// Numbas version: finer_feedback_settings
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"{minv11}", "maxValue": "{minv11}", "marks": 0.25, "showPrecisionHint": false}], "type": "gapfill", "prompt": "<p>&nbsp;</p>\n<table border=\"1\">\n<tbody>\n<tr><th>&nbsp;</th><th>$\\mathbb{Z}_{3}$</th><th>$\\mathbb{Z}_{5}$</th><th>$\\mathbb{Z}_{7}$</th><th>$\\mathbb{Z}_{11}$</th></tr>\n<tr>\n<td>$\\simplify[!basic]{{a1}*{b1}+{c1}*{d1}-{f1}}$</td>\n<td>[[0]]</td>\n<td>[[1]]</td>\n<td>[[2]]</td>\n<td>[[3]]</td>\n</tr>\n<tr>\n<td>$\\var{a2}$</td>\n<td>[[4]]</td>\n<td>[[5]]</td>\n<td>[[6]]</td>\n<td>[[7]]</td>\n</tr>\n<tr>\n<td>$ \\displaystyle{ \\frac{1}{\\var{b3}} } $</td>\n<td>[[8]]</td>\n<td>[[9]]</td>\n<td>[[10]]</td>\n<td>[[11]]</td>\n</tr>\n<tr>\n<td>$ \\displaystyle{ \\frac{\\var{c3}}{\\var{b3}} } $</td>\n<td>[[12]]</td>\n<td>[[13]]</td>\n<td>[[14]]</td>\n<td>[[15]]</td>\n</tr>\n</tbody>\n</table>", "showCorrectAnswer": true, "marks": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "<p>Simplify the following in each of $\\mathbb{Z}_{3}, \\; \\mathbb{Z}_{5}, \\; \\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.</p>\n<p>For the last two questions, recall that for a prime $p$, $\\displaystyle{\\frac{1}{b} \\bmod{p}}$ is the unique solution to $bx \\equiv 1 \\bmod{p}$.</p>\n<p>You must input all values as positive integers; negative integers are not accepted.</p>", "tags": ["checked2015", "MAS3214"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "", "licence": "Creative Commons Attribution 4.0 International", "description": "<p>Express various integers and rationals mod $\\mathbb{Z}_3, \\;\\mathbb{Z}_5,\\;\\mathbb{Z}_7,\\;\\mathbb{Z}_{11}$</p>"}, "advice": "<p>For the first row, it may help to simplify each of the numbers modulo the number corresponding to each column, and then perform the calculation.</p>\n<p>For example,</p>\n<p>\\begin{align}<br/> \\simplify[!basic]{{a1}*{b1}+{c1}*{d1}-{f1}} &amp;\\equiv \\simplify[!basic]{{mod(a1,3)}*{mod(b3,3)}+{mod(c1,3)}*{mod(d1,3)}-{mod(f1,3)}} &nbsp;\\mod{3} \\\\<br/>&amp;\\equiv \\var{ans13} \\mod{3}<br/>\\end{align}</p>\n<p>In the second row, you just have to work out the remainder when dividing $\\var{a2}$ by each of $3$, $5$, $7$ and $11$.</p>\n<p>In the third row we have to find the inverse of $\\var{b3}$ in each of $\\mathbb{Z}_{3}$, $\\mathbb{Z}_{5}$, $\\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.</p>\n<p>It should be clear to you that $\\var{b3}$ is coprime to each of $3$, $5$, $7$ and $11$, hence we can find an inverse in each of $\\mathbb{Z}_{3}$, $\\mathbb{Z}_{5}$, $\\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.</p>\n<p>You can do this in $\\mathbb{Z}_{3}$ by finding $x$ such that $\\var{b3}x \\equiv 1 \\bmod{3}$, similarly for $5$, $7$, $11$.</p>\n<p>You can do this by trying values to find one that works; this is easy for $\\mathbb{Z}_{3}$ since there are only two cases to check, and you will find that</p>\n<p>$b = \\var{inv3}$ satisfies $\\var{inv3} \\times \\var{b3} = \\var{inv3*b3} \\equiv 1 \\bmod{3}.$</p>\n<h4>Inverse in $\\mathbb{Z}_{5}$</h4>\n<p>If you cannot immediately find an inverse in $\\mathbb{Z}_{5}$, you can use the Euclidean algorithm to find $a$ and $b$ such that</p>\n<p>\\[\\var{b3}b +5a = \\operatorname{gcd}(\\var{b3},5) = 1\\]</p>\n<p>as $\\var{b3}$ and $5$ are coprime.</p>\n<p>It follows that $\\var{b3}b \\equiv 1 \\bmod{5}$, and hence $b \\bmod{5}$ is the inverse of $\\var{b3}$ in $\\mathbb{Z}_{5}$.</p>\n<p>Using the Euclidean algorithm I find that</p>\n<p>\\[\\simplify[!basic]{ {max(5,b3)}*{p5} + {min(5,b3)}*{q5} = 1 }.\\]</p>\n<p>Hence the inverse is</p>\n<p>\\[ b = \\var{rinv5} \\equiv \\var{inv5} \\bmod{5}.\\]</p>\n<h4>Inverse in $\\mathbb{Z}_{7}$</h4>\n<p>Once again if you cannot spot the solution:</p>\n<p>Using the Euclidean algorithm I find that</p>\n<p>\\[\\simplify[!basic]{ {max(7,b3)}*{p7} + {min(7,b3)}*{q7} = 1 }.\\]</p>\n<p>Hence the inverse of $\\var{b3}$ in $\\mathbb{Z}_{7}$ is</p>\n<p>\\[b = \\var{rinv7} \\equiv \\var{inv7} \\bmod{7}.\\]</p>\n<h4>Inverse in $\\mathbb{Z}_{11}$</h4>\n<p>Using the Euclidean algorithm I find that</p>\n<p>\\[\\simplify[!basic]{ {max(11,b3)}*{p11} + {min(11,b3)}*{q11} = 1}. \\]</p>\n<p>Hence the inverse of $\\var{b3}$ in $\\mathbb{Z}_{11}$ is</p>\n<p>\\[ b = \\var{rinv11} \\equiv \\var{inv11} \\bmod{11}.\\]</p>\n<h4>Last Question</h4>\n<p>The last question's answers are found by, for example in $\\mathbb{Z}_{7}$, considering (all $\\bmod 7$)</p>\n<p>\\[\\frac{\\var{c3}}{\\var{b3}} = \\var{c3} \\times \\frac{1}{\\var{b3}} \\equiv \\var{c3} \\times \\var{inv7} \\equiv \\var{c3*inv7} \\equiv \\var{minv7} \\bmod{7}.\\]</p>"}]}], "contributors": [{"name": "Blathnaid Sheridan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/447/"}], "extensions": [], "custom_part_types": [], "resources": []}