Same with me pointing out in Bruno Mars' video " It will Rain", that morphine is used for physical, not emotional, pain. No wonder I'm writing this comment from my home on Friday night , while all others are out in bats, clubs.
Well, part of the issue is that exp(z) plays a role when dealing with Complexes, e.g., Polar representation, but it's not 1-1, and as such has no global inverse log. This requires you to patch together local inverses. A mess.
Edit: Here the square root too, may involve arguments, branches, and...
Sure,
https://en.wikipedia.org/wiki/Compactness_theorem
Through the concept of Elementary Equivalence , aka, the Transfer Principle, 1st -order properties are preserved "upwardly", from lower to higher cardinality, between models of different cardinality.
Edit: Is that what you were asking?
You only need to check through ## \aleph_0## models of the Reals , per Lowenheim -Skolem. Outside of the standard Reals, you lose " standard " metrizability, since metric are Real-valued.
Or, to be pedantic, use that the Complexes can be seen as a field extension by ## x^2+ 1 ##. Then, in the extension ## \mathbb R[x]/(x^2 +1) ; x^2 +1 =0 ## You designate this new element by ##i ##..
Eigenvalues ##\lambda## for a matrix ##A##are defined to satisfy ##Det(A-\lambda I)=0##. This comes from ##Ax=\lambda x ##, so that ##(A-\lambda I )x=0 ##.
One result I remember is that given ##G## a Lie Group; here ## \mathbb T^2##, a subset is a Lie Group if it's both a subgroup and topologically closed in ##G##.
Worse comes to worse, time is running out on the exam, cube both sides. Only one term will remain to a non-unity power p/q. Leave it alone in one side and raise both sides to the qth power. Hardly elegant, but at least you'll get some credit.
Edit: In this exercise, you first cube both sides...
Had to correct my friend . I told him to put a potato in his bathing suit at the beach to help him impress women. But In the front, not in the back, as he did last time.
This works, but needs rescaling to fit the OP:
Assume there are numbers S between 0 and 1. Then since the set is bounded, it has a least upper bound L, so that:
0<L<1
Now multiply thtough by L:
0< L^2<L
Then L^2 is a bound lower than L. Contradiction.