Question

find the value of
$(\sqrt{2} ) \times ( \sqrt{2})$
​

Answer 1

Answer:

Math: HSA.REI.A.2. Practice some ... You are finding the possible solutions of x. ... You're multiplying each side by a different amount!

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Answer 2

$\huge \mathfrak {Answer 》}$

Only finite exponents are strictly speaking defined. To make sense of this expression, we have to interpret in terms of some sort of limit. Thankfully, it is pretty clear what we want to do. Consider the sequence where a1=2–√ and

an+1=2–√an.

Then we have

2–√2√2√⋯=limn→∞an.

Does this limit even converge? It isn’t obvious that it does—for example, if we replaced 2–√ with 2 , then it definitely wouldn’t. However, we shall show that

2–√=a1<a2<a3<…<2,

which will prove that a) the limit converges, and b) it converges to a value between 2–√ and 2 .

We will prove this by induction—that is, we shall show that 2–√≤ak<ak+1<2 for all integers k . This is obviously true if k=1 , so it shall suffice to assume that it is true for k=n−1 and prove it for k=n .

That is, we are given that 2–√≤an−1<an<2 , and we need to show that it follows that 2–√≤an<an+1<2 . First, we shall show that

an+1=2–√an>an.

This will follow if we can show that

2an>a2n,

but this is true, since we know 0<an<2 (I won’t provide a proof here, but it can be shown by looking at the second derivative of 2x−x2 ).

Secondly, we note that since 1<an<2 , it follows that

2–√1=2–√<2–√an=an+1<2–√2=2.

Our induction is complete—we know for certain that

2–√=a1<a2<a3<…<2,

as desired.

From here, we have the problem of actually determining the value of this limit. Thankfully, this isn’t too hard. Let

2–√2√2√⋯=limn→∞an=a.

Then,

2–√a=limn→∞2–√an=limn→∞an+1=a,

so it remains to determine what are the solutions to a=2–√a given that 2–√<a≤2 . This is a bit easier if we take a logarithm, yielding

log(a)=a2log(2),

or even better

alog(a)=2log(2).

This has the obvious solution a=2 , but we have to be certain that there aren’t any others. Thankfully, it is easy to check that the function a/log(a) is strictly decreasing on the interval 2–√<a≤2 , and therefore there can only be one solution in that range. We conclude that

2–√2√2√⋯=2.