Harriot, Thomas, Mss. 6785

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page |< < (204) of 882 > >|
    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6785_f204" o="204" n="407"/>
          <div type="page_commentary" level="0" n="0">
            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> The reference at the top of the page is to Zetetic IV.1 from
                <emph style="it">Zeteticorum libri quinque</emph>
                <ref id="Viete 1593a" target="http://www.e-rara.ch/zut/content/pageview/2683942"> (Viète 1593a, Zet </ref>
              . This is also Proposition II.8 from the
                <emph style="it">Arithmetica of Diophantus</emph>
                <ref id="diophantus_1575" target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/mpiwg/online/permanent/library/W770Y3H9/&start=51&viewMode=images&pn=56"> (Diophantus 1575, </ref>
              . In Zeteticum IV.1, Viète wrote as follows: </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Zeteticum I
                  <lb/>
                Invenire numero duo quadrata, aequalia dato quadrato.
                  <lb/>
                […]
                  <lb/>
                Eoque recidit Analysis Diophantæa, secundum quam oporteat B quadratum, in duo quadrata dispescere. Latus primi quadrati esto A, secundi B – S in A/R. Primi lateris in quadratum, est A quadratum. Secundi, B quad. – S in A in B2/R + S quad. in A quad./R quad. Quae duo quadrata ideo aequalia sunt B quadrato.
                  <lb/>
                Aequalitas igitur ordinetur. S in R in B2/S quad. + R quadr. aequabitur A lateri primi singularis quadrati. Et latus secundi fit R quad. in B, – S quad. in B/S quad. + R quad. Nempe triangulum rectangulum numero effingitur a lateribus duobus S & & fit hypotenusa similis S quad + R quad. basis similis S quadrato – R quadrato. Perendiculum simile S in R2. Itaque ad dispectionem B quadrati fit, ut S quadr. + R quadr. ad B hypotenusam similis trianguli, ita R quadr. – S quadr. ad basim, latus unus singularis & ita S in R2 ad perpendiculum, latus </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> To find in numbers, two squares equal to a given square.
                  <lb/>
                […]
                  <lb/>
                The same is taught in the analysis of Diophantus, according to which it is required to divide the square of
                  <math>
                    <mstyle>
                      <mi>b</mi>
                    </mstyle>
                  </math>
                into two other squares. Let the side of the first be
                  <math>
                    <mstyle>
                      <mi>a</mi>
                    </mstyle>
                  </math>
                , of the second
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mo>-</mo>
                      <mfrac>
                        <mrow>
                          <mi>s</mi>
                          <mi>a</mi>
                        </mrow>
                        <mrow>
                          <mi>r</mi>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                . The square of the first side is
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>a</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                , of the second
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>b</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>-</mo>
                      <mfrac>
                        <mrow>
                          <mn>2</mn>
                          <mi>s</mi>
                          <mi>a</mi>
                          <mi>b</mi>
                        </mrow>
                        <mrow>
                          <mi>r</mi>
                        </mrow>
                      </mfrac>
                      <mo>+</mo>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mrow>
                        <msup>
                          <mi>a</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                . Which two squares are therefore equal to
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>b</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                .
                  <lb/>
                The equalisation is carried out. [We obtain]
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mo>=</mo>
                      <mfrac>
                        <mrow>
                          <mn>2</mn>
                          <mi>s</mi>
                          <mi>r</mi>
                          <mi>b</mi>
                        </mrow>
                        <mrow>
                          <mrow>
                            <msup>
                              <mi>s</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                          <mo>+</mo>
                          <mrow>
                            <msup>
                              <mi>r</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                . And the second side is
                  <math>
                    <mstyle>
                      <mfrac>
                        <mrow>
                          <mrow>
                            <msup>
                              <mi>r</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                          <mi>b</mi>
                          <mo>-</mo>
                          <mrow>
                            <msup>
                              <mi>s</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                          <mi>b</mi>
                        </mrow>
                        <mrow>
                          <mrow>
                            <msup>
                              <mi>s</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                          <mo>+</mo>
                          <mrow>
                            <msup>
                              <mi>r</mi>
                              <mn>2</mn>
                            </msup>
                          </mrow>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                . That is, a right-angled triangle in numbers may be constructed from the two terms
                  <math>
                    <mstyle>
                      <mi>s</mi>
                    </mstyle>
                  </math>
                and
                  <math>
                    <mstyle>
                      <mi>r</mi>
                    </mstyle>
                  </math>
                , and the hypotenuse will be proportional to
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>+</mo>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                , the base to
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>-</mo>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                [actually
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>-</mo>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                ], the perpendicular to
                  <math>
                    <mstyle>
                      <mn>2</mn>
                      <mi>r</mi>
                      <mi>s</mi>
                    </mstyle>
                  </math>
                . Thus, the division of
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>b</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                [into two other squares] gives a triangle with hypotenuse proportional to
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>+</mo>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                , base proportional to
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>-</mo>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                , [axtually
                  <math>
                    <mstyle>
                      <mrow>
                        <msup>
                          <mi>r</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                      <mo>-</mo>
                      <mrow>
                        <msup>
                          <mi>s</mi>
                          <mn>2</mn>
                        </msup>
                      </mrow>
                    </mstyle>
                  </math>
                ], the side of one square, and perpendicular proportional to
                  <math>
                    <mstyle>
                      <mn>2</mn>
                      <mi>s</mi>
                      <mi>r</mi>
                    </mstyle>
                  </math>
                , the side of the other. </s>
              </quote>
              <lb/>
              <s xml:space="preserve"> For the first time in this run of pages, Harriot refers directly to Diophantus, not only in the heading but also in the course of his working. It seems likely that he had now turned directly to Problem II.8 of the
                <emph style="it">Arithemtica</emph>
              in the edition by Wilhelm
                <emph style="it">Diophanti Alexandrini rerum arithmeticarum libri sex</emph>
                <ref id="diophantus_1575" target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/mpiwg/online/permanent/library/W770Y3H9/&start=51&viewMode=images&pn=56"> (Diophantus 1575, </ref>
              . There he would have found that Diophantus gave only a single numerical example, with none of the generality that Viète had introduced. </s>
              <lb/>
              <s xml:space="preserve"> Supposing the initial square was 16, Diophantus took the side of the first square to be some number which, following Xylander, we may call
                <math>
                  <mstyle>
                    <mi>N</mi>
                  </mstyle>
                </math>
              , with side
                <math>
                  <mstyle>
                    <mi>Q</mi>
                  </mstyle>
                </math>
              , and the side of the second square to be
                <math>
                  <mstyle>
                    <mn>2</mn>
                    <mi>N</mi>
                    <mo>-</mo>
                    <mn>4</mn>
                  </mstyle>
                </math>
              . In Viète's more general notation, 16 was replaced by
                <math>
                  <mstyle>
                    <mi>B</mi>
                  </mstyle>
                </math>
              and
                <math>
                  <mstyle>
                    <mi>N</mi>
                  </mstyle>
                </math>
              by
                <math>
                  <mstyle>
                    <mi>A</mi>
                  </mstyle>
                </math>
              . The side of the second square, in Diophantus's method, was thus
                <math>
                  <mstyle>
                    <mn>2</mn>
                    <mi>A</mi>
                    <mo>-</mo>
                    <mi>B</mi>
                  </mstyle>
                </math>
              . Harriot was therefore correct in his assertion that Viète, who wrote the second side as
                <math>
                  <mstyle>
                    <mi>B</mi>
                    <mo>-</mo>
                    <mfrac>
                      <mrow>
                        <mi>S</mi>
                        <mi>A</mi>
                      </mrow>
                      <mrow>
                        <mi>R</mi>
                      </mrow>
                    </mfrac>
                  </mstyle>
                </math>
              , had actually proceeded differently from Diophantus. </s>
              <lb/>
              <s xml:space="preserve"> Diophantus was working purely in numbers, but for Viète, who was relating the problem to lengths of sides of a triangle, it was clearly more natural to take the side of the second square to be
                <math>
                  <mstyle>
                    <mi>B</mi>
                    <mo>-</mo>
                    <mfrac>
                      <mrow>
                        <mi>S</mi>
                        <mi>A</mi>
                      </mrow>
                      <mrow>
                        <mi>R</mi>
                      </mrow>
                    </mfrac>
                  </mstyle>
                </math>
              . Viete did not specify the relative sizes of
                <math>
                  <mstyle>
                    <mi>R</mi>
                  </mstyle>
                </math>
              and
                <math>
                  <mstyle>
                    <mi>S</mi>
                  </mstyle>
                </math>
              , but it must be the case that
                <math>
                  <mstyle>
                    <mfrac>
                      <mrow>
                        <mi>S</mi>
                        <mi>A</mi>
                      </mrow>
                      <mrow>
                        <mi>R</mi>
                      </mrow>
                    </mfrac>
                    <mo>&</mo>
                    <mi>l</mi>
                    <mi>t</mi>
                    <mo>;</mo>
                    <mi>B</mi>
                  </mstyle>
                </math>
              if the triangle is to be constructed. In the case where
                <math>
                  <mstyle>
                    <mi>S</mi>
                    <mo>></mo>
                    <mi>R</mi>
                  </mstyle>
                </math>
              , Harriot argues that one should instead use the method of Diophantus (who gave the example where
                <math>
                  <mstyle>
                    <mi>S</mi>
                    <mo>:</mo>
                    <mi>R</mi>
                    <mo>=</mo>
                    <mn>2</mn>
                    <mo>:</mo>
                    <mn>1</mn>
                  </mstyle>
                </math>
              ), in which case one requires
                <math>
                  <mstyle>
                    <mfrac>
                      <mrow>
                        <mi>S</mi>
                        <mi>A</mi>
                      </mrow>
                      <mrow>
                        <mi>R</mi>
                      </mrow>
                    </mfrac>
                    <mo>></mo>
                    <mi>B</mi>
                  </mstyle>
                </math>
              . </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> 3.) Diophantus. lib. 2. 8. Zet. 4.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Diophantus, Book II, Proposition 8 Zetetica, Book IV, Zetetic 1</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Dividere
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            in duo quadrata numeri.
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Divide
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>b</mi>
                </mstyle>
              </math>
            into two square numbers. </s>
            <s xml:space="preserve"> Sit 1.
              <reg norm="quadratus" type="abbr">quad</reg>
            .
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Let the first square be
              <math>
                <mstyle>
                  <mi>a</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
            . </s>
            <s xml:space="preserve"> Erit: 2.
              <reg norm="quadratus" type="abbr">quad</reg>
            .
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mi>a</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            The second square will be
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mi>a</mi>
                  <mi>a</mi>
                </mstyle>
              </math>
            .
              <lb/>
            </s>
            <s xml:space="preserve"> Qæritur latus
              <lb/>
            huius 2
              <emph style="super">i</emph>
            et fit:
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            vel secundi
              <reg norm="Diophantus" type="abbr">Diophant</reg>
            :
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>-</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            The side of this second square is sought; and it will be
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            , or according to Diophantus
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>-</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> latus secundi fuerit.
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            the side of the second will be
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            </s>
            <s xml:space="preserve"> (sive ex hypothesi quod
              <lb/>
            est
              <math>
                <mstyle>
                  <mi>r</mi>
                  <mo>></mo>
                  <mi>s</mi>
                </mstyle>
              </math>
            )
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            formeretur pro lateri 2
              <emph style="super">i</emph>
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            (or from the hypothesis that
              <math>
                <mstyle>
                  <mi>r</mi>
                  <mo>></mo>
                  <mi>s</mi>
                </mstyle>
              </math>
            ), the side of the second square will be
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mo>-</mo>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            </s>
            <s xml:space="preserve"> Et in illa
              <lb/>
            forma
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>b</mi>
                      <mi>s</mi>
                      <mi>s</mi>
                      <mo>+</mo>
                      <mi>b</mi>
                      <mi>r</mi>
                      <mi>r</mi>
                    </mrow>
                    <mrow>
                      <mi>s</mi>
                      <mi>s</mi>
                      <mo>+</mo>
                      <mi>r</mi>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>=</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            . latus
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            And in that form
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>b</mi>
                      <mi>s</mi>
                      <mi>s</mi>
                      <mo>+</mo>
                      <mi>b</mi>
                      <mi>r</mi>
                      <mi>r</mi>
                    </mrow>
                    <mrow>
                      <mi>s</mi>
                      <mi>s</mi>
                      <mo>+</mo>
                      <mi>r</mi>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>=</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            is the side of the given square. </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Sed si ponatur quod
              <math>
                <mstyle>
                  <mi>s</mi>
                </mstyle>
              </math>
            sit
              <math>
                <mstyle>
                  <mo>></mo>
                  <mi>r</mi>
                </mstyle>
              </math>
            ;
              <emph style="st">[???]</emph>
            ut posuit Vieta
              <lb/>
            latus 2
              <emph style="super">i</emph>
            [???] est
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>-</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            . ut
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            But if we put
              <math>
                <mstyle>
                  <mi>s</mi>
                  <mo>></mo>
                  <mi>r</mi>
                </mstyle>
              </math>
            , as Viète supposed, the side of the second square is
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mi>s</mi>
                      <mi>a</mi>
                    </mrow>
                    <mrow>
                      <mi>r</mi>
                    </mrow>
                  </mfrac>
                  <mo>-</mo>
                  <mi>b</mi>
                </mstyle>
              </math>
            , as in Diophantus. </s>
            <s xml:space="preserve"> Vieta igitur
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Viete is therefore to be ]</s>
          </p>
        </div>
      </text>
    </echo>
Impressum