## Frequently Asked Questions

#### Which constructions are impossible?

**Theorem**

Any geometrical construction involving straight-edge and compass that can be shown to be equivalent to a cubic equation containing rational coefficients and irrational roots is impossible.

**Proof**

Starting with a unit length it is possible to produce any integral length. From the work of Descartes we have seen how it is possible to construct multiples of any integral lengths. By using the multiplying method in reverse we were able to divide, hence we can produce any rational length. We shall say that constructed lengths of this type exist in the first number field, `F`_{1}.

By using a constructed length in `F`_{1} we are able to find the roots of quadratics or by using the square root construction method we can form any length of the form `a` + `k`√`b`, where `a`, `b` and `k` are rational values. Clearly any length of this form cannot be found in `F`_{1}, so we have created a new set of construction lengths. We shall say that these lengths exist in the second number field, `F`_{2}.

By a further application we shall form a new type of length √(`a` + `k`√`b`). This is, in effect, the second root of a second root, i.e. a fourth root. Also it will be noted that this length is part of a new number field, `F`_{3}; that is, it does not exist in `F`_{1} or `F`_{2}. By successive applications of the square root construction method we will produce new lengths that could not exist in previous number fields and so we continue to create new number fields.

Clearly we cannot perform an infinite number of construction steps, so the number of possible number fields must be finite. This point is important.

It should be clear that `F`_{1} contains rational lengths, `F`_{2} contains lengths that solve a quadratic (2nd roots) and `F`_{3} contains lengths that solve quartics (4th roots), i.e. a second root of a second root. This seems to suggest that a cube root (3rd root) does not exist in any of the number fields possible by construction. A further suggestion may be implied: any construction method that has an algebraic equivalence of a cubic nature cannot be completed with compass and straightedge.

The astute reader will realise that not all cubic equations yield cubic roots. Clearly this is true, the cubic equation, (`x` − 1)(`x` + 2)(`x` − ½) = 0, has roots 1, −2 and ½, and so we immediately dispel the assertion that any construction method that lends itself to the algebraic equivalence of a cubic equation cannot be completed; as these roots are all possible constructions. However we will now show that any construction, whose algebraic equivalent is a cubic with rational coefficients and its roots are not rational, is impossible.

Let us examine a cubic of the form `x`³ + `ax`² + `bx` + `c` = 0, where `a`, `b` and `c` are rational values and the roots of the equation are the non-rational values α_{1}, α_{2} and α_{3}.

Hence (`x` − α_{1})(`x` − α_{2})(`x` − α_{3}) = 0.

If the cubic equation had rational roots we would locate them in the first number field, `F`_{1}, but as the roots are not rational we need to construct into further number fields. By considering constructible lengths we know that `F`_{2} generates lengths of the form `a` + `k`√`b`. Any further number fields will be of the form `p` + `q`√`r`, where `p`, `q` and `r` are all lengths that exist in a previous number field.

Let us imagine that `F _{n}` is the least such number field in which a root of the cubic equation exists. Without loss of generality, let α

_{1}=

`p`+

`q`√

`r`.

Now if one of the roots, α_{1}, is of the form `p` + `q`√`r`, substituting back into the factorised form of the cubic equation we get,

(`x` − (`p` + `q`√`r`))(`x` − α_{2})(`x` − α_{3}) = 0

However, if we wish to expand this and regain rational coefficients, one of the other roots must be the conjugate of α_{1}, in order to eliminate √`r`. So let α_{2} = `p` − `q`√`r`.

∴ (`x` − (`p` + `q`√`r`))(`x` − (`p` − `q`√`r`))

= `x`² − (`p` + `q`√`r`)`x` − (`p` − `q`√`r`)`x` + (`p` + `q`√`r`)(`p` − `q`√`r`)

= `x`² − 2`px` + `p`² − `q`²`x`

Then,

(`x`² − 2`px` + `p`² − `q`²`x`)(`x` − α_{3})

= `x`³ + (−2`p` − α_{3})`x`² + (2`p`α_{3} − `q`²`r`)`x` + `q`²`r`α_{3}

Comparing this with `x`³ + `ax`² + `bx` + `c` = 0 we we deduce that `a` = −2`p` − α_{3}

Therefore α_{3} = −2`p` − `a`, which exists in one of the number fields already created. But note that it does not require the square root that α_{1} required to make its construction necessary. Hence it must exist in a number field before `F _{n}`. However we set out initially with the supposition that

`F`was the least such number field in which a root lies, so we encounter an unavoidable contradiction.

_{n}Hence we prove that any geometrical construction whose algebraic equivalence is a cubic equation, with rational coefficients and no rational roots, will be impossible.