0 title:

begin quote: " § 4.   The Existence of Square Roots. " end quote:

See ita section 4 of chapter 2.

1 introduction:

If x is any real number, then the square of x is the second power of x:   x² .

begin quote: " A square root of a given number is a number whose square is the given number. Since " end quote: [by the sign rules of arithmetic] begin quote: " the square of any nonzero number is positive, only non-negative numbers can have square roots. The number zero has one square root, which is zero itself. " end quote: See ita-2-4-2-pg28.

2 proposition statement and preliminary analysis:

begin quote: " Proposition. Every positive number has a unique positive square root. " end quote: See ita-2-4-3-pg28.

[By O4]: begin quote: " If then . That is, bigger positive numbers have bigger squares. Thus any given real number can have at most one positive square root. It remains to show that if   aR,   a > 0,   then   a   has at least one positive square root. " end quote: See ita-2-4-4-pg28.

begin quote: " For this purpose consider the set   S = { xR :   x ≥ 0,   x² ≤ a }.   This set is nonempty, since   0 ∈ S,   and bounded from above, since if   x > max{a,1}   we have   x²   =   x · x   >   x · 1   =   x  >  a.   Hence   y = l.u.b. S   exists. " end quote: See ita-2-4-4-pg28.

3 banner for proof:

begin quote: " We proceed to show that   y² = a . " end quote: See ita-2-4-4-pg28.

4 proof:

begin quote: " First,   y > 0,   for   min{1, a} ∈ S,   since [ if   1 < a   then   1 ∈ S,   while if   a < 1   then   aS ] .

Next, for any   ε   such that   0 < ε < y   we have   0   <   y - ε   <   y   <   y + ε ,   so   (y - ε)²   <   y²   <   (y + ε)² ,   since bigger positive numbers have bigger squares.

By the definition of   y   there are numbers greater than   y - ε   in   S,   but   y + ε   ∉   S . [Then, using the definition of S, we get] (y - ε)²   ≤   a   <   (y + ε)² .

Hence   (y - ε)² - (y + ε)²   <   y² - a   <   (y + ε)² - (y - ε)² ,   so   |y² - a|   <   (y + ε)² - (y - ε)²   =   4yε .

The inequality   |y² - a| < 4yε   holds for any   ε   such that   0 < ε < y,   and by choosing   ε   small enough we can make   4yε   smaller than any preassigned positive number. Thus   |y² - a|   is less than any positive number. Since   |y² - a| ≥ 0,   we must have   |y² - a| = 0,   proving   y² = a .

" end quote: See ita-2-4-4-pg28.