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"> On this page, Harriot works a problem from
                <emph style="it">De regula aliza liber</emph>
                <ref id="cardano_1570b"> (Cardano </ref>
              , Chapter 41, pages 82–83. Cardano states the problem as follows: </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> De difficillimo problemate quod facillimum uidetur. CAP. XLI.
                  <lb/>
                Nihil est admirabilius quam cum sub facili quaestione latet difficillimus scrupulus, huiusmodi est hic: quadratum
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
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                  </math>
                cum latere
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                est 10, & quadratum
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                cum latere
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>d</mi>
                    </mstyle>
                  </math>
                est 8, quaeritur quantum fit unum horum seu latus seu quadratum: Quia ergo
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                est 10, &
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
                    </mstyle>
                  </math>
                1 quad. erit
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                10 m: 1 quad. igitur bc 100 m: 20 quad. p: 1 pos. p: 1 quad. quad. & hoc est aequale 8, quare 1 quad. quad. p: 92, aequatur 20 quad. m: 1 pos. adde 19 quad. utrinque fient 1 quad. quad. p: 19, quadrat. p: 92, aequalia 39. quad, m: 1 pos. detrahe
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                      <mn>1</mn>
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                        </mrow>
                        <mrow>
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                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                , erunt 1 quad. quad. p: 19 quad. p:
                  <math>
                    <mstyle>
                      <mn>9</mn>
                      <mn>0</mn>
                      <mfrac>
                        <mrow>
                          <mn>1</mn>
                        </mrow>
                        <mrow>
                          <mn>4</mn>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                aequalia 39 quad. m: 1 pos. m:
                  <math>
                    <mstyle>
                      <mn>1</mn>
                      <mfrac>
                        <mrow>
                          <mn>1</mn>
                        </mrow>
                        <mrow>
                          <mn>4</mn>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                , inde adde 2 pos. p: 1 quad. utrinque ut in Arte magna, & videbis difficillimam quaestionem. </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> Chapter 41. On a very difficult question that seems easy.
                  <lb/>
                Nothing is more remarkable than when under an easy question there lies a very hard stone, of which kind is this: the square of
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
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                  </math>
                with the side
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                is 10, and the square of
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                with the side
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>d</mi>
                    </mstyle>
                  </math>
                is 8; it is required to find one of those sides or squares. Therefore because
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                is 10, and
                  <math>
                    <mstyle>
                      <mi>a</mi>
                      <mi>b</mi>
                    </mstyle>
                  </math>
                is one square,
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                will be 10 minus a square. Therefore the square of
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                    </mstyle>
                  </math>
                is 100 minus 20 squares plus 1 square-square. Therefore
                  <math>
                    <mstyle>
                      <mi>b</mi>
                      <mi>c</mi>
                      <mi>d</mi>
                    </mstyle>
                  </math>
                will be 100 minus 20 squares plus 1 unknown plus 1 square-square, and this is equal to 8, whence 1 square-square plus 92 equals 20 squares minus 1 unknown. Add 19 squares to each side, making 1 square-square plus 19 squares plus 92 equal to 39 squares minus 1 unknown. Subtract
                  <math>
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                      <mn>1</mn>
                      <mfrac>
                        <mrow>
                          <mn>1</mn>
                        </mrow>
                        <mrow>
                          <mn>4</mn>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                , to give 1 square-square plus 19 squares plus
                  <math>
                    <mstyle>
                      <mn>9</mn>
                      <mn>0</mn>
                      <mfrac>
                        <mrow>
                          <mn>1</mn>
                        </mrow>
                        <mrow>
                          <mn>4</mn>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                equal to 39 squares minus 1 unknown minus
                  <math>
                    <mstyle>
                      <mn>1</mn>
                      <mfrac>
                        <mrow>
                          <mn>1</mn>
                        </mrow>
                        <mrow>
                          <mn>4</mn>
                        </mrow>
                      </mfrac>
                    </mstyle>
                  </math>
                , whence add 2 unknowns plus one square to each side as in the Ars magna, and you will see this very difficult question. </s>
              </quote>
              <lb/>
              <s xml:space="preserve"> Harriot lets the side of
                <math>
                  <mstyle>
                    <mi>b</mi>
                    <mi>c</mi>
                  </mstyle>
                </math>
              be
                <math>
                  <mstyle>
                    <mi>a</mi>
                  </mstyle>
                </math>
              and then works through these instructions precisely. </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve" xml:lang="lat"> Cardan. de Aliza. pa.
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Cardano,
            <emph style="it">De regula aliza liber</emph>
          , page 83 </head>
          <p xml:lang="lat">
            <s xml:space="preserve">
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            ]</s>
            <lb/>
            <s xml:space="preserve"> Quæritur
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            vel
              <math>
                <mstyle>
                  <mi>f</mi>
                  <mi>g</mi>
                </mstyle>
              </math>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Sought,
              <math>
                <mstyle>
                  <mi>d</mi>
                  <mi>f</mi>
                </mstyle>
              </math>
            and
              <math>
                <mstyle>
                  <mi>f</mi>
                  <mi>g</mi>
                </mstyle>
              </math>
            . </s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve">
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            Another ]</s>
          </p>
          <p xml:lang="lat">
            <s xml:space="preserve"> Et per
              <lb/>
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            And by ]</s>
          </p>
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