# Tag Info

200

Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero". You must ...

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Many people believe that since humans have $10$ fingers, we use base $10$. Let's assume that the Martians have $b$ fingers and thus use a base $b$ numbering system, where $b \neq 10$ (note that we can't have $b=10$, since in base $10$, $x=8$ shouldn't be a solution). Then since the $50$ and $125$ in the equation are actually in base $b$, converting them to ...

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The magic of the number 10 comes from the fact that "1" is the multiplicative unit and "0" is the additive unit. The first two-digit-number in positional notation is always 10 and also always denotes the number of digits.

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Many languages have (at least relicts of) non-decimal counting, very often vigesimal (because we have 20 fingers plus toes), but also many other systems. I recommend an old Gutenberg project of mine, The Number Concept Note for example that the Danish word for 55 is femoghalvtreds "five more than half the third twenty-block"

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You may also find of interest some more general results besides the mentioned Frobenius Theorem. Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents ($\rm\:x^n = 0 \ \Rightarrow\ x = 0\:$) is isomorphic as a ring to a direct sum of copies of $\rm\:\mathbb R\:$ and ...

44

Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions. Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic ...

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The formal way to understand this is, of course, using the definition of real numbers. A real number is "allowed" to have infinite digits after the decimal point, but only a finite number of digits before. (http://en.wikipedia.org/wiki/Real_number) (if it interests you, there are numbers that have infinite digits before the decimal point, and only a finite ...

36

There are two terminologies that I'm familiar with. Sometimes, the part to the right of the decimal (cents) is called the mantissa, and the part to the left (dollars, in your metaphor), is called the characteristic. But I also like the generic terms integer-part and fractional-part. It's what I and those with whom I do research call them (who uses the work ...

33

One way to convert any decimal fraction to base $16$ is as follows (taking $\pi$ as an example).$$\pi=\color{blue}3.141592...$$ Take the whole number part and convert it to base $16$ as usual. In this case $\color{blue}3$ will remain as $3$. So we have so far got $3.14159..._{10}=\color{red}{3...._{16}}$ This now leaves us with $0.141592...$ - ...

30

It is unclear exactly what kind of "basis transformations" you want to allow, but presumably any such thing would be a bijection $\mathbb{R}\to\mathbb{R}$. But there is no bijection $\mathbb{R}\to\mathbb{R}$ which exchanges the irrationals and the rationals, because it would restrict to a bijection between the rationals and the irrationals, and the ...

29

Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then $$10^mx=10^ma+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and $$10^{m+p}x=10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots ... 28 The short answer to your question is that by definition we only allow real numbers to have finitely many digits before the decimal point. There are very good reasons for this: Formally, we can think of a number as a finite sequence of digits x_0,\ x_1, \ \ldots , x_N, where the number x is equal to$$x=\sum_{n=0}^Nx_n10^n$$For example, the number 126 ... 28 If we are working in base b (we must have b\gt3), then 0.3333\ldots is$$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$Since$$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$then... 28 In base 6, 14641 is not a valid representation of a number, because 6 is not a valid digit in base 6. It is true that 1+4\cdot 6 + 6\cdot 6^2+4\cdot 6^3+6^4 is still equal to 7^4. But the proper representation for this number in base 6 is 15041_6. 28 Expanding on the comment by J.M., let me quote from the (highly recommended) book by Georges Ifrah The Universal History of Numbers (Wiley, 2000, pp. 21-22): Traces of the anthropomorphic origin of counting systems can be found in many languages. In the Ali language (Central Africa), for example, "five" and "ten" are respectively moro and mbouna: moro ... 28 13 fingers. Translate 5x^2-50x+125 into base-b:$$ 5x^2-(5b)x+(b^2+2b+5) $$Since this has roots x=5 and x=8 we must have$$ 5x^2-(5b)x+(b^2+2b+5)=k(x-5)(x-8)=kx^2-13kx+40k $$so, equating coefficients,$$ 5=k,\quad 5b=13k,\quad b^2+2b+5=40k $$and so b=13. It's easy to check that the last equation is satisfied as well. Perhaps the Martians had ... 27 actually numbers from 11 to 16 are quite regular in French (and in Italian) too: they just are a derivation from Latin. | French | Italian | Latin un | on·ze | un·dici | un·decim deux | dou·ze | do·dici | duo·decim trois | trei·ze | tre·dici | tre·decim quatre | quator·ze | quattor·dici | ... 27 Hint: if we multiply 0.33333\ldots by 5 then we get 0.(15)(15)(15)(15)(15)\ldots. Compare that to what happens when we multiply the same by 3: 0.99999\ldots, and its interpretation in decimal. 26 Your intuition is correct for instance for all b > 2, \frac{1}{b-1} is not going to have a finite representation, and will have the representation \frac{1}{b-1} = 0.1111111...._b. Eg, \frac{1}{9} = .11111111 in base 10. 26 Actually, if you go back in time a bit in English, you'll realise that English was 'strange' too: Four score and seven years ago our fathers brought forth on this continent a new nation, conceived in liberty, and dedicated to the proposition that all men are created equal. (The Gettysberg Address, 1863) Now if you were to translate that into French in ... 23 As you mentioned,$$6 = {\color{red}1}\cdot 2^2+ {\color{red}1}\cdot 2^1+{\color{red}0}\cdot 2^0 = {\color{red}{110}}_B.$$Analogously$$\frac{1}{4} = \frac{1}{2^2} = {\color{red}0}\cdot2^0 + {\color{red}0}\cdot 2^{-1} + {\color{red}1}\cdot 2^{-2} = {\color{red}{0.01}}_B.$$Edit: These pictures might give you some more intuition ;-) Here \frac{5}{16} = ... 23 Every finite-dimensional division algebra over \mathbb{R} is one of \mathbb{R}, \mathbb{C} or \mathbb{H}. This is what is called the Frobenius Theorem. You may refer to here for details. 23 In an interview, you can impress the interviewer, by mentally calculating and determining the result. As other answers have mentioned, you need to express the equation in base different from 10 and then equate it with the roots of the equation. From the options, its clear that the base is greater than 10. That means 5 and 8 are unit digits in some ... 21 Of course it seems natural to you; you grew up in the modern world, where everyone accepts zero. More importantly, people now accept the abstract concept of numbers and are capable of divorcing them from the things that they represent. This is a sophisticated point of view. From a more naive point of view, a number is a property of a collection of ... 21 I would like to expand on Trevor Wilson's answer. Base-b representation of integers is rooted in the fact that, for any non-negative integer n, there is a unique representation of n in the form$$n = \sum_{i=0}^\infty a_ib^i where $0 \le a_i < b$. For example, when $b$ is 3, and $n$ is 47, the unique solution has $a_0 = 2, a_1 = 0, a_2 = 2, a_3 ... 21 The natural numbers can be defined by Peano's Axioms (sometimes called the Peano Postulates): Zero is a number. If n is a number, the successor of n is a number. zero is not the successor of a number. Two numbers of which the successors are equal are themselves equal. (induction axiom.) If a set S of numbers contains zero and also the successor of every ... 21 As far as I can see, no definition of scientific notation states that every number must be able to be expressed. As it so happens, each nonzero number can be expressed in scientific notation. Zero is a special case. There is no way to have$a\cdot 10^k = 0$for$1\leq |a|<10$and$k\in\mathbb{Z}$since$x\cdot y = 0\Leftrightarrow (x=0~ \text{or} ...

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As a professor who faces this issue every time I teach (cryptography and algorithms both tend to run into non-decimal bases), I have the following policy: If decimal, just say the number (with the word "decimal" if we're mixing contexts) If any other base, read the digits and say the name of the base So I might say, "therefore the answer is one-zero-one ...

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