A number of the form $$\pm\sqrt{a}$$, where $$a$$ is a positive rational number which is not the square of another rational number, is called a pure quadratic surd. A number of the form $$a \pm \sqrt{b}$$, where $$a$$ is rational, and $$\sqrt{b}$$ is a pure quadratic surd, is sometimes called a mixed quadratic surd.

The two numbers $$a \pm \sqrt{b}$$ are the roots of the quadratic equation $x^{2} – 2ax + a^{2} – b = 0.$ Conversely, the equation $$x^{2} + 2px + q = 0$$, where $$p$$ and $$q$$ are rational, and $$p^{2} – q > 0$$, has as its roots the two quadratic surds $$-p \pm \sqrt{p^{2} – q}$$.

The only kind of irrational numbers whose existence was suggested by the geometrical considerations of § 3 are these quadratic surds, pure and mixed, and the more complicated irrationals which may be expressed in a form involving the repeated extraction of square roots, such as $\sqrt{2} + \sqrt{2 + \sqrt{2}} + \sqrt{2 + \sqrt{2 + \sqrt{2}}}.$

It is easy to construct geometrically a line whose length is equal to any number of this form, as the reader will easily see for himself. That irrational numbers of these kinds only can be constructed by Euclidean methods ( by geometrical constructions with ruler and compasses) is a point the proof of which must be deferred for the present.1 This property of quadratic surds makes them especially interesting.

Example VII

1. Give geometrical constructions for $\sqrt{2},\quad \sqrt{2 + \sqrt{2}},\quad \sqrt{2 + \sqrt{2 + \sqrt{2}}}.$

2. The quadratic equation $$ax^{2} + 2bx + c = 0$$ has two real roots2 if $$b^{2} – ac > 0$$. Suppose $$a$$, $$b$$, $$c$$ rational. Nothing is lost by taking all three to be integers, for we can multiply the equation by the least common multiple of their denominators.

The reader will remember that the roots are $$\{-b \pm \sqrt{b^{2} – ac}\}/a$$. It is easy to construct these lengths geometrically, first constructing $$\sqrt{b^{2} – ac}$$. A much more elegant, though less straightforward, construction is the following.

Draw a circle of unit radius, a diameter $$PQ$$, and the tangents at the ends of the diameters.

Take $$PP’ = -2a/b$$ and $$QQ’ = -c/2b$$, having regard to sign.3 Join $$P’Q’$$, cutting the circle in $$M$$ and $$N$$. Draw $$PM$$ and $$PN$$, cutting $$QQ’$$ in $$X$$ and $$Y$$. Then $$QX$$ and $$QY$$ are the roots of the equation with their proper signs.4

The proof is simple and we leave it as an exercise to the reader. Another, perhaps even simpler, construction is the following.

Take a line $$AB$$ of unit length. Draw $$BC = -2b/a$$ perpendicular to $$AB$$, and $$CD = c/a$$ perpendicular to $$BC$$ and in the same direction as $$BA$$. On $$AD$$ as diameter describe a circle cutting $$BC$$ in $$X$$ and $$Y$$. Then $$BX$$ and $$BY$$ are the roots.

3. If $$ac$$ is positive $$PP’$$ and $$QQ’$$ will be drawn in the same direction. Verify that $$P’Q’$$ will not meet the circle if $$b^{2} < ac$$, while if $$b^{2} = ac$$ it will be a tangent. Verify also that if $$b^{2} = ac$$ the circle in the second construction will touch $$BC$$.

4. Prove that $\sqrt{pq} = \sqrt{p} \times \sqrt{q},\quad \sqrt{p^{2}q} = p\sqrt{q}.$

## 14. Some theorems concerning quadratic surds.

Two pure quadratic surds are said to be similar if they can be expressed as rational multiples of the same surd, and otherwise to be dissimilar. Thus $\sqrt{8} = 2\sqrt{2},\quad \sqrt{\tfrac{25}{2}} = \tfrac{5}{2}\sqrt{2},$ and so $$\sqrt{8}$$, $$\sqrt{\frac{25}{2}}$$ are similar surds. On the other hand, if $$M$$ and $$N$$ are integers which have no common factor, and neither of which is a perfect square, $$\sqrt{M}$$ and $$\sqrt{N}$$ are dissimilar surds. For suppose, if possible, $\sqrt{M} = \frac{p}{q}\sqrt{\frac{t}{u}},\quad \sqrt{N} = \frac{r}{s}\sqrt{\frac{t}{u}},$ where all the letters denote integers.

Then $${\sqrt{MN}}$$ is evidently rational, and therefore (Ex. II. 3) integral. Thus $$MN = P^{2}$$, where $$P$$ is an integer. Let $$a$$, $$b$$, $$c, \dots$$ be the prime factors of $$P$$, so that $MN = a^{2\alpha} b^{2\beta} c^{2\gamma}\ \dots,$ where $$\alpha$$, $$\beta$$, $$\gamma, \dots$$ are positive integers. Then $$MN$$ is divisible by $$a^{2\alpha}$$, and therefore either (1) $$M$$ is divisible by $$a^{2\alpha}$$, or (2) $$N$$ is divisible by $$a^{2\alpha}$$, or (3) $$M$$ and $$N$$ are both divisible by $$a$$. The last case may be ruled out, since $$M$$ and $$N$$ have no common factor. This argument may be applied to each of the factors $$a^{2\alpha}$$, $$b^{2\beta}$$, $$c^{2\gamma}, \dots$$, so that $$M$$ must be divisible by some of these factors and $$N$$ by the remainder. Thus $M = P_{1}^{2},\quad N = P_{2}^{2},$ where $$P_{1}^{2}$$ denotes the product of some of the factors $$a^{2\alpha}$$, $$b^{2\beta}$$, $$c^{2\gamma}, \dots$$ and $$P_{2}^{2}$$ the product of the rest. Hence $$M$$ and $$N$$ are both perfect squares, which is contrary to our hypothesis.

If $$A$$, $$B$$, $$C$$, $$D$$ are rational and $A + \sqrt{B} = C + \sqrt{D},$ then either $$A = C$$, $$B = D$$ or $$B$$ and $$D$$ are both squares of rational numbers.

For $$B – D$$ is rational, and so is $\sqrt{B} – \sqrt{D} = C – A.$ If $$B$$ is not equal to $$D$$ (in which case it is obvious that $$A$$ is also equal to $$C$$), it follows that $\sqrt{B} + \sqrt{D} = (B – D)/(\sqrt{B}- \sqrt{D})$ is also rational. Hence $$\sqrt{B}$$ and $$\sqrt{D}$$ are rational.

If $$A + \sqrt{B} = C + \sqrt{D}$$, then $$A – \sqrt{B} = C – \sqrt{D}$$ (unless $\sqrt{B}$ and $\sqrt{D}$ are both rational).

Example VIII

1. Prove ab initio that $$\sqrt{2}$$ and $$\sqrt{3}$$ are not similar surds.

2. Prove that $$\sqrt{a}$$ and $$\sqrt{1/a}$$, where $$a$$ is rational, are similar surds (unless both are rational).

3. If $$a$$ and $$b$$ are rational, then $$\sqrt{a} + \sqrt{b}$$ cannot be rational unless $$\sqrt{a}$$ and $$\sqrt{b}$$ are rational. The same is true of $$\sqrt{a}- \sqrt{b}$$, unless $$a = b$$.

4. If $\sqrt{A} + \sqrt{B} = \sqrt{C} + \sqrt{D},$ then either (a) $$A = C$$ and $$B = D$$, or (b) $$A = D$$ and $$B = C$$, or (c) $$\sqrt{A}$$, $$\sqrt{B}$$, $$\sqrt{C}$$, $$\sqrt{D}$$ are all rational or all similar surds. [Square the given equation and apply the theorem above.]

5. Neither $$(a + \sqrt{b})^{3}$$ nor $$(a – \sqrt{b})^{3}$$ can be rational unless $$\sqrt{b}$$ is rational.

6. Prove that if $$x = p + \sqrt{q}$$, where $$p$$ and $$q$$ are rational, then $$x^{m}$$, where $$m$$ is any integer, can be expressed in the form $$P + Q \sqrt{q}$$, where $$P$$ and $$Q$$ are rational. For example, $(p + \sqrt{q})^{2} = p^{2} + q + 2p\sqrt{q},\quad (p + \sqrt{q})^{3} = p^{3} + 3pq + (3p^{2} + q)\sqrt{q}.$ Deduce that any polynomial in $$x$$ with rational coefficients ( any expression of the form $a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n},$ where $$a_{0}$$, … $$a_{n}$$ are rational numbers) can be expressed in the form $$P + Q\sqrt{q}$$.

7. If $$a + \sqrt{b}$$, where $$b$$ is not a perfect square, is the root of an algebraical equation with rational coefficients, then $$a – \sqrt{b}$$ is another root of the same equation.

8. Express $$1/(p + \sqrt{q})$$ in the form prescribed in Ex. 6. [Multiply numerator and denominator by $$p – \sqrt{q}$$.]

9. Deduce from Exs. 6 and 8 that any expression of the form $$G(x)/H(x)$$, where $$G(x)$$ and $$H(x)$$ are polynomials in $$x$$ with rational coefficients, can be expressed in the form $$P + Q\sqrt{q}$$, where $$P$$ and $$Q$$ are rational.

10. If $$p$$, $$q$$, and $$p^{2} – q$$ are positive, we can express $$\sqrt{p + \sqrt{q}}$$ in the form $$\sqrt{x} + \sqrt{y}$$, where $x = \tfrac{1}{2}\{p + \sqrt{p^{2} – q}\},\quad y = \tfrac{1}{2}\{p – \sqrt{p^{2} – q}\}.$

11. Determine the conditions that it may be possible to express $$\sqrt{p + \sqrt{q}}$$, where $$p$$ and $$q$$ are rational, in the form $$\sqrt{x} + \sqrt{y}$$, where $$x$$ and $$y$$ are rational.

12. If $$a^{2} – b$$ is positive, the necessary and sufficient conditions that $\sqrt{a + \sqrt{b}} + \sqrt{a – \sqrt{b}}$ should be rational are that $$a^{2} – b$$ and $$\frac{1}{2}\{a + \sqrt{a^{2} – b}\}$$ should both be squares of rational numbers.

1. See Ch. II, .↩︎
2. I.e. there are two values of $$x$$ for which $$ax^{2} + 2bx + c = 0$$. If $$b^{2} – ac < 0$$ there are no such values of $$x$$. The reader will remember that in books on elementary algebra the equation is said to have two ‘complex’ roots. The meaning to be attached to this statement will be explained in Ch. III.When $$b^{2} = ac$$ the equation has only one root. For the sake of uniformity it is generally said in this case to have ‘two equal’ roots, but this is a mere convention.↩︎
3. The figure is drawn to suit the case in which $$b$$ and $$c$$ have the same and $$a$$ the opposite sign. The reader should draw figures for other cases.↩︎
4. I have taken this construction from Klein’s Leςons sur certaines questions de géométrie élémentaire (French translation by J. Griess, Paris, 1896).↩︎