Exercise 4.1.4

Exercise statement

Prove the remaining identities in Proposition 4.1.6.

Proposition 4.1.6 (Laws of algebra for integers). Let $x,y,z$ be integers. Then we have

$x+y = y+x$
$(x+y) + z = x + (y+z)$
$x+0 = 0+x = x$
$x + (-x) = (-x) + x = 0$
$xy = yx$
$(xy)z = x(yz)$
$x1 = 1x = x$
$x(y+z) = xy+xz$
$(y+z)x = yx + zx$

Hints

One can save some work by using some identities to prove others. For instance, once you know that $xy=yx$ , you get for free that $x1=1x$ , and once you also prove $x(y+z) = xy+xz$ , you automatically get $(y+z)x = yx + zx$ for free.

How to think about the exercise

This is the sort of exercise where, even if you know exactly what you are doing, you might slip up and make a mistake shuffling symbols around. I would say to not worry about making small errors like that, as long as you can get most of them right. I checked my work a couple of times, but it’s possible that I’ve made such errors below.

Model solution

Write $x = (a{{-}\!{-}}b)$ , $y=(c{{-}\!{-}}d)$ , and $z = (e{{-}\!{-}}f)$ .

First we show that $x+y = y+x$ . We have $x+y = (a{{-}\!{-}}b)+(c{{-}\!{-}}d) = (a+c){{-}\!{-}}(b+d)$ and $y+x = (c{{-}\!{-}}d)+(a{{-}\!{-}}b) = (c+a){{-}\!{-}}(d+b) = (a+c){{-}\!{-}}(b+d)$ , where the last equality follows from the commutativity of addition on the natural numbers. Thus one sees that $x+y=y+x$ .

Next we show that $(x+y) + z = x + (y+z)$ . We have

$\begin{aligned}(x+y) + z &= ((a{{-}\!{-}}b) + (c{{-}\!{-}}d)) + (e{{-}\!{-}}f) \\ &= ((a+c){{-}\!{-}}(b+d)) + (e{{-}\!{-}}f) \\ &= (a+c+e){{-}\!{-}}(b+d+f)\end{aligned}$

$\begin{aligned}x + (y+z) &= (a{{-}\!{-}}b) + ((c{{-}\!{-}}d) + (e{{-}\!{-}}f)) \\ &= (a{{-}\!{-}}b) + ((c+e){{-}\!{-}}(d+f)) \\ &= (a+c+e){{-}\!{-}}(b+d+f)\end{aligned}$

Thus we see that $(x+y) + z = x + (y+z)$ .

Next we show that $x+0 = 0+x = x$ . We showed the commutativity of addition above, so $x+0 = 0+x$ . So we just need to show $0+x = x$ . We have $0+x = (0{{-}\!{-}}0)+(a{{-}\!{-}}b) = (0+a){{-}\!{-}}(0+b) = a{{-}\!{-}}b = x$ .

Next we show $x + (-x) = (-x) + x = 0$ . By commutativity of addition, we have $x + (-x) = (-x) + x$ . So we just need to show that $(-x) + x = 0$ . We have

$\begin{aligned}(-x) + x &= (b{{-}\!{-}}a) + (a{{-}\!{-}}b) \\ &= (b+a){{-}\!{-}}(a+b) \\ &= (a+b){{-}\!{-}}(a+b)\end{aligned}$

But $(a+b){{-}\!{-}}(a+b) = (0{{-}\!{-}}0)$ since $(a+b)+0 = 0+(a+b)$ .

Next we show that $xy = yx$ . We have $xy = (a{{-}\!{-}}b)(c{{-}\!{-}}d) = (ac+bd){{-}\!{-}}(ad+bc)$ and

$\begin{aligned}yx &= (c{{-}\!{-}}d)(a{{-}\!{-}}b) \\ &= (ca+db){{-}\!{-}}(cb+da) \\ &= (ac+bd){{-}\!{-}}(ad+bc)\end{aligned}$

Thus the two sides are equal, and we have $xy=yx$ .

The associativity of multiplication, $(xy)z = x(yz)$ , was already shown in the book.

Next we show that $x1 = 1x = x$ . We have $x1 = 1x$ by the commutativity of multiplication, which was shown above. Thus we can just show $1x=x$ . We have $1x = (1{{-}\!{-}}0)(a{{-}\!{-}}b) = (a{{-}\!{-}}b) = x$ .

Next we show $x(y+z) = xy+xz$ . We have

$\begin{aligned}x(y+z) &= (a{{-}\!{-}}b)((c{{-}\!{-}}d) + (e{{-}\!{-}}f)) \\ &= (a{{-}\!{-}}b)((c+e){{-}\!{-}}(d+f)) \\ &= (a(c+e) + b(d+f)){{-}\!{-}}(a(d+f) + b(c+e)) \\ &= (ac+ae+bd+bf){{-}\!{-}}(ad+af+bc+be)\end{aligned}$

$\begin{aligned}xy+xz &= (a{{-}\!{-}}b)(c{{-}\!{-}}d) + (a{{-}\!{-}}b)(e{{-}\!{-}}f) \\ &= ((ac+bd){{-}\!{-}}(ad+bc)) + ((ae+bf){{-}\!{-}}(af+be)) \\ &= (ac+bd + ae+bf){{-}\!{-}}(ad+bc + af+be) \\ &= (ac+ae+bd+bf){{-}\!{-}}(ad+af+bc+be)\end{aligned}$

The two sides are equal, so $x(y+z) = xy+xz$ .

Finally, we show that $(y+z)x = yx + zx$ . We have $(y+z)x = x(y+z) = xy+xz = yx+zx$ by the commutativity of multiplication, the distributive law, and commutativity again, all of which were shown earlier in the proof.

Tao Analysis Solutions

Exercise 4.1.4

Exercise statement

Hints

How to think about the exercise

Model solution

2 thoughts on “Exercise 4.1.4”

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Exercise statement

Hints

How to think about the exercise

Model solution

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2 thoughts on “Exercise 4.1.4”

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