We showed in class that sqr root 2 is not rational, i.e., that there
is no rational number y such that y^2= 2. The aim of this
problem is to show that there exists a positive real number
y such that y^2= 2.
Let A = {x is in R | x^2< 2}.
(a) Show that A is subset of [−2, 2].
(b) Suppose that y is the least upper bound of A. Show
that
i) y >= 1;
ii)y^2 >= 2;
iii)y^2 <= 2;
iv)y^2 = 2.
[Hint: Use proofs by contradiction for parts ii)–iii).]
we denote the solution toOriginally Posted by alexmin
to be
. then clearly, for all
,
it can be shown (by examples if necessary) that. thus
(am i begging the question here? i hate doing proofs for things that seem "obvious")
well, if you accept what is in (a), this is immediate. i guess you can show it by example, or come up with some approximation to(b) Suppose that y is the least upper bound of A. Show
that
i) y >= 1;or just some number greater than 1 but less than
assume to the contrary that y^2 < 2. then by the denseness of Q, we can find some rational x in A such that y^2 < x^2 < 2. a contradictionii)y^2 >= 2;
assume to the contrary that y^2 > 2. clearly we have that y^2 > 2 > x^2 for all x in A. but that means that there is some upper bound of the set A that is less than y, therefore, y cannot be the supremum as stated. there's our contradictioniii)y^2 <= 2;
i'm not in the mood for any fancy arguments at this point, i never liked this question to begin with. so i'd just slap on the abstract property of real numbers that says: if a and b are real numbers, with a <= b and b <= a then a = biv)y^2 = 2.
(b)(i) is straightforward: use the definition of upper bound.is an upper bound of
. In particular,
since
.
b(ii)
If, we can find a real number
such that
. Since
for some positive real number
. Thus
. But
(given that
)
. That’s our contradiction.
b(iii)
If, take a real number
. As before we can write
, so
.
is not an upper bound of
such that
. But
; then
. Contradiction again.
Note that b(ii) is true for all upper bounds of
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