Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 36 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 36 / 4 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 4 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 4 / 5 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 12F / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 12F / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 2F / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 1 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 1 / 5 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 4 / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 4 / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 27 / 1 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 27 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 27 / 4 / Solution

{wiki=Field_(mathematics)}

A field is a <commutative ring> with $0\ne1$ in which every nonzero element has a multiplicative inverse. Examples include the <rational numbers>, <real numbers>, and <complex numbers>. The <integers> are a <ring> but not a field: dividing $1$ by $2$ leaves them. For a <prime number> $p$, the <prime field> $\mathbb F_p=\mathbb Z/p\mathbb Z$ is a finite example, because every nonzero residue is <coprime> to $p$ and hence has a <modular inverse>.

In $\mathbb F_5$, the nonzero inverse pairs are $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Thus $2x=3$ has solution $x=3\cdot3=4$ in this field; the <real number> solution would be $3/2$. Dividing by a nonzero coefficient is valid in both settings, but it means division according to the chosen field's operations.

A field has no <zero divisors>: if $ab=0$ and $a\ne0$, multiplying by $a^{-1}$ gives $b=0$. This short argument supports <polynomial> factorization arguments and pivot division in <Gaussian elimination>. It does not hold in every <commutative ring>: modulo $6$, the nonzero residues $2$ and $3$ multiply to zero. The distinction is visible in the figure under <algebraic structure>.

Field

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