no common factor.

In the same way, if p = 26, then there is 26 3 = 26q 3, that is, the cube of q must be a factor of 26, but the cube of 26 only has two factors 1 and 26 2, so q can only be 1 or 26 2, which also contradicts the assumption that p and q have no common factor. Therefore, if the assumption is not true, the cube root of 26 is not a rational number, but an irrational number.

To sum up, the cube root of 26 is not a rational number, but an irrational number.

Yes, the cube root of 26 is a rational number. The cubic root is the 3rd root of a number, that is to say, the vertical band height of a number is equal to the stupid deed itself multiplied by its own multiplication, i.e., a 3=a*a*a.

Rational numbers are all numbers that can be represented by a finite number of integers or fractions, and the cube root of a rational number is also a rational number. The cube root of 26 can be represented by a finite number of integers or fractions, and its cube root is also a rational number. The cube root of 26 is approximately equal to.

The cube root of 26 is not a rational number. As can be seen from the question, the result of the square root of Lifeng Town of 26 belongs to the infinite non-volcanic cyclic decimals, which are irrational numbers, so they are not rational numbers.

Irrational numbers are infinite non-circular decimals, and the number of digits is naturally not exhausted, but I always have to have a certain degree of precision, which requires a limited number of digits. So they can be expressed as a decimal of a locating number and then an open operation. According to the shout and according to the is:

a+b) 2=a 2+2ab+b 2=a 2+(2a+b)*bb=[(a+b) 2-a 2] (2a+b) Generally in junior high school, just do so.

The sqrt (3) is the root of the three cherry blossoms.

In the counter-proof method, sqrt(3) is a rational number, which is counted as p q, p, q coprime.

So 3=p*p q*q

3q*q=p*p

So p is a multiple of 3 and counts as 3l

Substitution has q*q=3*l*l

Therefore, q is also a spine multiple of 3, which contradicts p p and q.

First of all, we must know what irrational numbers are.

Irrational numbers, i.e., real numbers that are not rational numbers, cannot be written as a ratio of two integers. If you write it as a decimal form, there are an infinite number of numbers after the decimal point and it does not circulate.

After 3 open squared, after the decimal point is an infinite non-cyclic decimal number, so that is: why is the square root of 3 open irrational number.

You use the counter-proof method, assuming that it is a rational number, and you can eventually get a contradiction, which is similar to proving that the root number 2 is an irrational number.

There is no rational number that is squared to 3