Harriot, Thomas, Mss. 6785

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            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> This question is a slight variation on a question from Girolamo
                <emph style="it">De arithmetica liber X</emph>
              (=
                <emph style="it">Ars magna</emph>
              )
                <ref id="cardano_1570b"> (Cardano </ref>
              , page 143, Chapter 39, Question I: </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Quaestio I.
                  <lb/>
                Exemplum. Inuenias tres numeros in continua proportione, quorum quadratum primi fit aequale secundo & & quadratum tertij fit aequale quadratis primi & </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> Question 1.
                  <lb/>
                Example. Find three numbers in continual proportion, such that the square of the first is equal to the second and third, and the square of the third is equal to the square of the first and </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> Cardan. Arith, lib. 10. cap. 39. pag.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Cardano,
            <emph style="it">De arithmetica liber X</emph>
          , Chapter 39, page 143. </head>
          <p xml:lang="lat">
            <s xml:space="preserve"> Invenire tres numeros
              <lb/>
            continue proportionales
              <lb/>
            quorum tertius sit aequalis
              <lb/>
            primo et secundo, et
              <lb/>
            quadratum primi sit aequale
              <lb/>
            aggregato secundi et
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Find three numbers in continual proportion of which the third is equal to the first and second, and the square of the first is equal to the sum of the second and third.</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Argumentatio
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            First ]</s>
            <lb/>
            <s xml:space="preserve">[…]</s>
            <lb/>
            <s xml:space="preserve"> Ergo
              <emph style="st">prima</emph>
            dispositio terminorum proportionalium ex prima conditione illata
              <lb/>
            ita se
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Therefore the arrangement of proportional terms according to the first condition is had ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Argumentatio secunda, seu de secunda
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Second argument, or from the second ]</s>
            <lb/>
            <s xml:space="preserve">[…]</s>
            <lb/>
            <s xml:space="preserve"> Unde tres proportionales
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Whence the three proportionals ]</s>
            <lb/>
            <s xml:space="preserve"> Examen fit in altera
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            To be tested in another ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Aggregatum I + II + III
              <math>
                <mstyle>
                  <msqrt>
                    <mrow>
                      <mn>1</mn>
                      <mn>2</mn>
                      <mn>5</mn>
                    </mrow>
                  </msqrt>
                  <mo>+</mo>
                  <mn>1</mn>
                  <mn>1</mn>
                </mstyle>
              </math>
            vel duplum
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            The sum of the first, second, and third is
              <math>
                <mstyle>
                  <msqrt>
                    <mrow>
                      <mn>1</mn>
                      <mn>2</mn>
                      <mn>5</mn>
                    </mrow>
                  </msqrt>
                  <mo>+</mo>
                  <mn>1</mn>
                  <mn>1</mn>
                </mstyle>
              </math>
            , or twice the third. </s>
          </p>
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