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user21820
user21820
7:00 AM
0
A: What are some simple problems that calculators get wrong?

user21820Here is a simple example. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $   $\color{green}{\Big( 99 · \big( 100·\sum_{k=0}^{10} (-\lfrac1{99})^k-99 \big)^\lfrac1{10} - 1 \Big)^\lfrac1{9} = 0}$. Windows Calc ( 99*( 100*(1-1/99+1/99y2-1/99y3+1/99y4-1/99y5+1/99y6-1/99y7+1/99y8-1/99y9+1/99y10)-99 )y(1/10) ...

 
 
5 hours later…
F. Zer
F. Zer
11:45 AM
Hi @user21820. Can I ask you a question ? I realize that there is something basic I do not understand.

I have the following binary digits:

111
If I multiply the digits by the weights of each of the numbers and sum:

111 = 1 * 2^2 + 1 * 2^1 + 1 * 2^0
is 7 in decimal numbering system. Now…

Why the result is always in decimal notation ? Is this because decimal is our default numbering system or there is another math trick I do not know about ?
So, really my question is, should I write instead:
Indicating what’s the equivalence between the LHS and RHS ? And that I should use the multiplication operator for decimal representation ?
 
 
2 hours later…
user21820
user21820
2:08 PM
@F.Zer Simply because we all agree to use decimal!
It's meaningless to ask to write the base everywhere, otherwise you can't write anything down.
For example, you wrote "111[2]", where the "2" is in (guess what?) base 10!
And you wrote "2[10]", where again the "10" is in.. base 10!
Same for the exponent "2"!
If you don't want to use decimal, that's your choice. After all, if you have reached PA you know that you only need ⟨0,1,+,·⟩ to express any natural number.
But it's inconvenient for everyday mathematics without base 10.
@F.Zer So to express what you want all you have to write is "111[2] = 1·2^2+1·2^1+1·2^0".
 
soupless
soupless
2:54 PM
I forgot how to do uniqueness proofs. Can I ask if this is correct? 
Given x,y∈ℕ:
    (x+1)(y+1) = (x+1)(y+1).
    ∃c∈ℕ (c = (x+1)(y+1)).
    Let P(n) ≡ n = (x+1)(y+1) for each n∈ℕ.
    Let d∈ℕ such that P(d).

    Given z∈ℕ:
        If P(z):
            z = d.
        P(z) ⇒ z = d.
    ∀z∈ℕ (P(z) ⇒ z = d).
    P(d) ∧ ∀z∈ℕ (P(z) ⇒ z = d).
    ∃c∈ℕ (P(c) ∧ ∀z∈ℕ (P(z) ⇒ z = c)).
    ∃!c∈ℕ (P(c)).

    ∃!c∈ℕ (c = (x+1)(y+1)).
∀x,y∈ℕ ∃!c∈ℕ (c = (x+1)(y+1)).
 
soupless
soupless
3:11 PM
@user21820 If you define multiple infix binary function-symbols, how will you say which operator precedes the other?
 
 
8 hours later…
F. Zer
F. Zer
11:30 PM
@user21820 Thank you very much !
 

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