After a lot of time spent trying to prove all five of Euclid's postulates, I now believe I have succeeded. I reproduce here my 'proofs' asking anyone to find a mistake in my use of rigour or logic, as I wish to verify my proofs.

As some of these proofs use other postulates, the proofs I am writing down here will be out of order. I start with the fifth postulate, as this is in most need of verification. I use Euclid's original version before going on to show that it is equally valid for Playfair's equivalent version.

Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

Proof: This means to say that the two lines meet on the side whose sum is less than 180 degrees.

From the diagram, we see that if we extend the lines indefinitely, we eventually get a triangle.

It has been proved that the sum of the angles of a triangle sum to 180 degrees.

This implies that the two angles formed by the third line which goes through the other two lines) cannot be equal to 180 degrees, as it would then violate the angle sum property of that triangle.

The triangle is only possible if the two angles are not equal to 180 degrees or more. This implies that the lines may only meet on those sides where the angles together sum up to less than 180 degrees.

Hence, Euclid's version is proved.

For Playfair's version, which states that given a lineland a pointP, there is one and only one line throughPparallel tol,

We take the line and the point. We must then construct a line from the point to the line perepndicular to linel. Now, apart from the special case where the line throughPis perpendicular to tol, we see that the question has now reverted to the Euclid's original version.

Now, any line whose angle with the perpendicular produces a sum of the interior angles on the same side which is less than 180 degrees must go throughl.

Only those lines whose angle when taken with the angle formed bylwith the perpendicular produce a sum of 180 degrees may not meet the linel.

An obtuse angle implies that the line meets on the other side, where the angle formed is acute, as in order to obey the rule that the angles on a line must be equal to 180 degrees, the other angle must always be acute. An acute angle implies that the line meets on that side, as an acute angle added to a right angle is always less than 180 degrees.

Therefore, this implies that no line which produces an obtuse or acute angle may be parallel tol. Therefore, the only other angle possible is a right angle, which taken in conjunction with the other interior angle forms 180 degrees, implying that any line which produces 90 degrees with the perpendicular is always parallel to linel.

As only one such line is possible, this proves Playfair's equivalent version.

I apologise if the language may make the text a little hard to understand.

Postulate 4: All right angles are equal to each other.

Proof: Let us assume that this is not true, and all right angles are not equal to each other.

This instantly leads to a contradiction, as it implies that a triangle may have more than one right angle.

Therefore, by reducio ad absurdum, we see that all right angles must be equal to each other.

Another proof is by looking at the definition of a right angle. A right angle is any angle equal to 90 degrees.

We know that 90 = 90 = 90 ...

We see therefore that all right angles are equal to 90 degrees and as 90 degrees is equal to 90 degrees,

this implies that all right angles are equal to each other.

Hence proved.

Postulate 2: A finite straight line may be extended indefinitely.

Proof: There are an infinite number of points in a region.

This implies that there are an infinite number of collinear points, as any operation with infinity that does not involve another infinity results in infinity. By collinear, I mean points between which a straight line may be drawn. ( I clarify this in order to prevent accusations of using a circular argument with the first postulate)

This implies that a line may be extended infinitely.

Postulate 3: A circle may be drawn with any center and any radius.

Note: By the term "collinear", I mean that it is possible to draw straight line from it another specific point.

Proof: This is a little trickier to prove, so I divided the problem down into two parts. I will first prove that a circle may have any radius.

Taking point A as centre, we may look at the radius as a line. By Postulate 2, we know that line may be extended indefinitely.

Therefore, the radius may be extended indefinitely.

This implies that a circle may have any radius.

The second part is to prove that a circle may have any center.

Taking any collinear point, we see that it is possible to draw a straight line between this and any other straight line.

By rotating the line by 360 degrees, we obtain a circle.

This implies that any collinear point may be the center of a circle, as the straight line that can be drawn may be considered a radius, and rotating the radius produces a circle.

Our next challenge is to show that all points are collinear, in order to fully prove this postulate. Fortunately, that is also the next postulate.

Postulate 1: A straight line may be drawn from any point to any other point.

Proof: Finally proved only yesterday, we must refer to the third and second postulate in order to fully prove this one. In order to prevent accusations of lack of rigor, I will use the still incomplete third postulate only in those cases where it may be applied.

Take any two collinear points A and B, where collinear means it is possible to draw a straight line between them. It is possible therefore to draw a straight line between them.

Now any points on the line AB must also be collinear, for otherwise a straight line could not have been drawn. Hence, it is also possible to draw a line from A to any point upon the line.

Now, let us rotate the line, such that the collinear point A is the centre of the circle so produced.

Now, it is possible to draw a straight line from A to any point in the circle. This is because the radius of the circle is a straight line, and upon rotation, it covers all the points in the circle, implying that a straight line can be drawn from all the points covered by the radius to the center of the circle, which is A.

Therefore all points in the circle are collinear to A i.e. they produce a straight line to A.

It is easy to show that all points in the plane are collinear: merely extend the radius infinitely, so the resultant circle encompasses the entire region.

Repeating the above for any point in the circle, we see that it is possible to draw a straight line from that point to any other point in its circle, and so on.

From the information above, we can deduce that all points are collinear to each other, or

It is possible to draw a straight line from any one point to any other point.

Hence proved.

I apologise to anyone who has had difficulty understanding me through my use of language, and I am willing to explain it again to anyone who has had difficulty understanding.

Thank you for taking the time to read this.