0

Is my understanding correct that a number is in a scientific normalized form, a negative exponent always signifies a fraction? I'm thinking that if in a normalized form only one digit is allowed before the radix point, it's enough to have $-1$ in the exponent to have a fraction

Max Koretskyi
  • 761
  • Depends what you mean by "scientific normalized form", and what you mean by fraction. Are you using "fraction" just to mean any real number between zero and one? – Gerry Myerson Sep 29 '16 at 07:25
  • @GerryMyerson, scientific normalized form. By fraction I mean any real number that is not an integer – Max Koretskyi Sep 29 '16 at 08:01
  • Yes. But you are badly abusing terminology and the word "fraction". As you now scientific notation is $N=a.bc \times 10^k $. If $k \ge 0$ then $10^k \ge 1$ and $N\ge 1$. If $k <0$ then $10^k \le 1/10$ and $N <1$. But no mathematician would call a number less than 1 a fraction, or a number greater than 1 a non-fraction. – fleablood Sep 29 '16 at 09:02
  • A fraction is a fairly useless mathematical term. It means anything written as a/b. So 27/1 is a fraction but 27 is not. 1/4 is a fraction but .25 is not. Anyway.... 7.345 x $10^2$ = 734.5 and has 3 leading terms. w x $10^k;k \ge k $ will have k+1 leading terms. 7.345 x $10^{-3} $=0.007345$ which is.... OH! I get what you meant! You meant fractional part! Every real x = [x] + {x} where [x] is an integer and 0$\le $ {x} < 1. {x} is called the fractional part. It is true that if the exponent is negative, the number is "purely fractional" as the integer part is 0. – fleablood Sep 29 '16 at 09:15
  • ....so 7.345 x 10^2 is a fraction? Well, fine. w x 10^-k = w/10^k < 1 is not an integer. So you are correct. – fleablood Sep 29 '16 at 09:19
  • @fleablood, I see, thanks a lot for your insights – Max Koretskyi Sep 29 '16 at 09:49
  • Take what I say with a grain of salt as pure mathematicians seem to speak a different language than scientists. But the upshot is.... what you thought a negative exponent meant is what a negative meant. (I.e. less than 1) – fleablood Sep 29 '16 at 09:54

1 Answers1

1

OK, so, we're talking about numbers of the form $\pm d_0.d_1d_2\dots\times10^n$ where the $d_i$ are digits, $d_0\ne0$, and $n<0$. Well, any such number is strictly between $-1$ and 1, and is not zero, so it's certainly not an integer.

Gerry Myerson
  • 179,216