« ΠροηγούμενηΣυνέχεια »
8a. (Not given by Euclid, but assumed by him). Lines and angles which are equal to one another, coincide with one another.
9. The whole is greater than its part.
10. Two straight lines cannot enclose a space. 11. All right angles are equal to each other.
12. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles that are less than two right angles (See page 100, Part III.).
A CHAIN OF SYLLOGISMS.
The conclusion of one syllogism may become the premiss of a second syllogism. The conclusion of the second syllogism may in its turn become the premiss of a third, and so on. Such a Chain of Syllogisms is used in proving the truth of the following proposition.
Given. The line AB equal to the line CD.
Required. To prove that if the line A B be placed upon the line CD, so that the point A falls upon C, then point B will coincide with point D.
Proof.-Let the point A fall on C, and the line AB on CD.
Lines which are equal coincide (Axiom 8a).
(a).. AB and C D coincide (1st conclusion).
When lines coincide their points coincide (assumed as an axiom).
A B and CD coincide (1st conclusion (a) become a premiss).
(b). The points of A B and C D coincide (2nd conclusion).
The points of AB and CD coincide (2nd conclusion (b) become a premiss).
B is a point in AB, and D is a point in CD (self-evident).
(c).. B coincides with D (conclusion required).
Here you see that which was given became a premiss of the 1st syllogism. The conclusion drawn from that syllogism became a premiss of the 2nd syllogism; and so on, step by step, till we reached the conclusion required.
But, as we have already seen, it is not necessary always to give both premisses of each syllogism. One of the two may be left out, but must always be understood to make the conclusion true.
Thus, in the first syllogism, we might leave out the major premiss (lines which are equal coincide), and merely give the minor premiss and the conclusion.
EXERCISES.-I. Write out the proof of the last proposition, omitting, in each syllogism, the premiss which might be left to be understood.
II. Write out the whole of the proposition, leaving out premisses, as above, and using the letters affixed to these lines instead of A, B, C, D.
PROBLEM (Euclid I. 3).
Definitions. A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal. And that point is called the centre of the circle. General Enunciation.
From the greater of two given straight lines to cut off a part equal to the less. Particular Enunciation. Given. Two straight lines, A B and C, of which AB is greater (that is, longer) than C. Required. To cut off from A B a part equal to C.
Construction.-From the point 4, in the line AB,
draw the line AD equal
to C. (Euc. I. 2 shows A
how to do this.)
From the centre A, at the distance A D, describe the circle D E F, cutting A D at E.
The line A E shall be equal to C.
(If AE be equal to C, we have done what was required that is, cut off from A B a part equal to C. But it is necessary to prove that A E is equal to C.)
Ist Proof (with syllogisms given in full).
Ist Syllogism.-All lines drawn from the centre of a circle to the circumference are equal. (Definition of circle).
A is the centre of a circle, and A E and AD are drawn from it to the circumference (Construction). (a). A E and AD are equal. (1st conclusion.) 2nd Syllogism.—(a) AE is equal to AD. (1st conclusion become a premiss.)
C is equal to AD (by construction).
(8).·. AE and C are each equal to A D. (2nd conclusion.)
3rd Syllogism.—(b) A E and Care each equal to A D. (2nd conclusion become a premiss.) Things equal to the same thing are equal to one another. (Axism I.)
.. AE is equal to C. (Conclusion wanted.)
Result.-Wherefore from A B, the greater of two given straight lines, a part A E has been cut off equal to C the less.
Q. E. F.
(The letters Q. E. F. stand for Quod erat faciendum-Which was to be done.)
2nd Proof (with contracted syllogisms—that is, with those premisses omitted which are capable of being understood instead of expressed.)
Ist Syllogism (compare these with the syllogisms in the 1st Proof.)
Because A is the centre of the circle D E F (a).. AE is equal to AD. (Def. 15 in Euc.) 2nd Syllogism.
But C is equal to A D (by construction). .. AE and C are each equal to AD. (a). 3rd Syllogism (both premisses understood). (b) .. AE is equal to C. (Axiom I.)
Wherefore from AB, the greater of two given straight lines, a part A E, has been cut off equal to C the less.
EXERCISE.-Write out the whole of this problem, using the contracted syllogisms for the proof, and these lines and figures.
Q. E. F.
(c) Repeat the definition of a circle learned with
the last problem.