1.4 Problems on Arithmetic Operations and Inequalities in a Field

(a)

\(2+3=5\);

(b)

\(3+4=7\);

(c)

\(2\cdot 2=4\);

(d)

\(3\cdot 2=6\);

(a)

Using the preliminary definition, e.g.: \(2=1+1\), and the Field Axioms: \begin {align*} 2+3 &= (1+1)+3 &\text {(definition of "2")}\\ &= 3+ (1+1) & \text {(Axiom II)}\\ &= (3+1)+1 &\text {(Axiom III)}\\ &= 4+1 &\text {(definition of "4")}\\ &=5 &\text {(definition of "5")} \end {align*}

(b)

Similarly, we have: \begin {align*} 3+4 &= (2+1)+4 &\text {(definition of "3")}\\ &= 4+(2+1) &\text {(Axiom II)}\\ &= 4+(1+2) &\text {(Axiom II)}\\ &= (4+1)+2 &\text {(Axiom III)}\\ &= 5+2 &\text {(definition of "5")}\\ &= 5+(1+1) &\text {(definition of "2")}\\ &= (5+1)+1 & \text {(Axiom III)}\\ &= 6+1 & \text {(definition of "6")}\\ &= 7 &\text {(definition of "7")} \end {align*}

(c)

Again, using the preliminary definition, and the Field Axioms we have: \begin {align*} 2\cdot 2 &= (1+1)\cdot 2 & \text {(definition of "2")}\\ &= 1\cdot 2 + 1 \cdot 2 & \text {(Axiom VI)}\\ &= 2+2 &\text {(Axiom IV)}\\ &= 2+(1+1) & \text {(definition of "2")}\\ &= (2+1)+1 & \text {(Axiom III)}\\ &= 3+1 & \text {(definition of "3")}\\ &= 4 & \text {(definition of "4")} \end {align*}

(d)

Similarly, \begin {align*} 3 \cdot 2 &= 3\cdot (1+1) & \text {(definition of "2")}\\ &= (1+1)\cdot 3& \text {(Axiom II)}\\ &= 3 \cdot 1+3 \cdot 1 & \text {(Axiom VI)}\\ &= 3+3 &\text {(Axiom IV)}\\ &= 3+ (2+1) &\text {(definition of "3")}\\ &= 3+(1+2) & \text {(Axiom II)}\\ &= (3+1)+2 & \text {(Axiom III)}\\ &= 4+2 & \text {(definition of "4")}\\ &= 4+(1+1) & \text {(definition of "2")}\\ &= (4+1)+1 & \text {(Axiom III)}\\ &= 5 +1 & \text {(definition of "5")}\\ &= 6 & \text {definition of "6")} \end {align*}

(i)

\(abcd=cbad = dacb\); similarly for addition.

(ii)

If \(x\neq 0\) and \(y \neq 0\), then \(xy\neq 0.\)

(iii)

\((xy)^{-1}= x^{-1}y^{-1}\), provided that \(y \neq 0\) and \(y \neq 0\). Why must one assume that neither \(x\) nor \(y\) are zero?

(iv)

If \(x\neq 0, \: y\neq 0\) and \(z \neq 0\), then \((xzy)^{-1}=x^{-1}y^{-1}z^{-1}\).

(v)

If \(x\neq 0\) and \(y \neq 0\), then \begin {align*} \frac {a}{x}\cdot \frac {b}{y}= \frac {ab}{xy} \;\; \text {and}\;\; \frac {a}{x}+\frac {b}{y} = \frac {ay+bx}{xy} \end {align*}

(vi)

\((a+b)(x+y)=ax+bx+ay+by\); (vi\('\)) \((a+b)^2=a^2+2ab +b^2.\)

(vii)

\((a+b)(x-y)=ax+bx-ay-by;\) (vii\('\)) \((a+b)(a-b)=a^2-b^2.\)

(i)

For addition, we have \(a+b+c+d\), which is equivalent to \(c+b+a +d\) and \(d+a+c+b\) \begin {align*} a+b+c+d &=(a+b+c)+d & \text {(Definition 3.)}\\ &= ((a+b)+c)+d & \text {(Definition 3.)} \\ &= (c+(a+b))+d & \text {(Axiom II)}\\ &= (c+(b+a))+d & \text {(Axiom II)}\\ &= ((c+b)+a))+d & \text {(Axiom III)}\\ &= (c+b+a)+d & \text {(Definition 3.)}\\ &= c+b+a+d & \text {(Definition 3.)} \end {align*}

Similarly, \begin {align*} a+b+c+d &=(a+b+c)+d & \text {(Definition 3.)}\\ &= d+(a+b+c) & \text {(Axiom II)}\\ &= d+((a+b)+c) &\text {(Definition 3.)}\\ &= d+(a+(b+c)) & \text {(Axiom III)}\\ &= d+(a+(c+b)) & \text {(Axiom II)}\\ &= d+((a+c)+b) & \text {(Axiom III)}\\ &= d+(a+c+b) & \text {(Definition 3.)}\\ &= d+a+c+b & \text {(Definition 3.')} \end {align*}

For multiplication, we have the expression \(abcd\), which is equal to \(cbad\) and \(dacb\). \begin {align*} abcd &= (abc)\cdot d & \text {(Definition 3.)}\\ &= ((ab)\cdot c)\cdot d & \text {(Definition 3.)}\\ &= (c\cdot (ab))\cdot d & \text {(Axiom II)}\\ &= (c\cdot (ba))\cdot d & \text {(Axiom II)}\\ &= ((cb)\cdot a)\cdot d & \text {(Axiom III)}\\ &= (cba)\cdot d & \text {(Definition 3.)}\\ &= cbad & \text {(Definition 3.)} \end {align*}

Similarly, \begin {align*} abcd &= (abc)\cdot d & \text {(Definition 3.)}\\ &= d\cdot (abc) &\text {(Axiom II)}\\ &= d\cdot ((ab)\cdot c) &\text {(Definition 3.)}\\ &= d\cdot (a\cdot (bc))) & \text {(Axiom III)}\\ &= d \cdot (a\cdot (cb)) & \text {(Axiom II)}\\ &= d\cdot ((ac)\cdot b) & \text {(Axiom III)}\\ &= d \cdot (acb) & \text {(Definition 3.)}\\ &= dacb & \text {(Definition 3.')} \end {align*}

(ii)

If \(x\neq 0\) and \(y \neq 0\), we have \(xy \neq 0\).
Assume, \(xy=0\) \begin {align*} xy&=0 & \text {(Assumption)}\\ (xy)\cdot y^{-1} &= 0\cdot y^{-1} & \text {(Corollary 2.)}\\ (xy) \cdot y^{-1} &= 0 & \text {(Corollary 5.)}\\ x\cdot (y\cdot y^{-1}) &=0 &\text {(Axiom III)}\\ x\cdot 1&=0 & \text {(Axiom V)}\\ x&=0 & \text {(Axiom IV)} \end {align*}

This contradicts our assumption that \(x\neq 0\), thus, \(xy\neq 0\)

(iii)

We need to show that \((xy)^{-1}=x^{-1}y^{-1}\), given that neither \(x\) nor \(y\) are zero. The equation implies that \(x^{-1}y^{-1}\) is the multiplicative inverse of \(xy\): \begin {align*} (xy)\cdot (x^{-1}y^{-1})= xyx^{-1}y^{-1} = xx^{-1}\cdot yy^{-1} =1\cdot 1=1 \end {align*}

Hence, \((xy)^{-1} =x^{-1}y^{-1}\). \(x\) or \(y\) cannot be \(0\), as in both cases, \(xy=0\), which does not have a multiplicative inverse.

(iv)

If \(x\neq 0, y \neq 0\) and \(z \neq 0\), then \((xyz)^{-1}=x^{-1}y^{-1}z^{-1}\). Similarly, the statement implies that \(x^{-1}y^{-1}z^{-1}\) is the multiplicative inverse of \(xyz\): \begin {align*} (xyz)(x^{-1}y^{-1}z^{-1}) = (x)\cdot (yz)\cdot (x^{-1}) \cdot (y^{-1}z^{-1})=(x) \cdot (x^{-1}) \cdot (yz)\cdot (y^{-1}z^{-1}) = 1 \cdot 1 \end {align*}

By (iii), and (i), thus we have the result.

(v)

If \(x \neq 0\) and \(y \neq 0\), we have: \begin {align*} \frac {a}{x}\cdot \frac {b}{y} &=ax^{-1} \cdot by^{-1} & \text {(Definition)}\\ &=ab x^{-1}y^{-1} & \text {(by (i))}\\ &= (ab)\cdot (xy)^{-1} & \text {(by (iii))}\\ &= \frac {ab}{xy} & \text {(Definition)} \end {align*}

Now, for addition, \begin {align*} \frac {a}{x} +\frac {b}{y} &=ax^{-1} + by^{-1} & \text {(Definition)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot x &= (ax^{-1}+by^{-1})\cdot x & \text {(Corollary 2)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot x &= ax^{-1}\cdot x + by^{-1}\cdot x &\text {(Axiom VI)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot x &= a\cdot 1 + by^{-1}\cdot x & \text {(Axiom V)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot x &= a + by^{-1}\cdot x & \text {(Axiom IV)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) &= (a+ by^{-1}x)\cdot y & \text {(Corollary 2)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) &= ay + by^{-1}xy & \text {(Axiom VI)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) &= ay + bxy^{-1}y & \text {(By (i))}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) &= ay + bx \cdot 1 & \text {(Axiom V)}\\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) &= ay+bx & \text {(Axiom IV)} \\ \left (\frac {a}{x} +\frac {b}{y}\right )\cdot (xy) \cdot (xy)^{-1} &= (ay+bx)\cdot (xy)^{-1} & \text {(Corollary 2)}\\ \left (\frac {a}{x} +\frac {b}{y}\right ) \cdot 1 &= (ay+bx)\cdot (xy)^{-1} & \text {(Axiom V)}\\ \frac {a}{x} +\frac {b}{y} &= (ay+bx)\cdot (xy)^{-1} & \text {(Axiom IV)}\\ \frac {a}{x} +\frac {b}{y} &= \frac {ay+bx}{xy} & \text {(Definition)} \end {align*}

(vi)

We need to prove that \((a+b)(x+y) =ax+bx+ay+by\): \begin {align*} (a+b)(x+y) &= a\cdot (x+y)+b\cdot (x+y) & \text {(Axiom VI)}\\ &= (x+y) \cdot a + (x+y)\cdot b & \text {(Axiom II)}\\ &= ax+ay+bx+by & \text {(Axiom VI)}\\ &= ax+bx+ay+by & \text {(By (i))} \end {align*}

When \(x=a\) and \(y=b\), we have: \begin {align*} (a+b)(a+b) &= a\cdot a + b\cdot a + a\cdot b+b\cdot b & \text {(By (vi))} \\ &= a^2+a\cdot b + a\cdot b + b^2 & \text {(Axiom II)}\\ &= a^2 + 1\cdot (a\cdot b) + 1\cdot (a\cdot b) + b^2 & \text {(Axiom IV)}\\ &= a^2 +(1+1)\cdot (a\cdot b)+b^2 & \text {(Axiom VI)}\\ &= a^2 + 2ab +b^2 & \text {(Definition of ``2'')} \end {align*}

(vii)

We need to prove that \((a+b)(x-y)=ax+bx-ay-by\). \begin {align*} (a+b)(x-y) &= (a+b)(x+(-y)) & \text {(Definition)}\\ &= a\cdot (x+(-y))+b\cdot (x+(-y)) & \text {(Axiom VI)}\\ &= (x+(-y)) \cdot a +(x+(-y)) \cdot b & \text {(Axiom II)}\\ &= xa + (-y)\cdot a + xb+(-y)\cdot b & \text {(Axiom VI)}\\ &= xa+ (- (y\cdot a))+xb +(-(y\cdot b)) & \text {(Corollary 6)}\\ &= ax+(-ay) +bx+(-by) & \text {(Axiom II)} \\ &= ax+ bx-ay-by & \text {(by (i))} \end {align*}

When \(x=a\) and \(y=b\), we have, by the current formula: \begin {align*} (a+b)(a-b) &= a\cdot a+ b\cdot a-a\cdot b-b\cdot b\\ &= a^2 + b\cdot a-a\cdot b-b^2 & \text {(Definition)}\\ &= a^2+ a\cdot b - a\cdot b - b^2 & \text {(Axiom II)}\\ &= a^2 +0-b^2 & \text {(Axiom V)}\\ &= a^2 -b^2 & \text {(Axiom IV)} \end {align*}

Assume for a \(k\) term expression we have \((a_1+a_2+...+a_k)x= a_1x+a_2x+...+a_kx\). Then, for the \(k+1\) term expression, we have: \begin {align*} (a_1+a_2+...a_k+a_{k+1})x &= ((a_1+a_2+...+a_k)+a_{k+1})x & \text {(Definition)}\\ &= (a_1+a_2+...+a_k)x + a_{k+1}x & \text {(Axiom VI)}\\ &= a_1x+a_2x+...+a_kx +a_{k+1}x & \text {(Inductive Hypothesis)} \end {align*}

Hence, the equation holds for \(k+1\) term expression if it holds for a \(k\) term expression, because the equation is shown to hold for a \(3\) term expression, by induction, it holds for all \(k\geq 3\) term expression including \(4\) and \(5\) term expressions.

(i)

If \(x>0\), then also \(x^{-1}>0.\)

(ii)

If \(x>y>z>u\), then \(x>u\).

(iii)

If \(x>y \geq 0\), then \begin {align*} x^2>y^2\geq 0\;\;\text {and}\;\; x^3>y^3 \geq 0\;\; \text {(where $x^3=x^2x$);} \end {align*}

similarly, \begin {align*} x^4>y^4\geq 0\;\;\text {(where $x^4=x^3x$).} \end {align*}

(iv)

If \(x>y>0\), then \(1/x<1/y\). What if \(x>0>y\) or \(0>x>y\)?

(v)

\(|a+b+c|\leq |a|+|b|+|c|\) and \(|a+b+c+d|\leq |a|+|b|+|c|+|d|.\)

(i)

We have to prove that when \(x>0\), \(x^{-1}>0\). Assume that \(x^{-1}<0\) instead but \(x>0\). We have: \begin {align*} x^{-1} &<0 \\ x^{-1}\cdot x &< 0\cdot x & \text {(Axiom IX)}\\ x^{-1}\cdot x &< 0 & \text {(Corollary 5)}\\ 1&<0 & \text {(Axiom V)} \end {align*}

But, if \(1<0\), then by the monotonicity of multiplication, we have \(x<0\), which contradicts our premise. Hence \(x^{-1}\) is either \(0\) or greater than \(0\), by the trichotomy of an ordered field. But, if \(x^{-1}=0\), then \begin {align*} x^{-1}&=0 \\ x^{-1}\cdot x &= 0 \cdot x & \text {(Corollary 2)}\\ x^{-1}\cdot x &=0 &\text {(Corollary 5)}\\ 1&=0 & \text {(Axiom V)} \end {align*}

But, if \(1=0\), then, by Corollary 2 and 5, we have \(x=0\), which again contradicts our premise. Hence, by the trichotomy of an ordered field, \(x^{-1}>0\) when \(x>0\).

(ii)

If \(x>y\) and \(y>z\), then by transitive property of an ordered field, \(x>z\). If in addition, \(z>u\), applying the transitive property once more, \(x>u\).

(iii)

Given \(x>y\geq 0\), we need to prove that \(x^2>y^2\geq 0\). We have \begin {align*} x &>y \\ x\cdot x &>y \cdot y & \text {(Corollary 8)}\\ x^2&>y^2 \geq 0 & \text {(Corollary 9)} \end {align*}

Similarly, applying Corollary 8, we get \(x^3>y^3\), Then because \(x^2>0\), \(x^3>0\) by Axiom IX and Corollary 5. Similarly \(y^3\geq 0\), Hence \(x^3>y^3\geq 0\). With the same reasoning, we also get \(x^4>y^4\geq 0\).

(iv)

If \(x>y>0\), then we need to prove that \(1/x<1/y\), and how this changes if \(x>0>y\) or \(0>x>y\). \begin {align*} x>y>0 \\ x^{-1},\:\: y^{-1} >0 \end {align*}

By, (i), Hence, Multiplying both sides of the inequality, \(x>y\) by \(x^{-1}\) and \(y^{-1}\), by the monotonicity of multiplication (Axiom IX), we get: \begin {align*} x&>y\\ 1&>yx^{-1}\\ y^{-1}&>x^{-1}. \end {align*}

If \(x>0>y\), \(x^{-1}>0\) by (ii), but, \(y^{-1}<0\), but by Corollary 7, \(-y^{-1}\) would be positive. Hence, by monotonicity of multiplication (Axiom IX), \begin {align*} x&>y\\ 1&>yx^{-1}\\ -y^{-1}&> -x^{-1}\\ x^{-1}&>y^{-1} \end {align*}

Similarly, if \(0>x>y\), then \(-x^{-1}\) and \(-y^{-1}\) are positive, hence by the monotonicity of multiplication, \begin {align*} x&>y\\ -1&>-yx^{-1}\\ y^{-1}&>x^{-1} \end {align*}

(v)

Now, we need to show that \(|a+b+c|\leq |a|+|b|+|c|\). \begin {align*} |a+b+c| &= |(a+b)+c| & \text {(Definition)}\\ &\leq |(a+b)|+|c| &\text {(Corollary 12)}\\ &\leq (|a|+|b|)+|c| & \text {(Corollary 12)}\\ &\leq |a|+|b|+|c| \end {align*}

The same reasoning applies for an \(k\) term sum.