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English sentences are given and for each the appropriate proposition involving quantifiers is to be chosen. Also choose whether the propositions are true or false.
", "licence": "Creative Commons Attribution 4.0 International"}, "extensions": [], "variables": {"select": {"templateType": "anything", "name": "select", "definition": "shuffle(list(0..length(all)-1))[0..3]", "group": "Part 0", "description": ""}, "marking_matrix": {"templateType": "anything", "name": "marking_matrix", "definition": "map(list(neg_marks[x])+[all[select[x]][2]]+[-1*all[select[x]][2]],x,0..2)", "group": "Part 0", "description": ""}, "select1": {"templateType": "anything", "name": "select1", "definition": "list(set(0..length(all)-1)-set(select))[0..3]", "group": "Part 1", "description": ""}, "all": {"templateType": "anything", "name": "all", "definition": "[['The square of any real number is greater than $0$.',\n '$\\\\forall x \\\\in \\\\mathbb{R}\\\\\\\\ \\\\;(x^2\\\\gt 0).$',-0.5],\n ['Any real number can become non-negative when raised to some integer power.',\n '$\\\\forall x \\\\in \\\\mathbb{R} \\\\;\\\\exists n \\\\in \\\\mathbb{N}\\\\\\\\ \\\\;(x^n\\\\geq 0).$',0.5],\n ['A subset of the natural numbers is a subset of the reals.',\n '$\\\\forall X \\\\subseteq \\\\mathbb{N}\\\\\\\\ \\\\;(X\\\\subseteq \\\\mathbb{R}).$',0.5],\n ['For every natural number $n$ there is a subset of $\\\\mathbb{N}$ with less than $n$ members.',\n '$\\\\forall n \\\\in \\\\mathbb{N}\\\\;\\\\exists X \\\\subseteq \\\\mathbb{N}\\\\\\\\ \\\\;(|X|\\\\lt n).$',-0.5],\n ['All subsets of the natural numbers have less than a fixed number of elements.',\n '$\\\\exists n \\\\in \\\\mathbb{N}\\\\; \\\\forall X \\\\subseteq \\\\mathbb{N}\\\\\\\\ \\\\;(|X|\\\\lt n).$',-0.5],\n ['All subsets of the natural numbers are finite.',\n '$\\\\forall X \\\\subseteq \\\\mathbb{N}\\\\;\\\\exists n \\\\in \\\\mathbb{Z}\\\\\\\\ \\\\;(|X|=n).$',-0.5],\n ['Given an integer $n$, there is a subset of the natural numbers with $n$ elements.',\n '$\\\\forall n \\\\in \\\\mathbb{Z}\\\\;\\\\exists X \\\\subseteq \\\\mathbb{N}\\\\\\\\ \\\\;(|X|=n).$',-0.5],\n ['Given an integer, then adding $5$ to it gives another integer.',\n '$\\\\forall n \\\\in \\\\mathbb{Z}\\\\; \\\\exists m \\\\in \\\\mathbb{Z}\\\\\\\\ \\\\; (m=n+5).$',0.5],\n ['There is an integer $m$ such that adding $5$ to any integer gives $m$.',\n '$\\\\exists m \\\\in \\\\mathbb{Z}\\\\; \\\\forall n \\\\in \\\\mathbb{Z}\\\\\\\\ \\\\; (m=n+5).$',-0.5]\n ]", "group": "Ungrouped variables", "description": ""}, "marking_matrix1": {"templateType": "anything", "name": "marking_matrix1", "definition": "map(list(neg_marks[x])+[all[select1[x]][2]]+[-1*all[select1[x]][2]],x,0..2)", "group": "Part 1", "description": ""}, "select2": {"templateType": "anything", "name": "select2", "definition": "list(set(0..length(all)-1)-(set(select) or set(select1)))", "group": "Part 2", "description": ""}, "neg_marks": {"templateType": "anything", "name": "neg_marks", "definition": "id(3)+matrix(map(map(-0.5,x,0..2),y,0..2))", "group": "Ungrouped variables", "description": ""}, "marking_matrix2": {"templateType": "anything", "name": "marking_matrix2", "definition": "map(list(neg_marks[x])+[all[select2[x]][2]]+[-1*all[select2[x]][2]],x,0..2)", "group": "Part 2", "description": ""}}, "advice": "$\\forall x \\in \\mathbb{R}\\;(x^2\\gt 0)$ is false, because there is a single counterexample when $x=0$ (in which case $x^2=0$, which is not allowed in the strict inequality).
\n$\\forall x \\in \\mathbb{R} \\;\\exists n \\in \\mathbb{N}\\;(x^n\\geq 0)$ is true, because it is always true for any even value of $n$. As seen in the previous example, $n=2$ will work here because the inequality is no longer strict. You might also consider the simpler case $n=0$, though $x^0$ is not well-defined when $x=0$.
\n$\\forall X \\subseteq \\mathbb{N}\\;(X\\subseteq \\mathbb{R})$ is true, because $\\mathbb{N} \\subset \\mathbb{R}$ itself - that is, all natural numbers are also real numbers.
\n$\\forall n \\in \\mathbb{N}\\;\\exists X \\subseteq \\mathbb{N}\\;(|X|\\lt n)$ is false, because there is a single counterexample when $n=0$ (in which case $|X|$ cannot possibly be negative). Notice $n=1$ is not a counterexample since $|\\emptyset| = 0 < 1$.
\n$\\exists n \\in \\mathbb{N}\\; \\forall X \\subseteq \\mathbb{N}\\;(|X|\\lt n)$ is false, because for any given natural number $n$, we could take $X$ to be the set $\\{0,1,2,\\ldots,n\\}$, which has size $n+1$ and so $|X| \\nless n$.
\n$\\forall X \\subseteq \\mathbb{N}\\;\\exists n \\in \\mathbb{Z}\\;(|X|=n)$ is false, because there exist subsets of $\\mathbb{N}$ which are infinite in size, and therefore cannot have cardinality equal to a natural number. For example, $\\mathbb{N}$ itself is an infinite subset of $\\mathbb{N}$, as is, for example, the set of all even natural numbers $\\{2k \\; | \\; k \\in \\mathbb{N}\\}$.
\n$\\forall n \\in \\mathbb{Z}\\;\\exists X \\subseteq \\mathbb{N}\\;(|X|=n)$ is false, because the statement breaks for any negative value of $n$, which is allowed since $\\mathbb{Z}$ includes the negative integers. No set can have negative cardinality.
\n$\\forall n \\in \\mathbb{Z}\\; \\exists m \\in \\mathbb{Z}\\; (m=n+5)$ is true, because $\\mathbb{Z}$ is closed under addition, and $5 \\in \\mathbb{Z}$. That is, the set of integers has the property that adding any two integers will always return an integer.
\n$\\exists m \\in \\mathbb{Z}\\; \\forall n \\in \\mathbb{Z}\\; (m=n+5)$ is false, because any two different values of $n$ will yield a different $m$ (e.g. $n=0 \\implies m=5$, but $n=1 \\implies m=6$), so $m=n+5$ for a fixed $m$ cannot hold for even two different values of $n$, let alone all integer values.
", "tags": [], "statement": "Choose the appropriate proposition for the following English sentences. Also choose whether they are true or false.
\nYou must choose two of the five options presented in each row. You may need to scroll over to see all five options.
\nNote also that every wrong answer takes away from your score. However, your minimum score is $0$.
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