We want to show, among other things, that if the number $\sin\alpha$ is a "constructible number," then it is possible to construct by ruler and compass an angle of measure $\alpha$. Note that "constructible" and "construct" have different meanings. The term "constructible" is fundamentally algebraic, while "construct" is purely geometric. This link between algebra and geometry is what makes it possible to prove, for example, that the $60^\circ$ angle is not ruler and compass trisectable.
We start with the assumption that we are given two points in the plane, distance $1$ apart. We need to prove the following lemma, which is not hard to prove, but requires a couple of pages of geometric work.
Lemma: Suppose that we can construct with ruler and compass two points that are distance $x$ apart, and also two points that are distance $y$ apart. Then we can construct two points that are (i) distance $x+y$ apart; (ii) distance $|x-y|$ apart; (iii) distance $xy$ apart; (iv) distance $\frac{x}{y}$ apart (if $y\ne 0$); (v) distance $\sqrt{xy}$ apart.
Now suppose that $\sin\alpha$ is a constructible number, where we assume $0\lt\sin\alpha\lt 1$. We want to construct the angle $\alpha$ between $0$ and $\frac{\pi}{2}$.
By the above lemma we can construct by ruler and compass points $P$ and $Q$ such that the distance from $P$ to $Q$ is $\sin\alpha$. Draw a circle with centre $P$ and radius $1$. Draw the line $\ell$ through $Q$ perpendicular to $Q$, and suppose the line $\ell$ meets the circle at point $R$. Then $\angle RPQ=\alpha$.
For the specific question you asked, which is easier, we need to work in the other direction. Suppose that we can construct, by ruler and compass, an angle $\alpha$, that is, a point $P$ and two rays through $P$ such that the angle between the rays is $\alpha$. Then (if $0\lt \alpha\lt \frac{\pi}{2}$) we can draw a right triangle $PQR$ which has $\alpha$ as one of its angles. Then $\sin\alpha=\frac{QR}{PR}$. Now we need to show that $QR$ and $PR$ are constructible numbers. For that we show that any segment $XY$ constructible by ruler and compass starting from a segment of length $1$ has length a constructible number. To do that we do an induction on the length of the construction. There are two facts of analytic geometry required. (a) Let $\ell_1$ be a line that passes through two points whose coordinates are constructible numbers, and let $\ell_2$ also be such a line. Then the coordinates of the intersection point of $\ell_1$ and $\ell_2$ are constructible numbers. (b) Let $\ell$ be a line with equation whose coefficients are constructible numbers, and $C$ a circle whose equation has constructible coefficients, then the points of intersection of $\ell$ and $C$ have constructible coordinates.
If you place one end of the compass at A and the other at B, you are allowed to draw either a circle centered at A or a circle centered at B - nothing else. You may be thinking that you can lift the compass off the plane and transfer it to another point while retaining the separation of the two points of the compass by the distance d. In classic straightedge-and-compass constructions, this is explicitly not allowed - the compass is considered to be 'collapsible', i.e the joint where the two arms of the compass meet is 'loose', so the moment you lift the compass, the plane is no longer able to hold the two arms at distance d.
In the case at hand: you can draw a line bisecting the angle between the X and Y axes, and mark the point P' where it intersects the circle about the origin of radius $\sqrt2$. It is not hard to see that this point is at a distance 1 from the X and Y axes, so you can find a point of distance 1 on (say) the Y axis by drawing a line parallel to the X axis through P'. You can then draw a circle of radius 1 centered at P', intersecting the first line at P''. A circle centered at the origin through P'' will then intersect the X axis at a distance $1 + \sqrt2$ from the origin.
Best Answer
There are many attractive geometric arguments for irrationality. Here is one for the Golden Ratio.
Construct a golden rectangle. A recipe for doing this can be found in Euclid, and was known to early Pythagoreans.
There should be a diagram to illustrate the idea. Perhaps you can draw it yourself. Say the golden rectangle is $ABCD$, with the vertices as usual enumerated counterclockwise, and let $AB$ be a long side of the rectangle.
We prove that the long side and the short side of a golden rectangle are incommensurable. Suppose to the contrary that sides $AB$ and $BC$ have a common measure $m$. Or else, in more modern language, suppose that $AB$ and $BC$ have lengths that are each an integer multiple of some common number $m$. Or, else, even more arithmetically, suppose that each side is an integer.
Cut off a square $AEFD$ from the rectangle, by finding the point $E$ on $AB$ such that $AE=AD$, and slicing straight up. That leaves a rectangle $EBCF$, which by the definition of golden rectangle, is itself golden. It is clear that $m$ is a common measure of the sides of $EBCF$.
Continue, by cutting off a square from $EBCF$, leaving an even smaller golden rectangle whose sides have common measure $m$. Clearly, this process can be continued forever. But after a while, each side of the little golden rectangle just produced will be less than the hypothesized common measure $m$, and we get our contradiction.
There has been speculation that this was the first irrationality proof. The only problem with this theory is the total lack of evidence.