Why the cube root of two is no existing number on the real number line?

11 Answers

  • Jim
    Lv 7
    4 weeks ago

    Cube root are "closed" (always possible) in Real Numbers.

  • sepia
    Lv 7
    4 weeks ago

    Given a number x, the cube root of x is a number a such that a^3 = x. 

    If x positive a will be positive, if x is negative a will be negative. 

  • 4 weeks ago

    Let x=2^(1/3)=1.2599210... is an irrational number, it must exist in the real

    axis theoretically & 1<x<1.26. Not like sqr(2) or sqr(3) which individually can

    be represented by a line segment & marked on the real axis. Maybe, x could

    be, I am not certain.

  • 4 weeks ago

    The real cube root of 2 = 1.2599210498948731647.... That's certainly on the real number line.

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  • 4 weeks ago

    In mathematics, the cube root of a number x is a number y such that y^3 = x. 

    All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots.

  • 4 weeks ago

    2¹/³ => 2¹/³[cos2nπ + isin2nπ]¹/³

    so, 2¹/³[cos(2nπ/3) + isin(2nπ/3)]

    With n = 0, we have, 2¹/³[cos(0) + isin(0)] = 2¹/³

    With n = 1, we have, 2¹/³[cos(2π/3) + isin(2π/3)]

    i.e. 2¹/³(-1/2 + i√3/2) => -0.63 + 1.09i

    With n = 2, we have, 2¹/³[cos(4π/3) + isin(4π/3)]

    i.e. 2¹/³(-1/2 - i√3/2) => -0.63 - 1.09i

    So 2¹/³ = 1.26 is indeed a real cube root of 2

    The other two cube roots are imaginary.


  • 4 weeks ago

    I'm not sure what you mean by that, the real-valued cube root of 2 is definitely a real number. There are also two more which are complex if you'd like?

  • 4 weeks ago

    ∛2 e⁰ ≈ 1.25992 (real, principal root)

    ∛2 e^((2 i π)/3) ≈ -0.6300 + 1.0911 i

    ∛2 e^(-(2 i π)/3) ≈ -0.6300 -1.0911 i

  • 4 weeks ago

    There are 3 cube roots of 2. One is real (but irrational) and the other two are complex.

    The principal cube root is real. Remember just because a number is irrational (can't be represented as the ratio of two integers) doesn't mean it isn't real. It is real and on the number line. It's between 1 and 2 which can be shown as follows:

    1 < 2 < 8

    ∛1 < ∛2 < ∛8

    1 < ∛2 < 2

    Btw, the link below contains a proof that the (principal) cube root of 2 is irrational, if you need that.

  • Baal
    Lv 6
    4 weeks ago

    My ancient Scientific Calculator says the cubed root of 2 is 1.25992105 and so does this online calculator.


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