Convert the Number -1 880.599 999 999 999 909 050 529 822 707 176 208 496 9 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number -1 880.599 999 999 999 909 050 529 822 707 176 208 496 9(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)
The first steps we'll go through to make the conversion:
Convert to binary (to base 2) the integer part of the number.
Convert to binary the fractional part of the number.
1. Start with the positive version of the number:
|-1 880.599 999 999 999 909 050 529 822 707 176 208 496 9| = 1 880.599 999 999 999 909 050 529 822 707 176 208 496 9
2. First, convert to binary (in base 2) the integer part: 1 880.
Divide the number repeatedly by 2.
Keep track of each remainder.
We stop when we get a quotient that is equal to zero.
- division = quotient + remainder;
- 1 880 ÷ 2 = 940 + 0;
- 940 ÷ 2 = 470 + 0;
- 470 ÷ 2 = 235 + 0;
- 235 ÷ 2 = 117 + 1;
- 117 ÷ 2 = 58 + 1;
- 58 ÷ 2 = 29 + 0;
- 29 ÷ 2 = 14 + 1;
- 14 ÷ 2 = 7 + 0;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
3. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
1 880(10) =
111 0101 1000(2)
4. Convert to binary (base 2) the fractional part: 0.599 999 999 999 909 050 529 822 707 176 208 496 9.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.
- #) multiplying = integer + fractional part;
- 1) 0.599 999 999 999 909 050 529 822 707 176 208 496 9 × 2 = 1 + 0.199 999 999 999 818 101 059 645 414 352 416 993 8;
- 2) 0.199 999 999 999 818 101 059 645 414 352 416 993 8 × 2 = 0 + 0.399 999 999 999 636 202 119 290 828 704 833 987 6;
- 3) 0.399 999 999 999 636 202 119 290 828 704 833 987 6 × 2 = 0 + 0.799 999 999 999 272 404 238 581 657 409 667 975 2;
- 4) 0.799 999 999 999 272 404 238 581 657 409 667 975 2 × 2 = 1 + 0.599 999 999 998 544 808 477 163 314 819 335 950 4;
- 5) 0.599 999 999 998 544 808 477 163 314 819 335 950 4 × 2 = 1 + 0.199 999 999 997 089 616 954 326 629 638 671 900 8;
- 6) 0.199 999 999 997 089 616 954 326 629 638 671 900 8 × 2 = 0 + 0.399 999 999 994 179 233 908 653 259 277 343 801 6;
- 7) 0.399 999 999 994 179 233 908 653 259 277 343 801 6 × 2 = 0 + 0.799 999 999 988 358 467 817 306 518 554 687 603 2;
- 8) 0.799 999 999 988 358 467 817 306 518 554 687 603 2 × 2 = 1 + 0.599 999 999 976 716 935 634 613 037 109 375 206 4;
- 9) 0.599 999 999 976 716 935 634 613 037 109 375 206 4 × 2 = 1 + 0.199 999 999 953 433 871 269 226 074 218 750 412 8;
- 10) 0.199 999 999 953 433 871 269 226 074 218 750 412 8 × 2 = 0 + 0.399 999 999 906 867 742 538 452 148 437 500 825 6;
- 11) 0.399 999 999 906 867 742 538 452 148 437 500 825 6 × 2 = 0 + 0.799 999 999 813 735 485 076 904 296 875 001 651 2;
- 12) 0.799 999 999 813 735 485 076 904 296 875 001 651 2 × 2 = 1 + 0.599 999 999 627 470 970 153 808 593 750 003 302 4;
- 13) 0.599 999 999 627 470 970 153 808 593 750 003 302 4 × 2 = 1 + 0.199 999 999 254 941 940 307 617 187 500 006 604 8;
- 14) 0.199 999 999 254 941 940 307 617 187 500 006 604 8 × 2 = 0 + 0.399 999 998 509 883 880 615 234 375 000 013 209 6;
- 15) 0.399 999 998 509 883 880 615 234 375 000 013 209 6 × 2 = 0 + 0.799 999 997 019 767 761 230 468 750 000 026 419 2;
- 16) 0.799 999 997 019 767 761 230 468 750 000 026 419 2 × 2 = 1 + 0.599 999 994 039 535 522 460 937 500 000 052 838 4;
- 17) 0.599 999 994 039 535 522 460 937 500 000 052 838 4 × 2 = 1 + 0.199 999 988 079 071 044 921 875 000 000 105 676 8;
- 18) 0.199 999 988 079 071 044 921 875 000 000 105 676 8 × 2 = 0 + 0.399 999 976 158 142 089 843 750 000 000 211 353 6;
- 19) 0.399 999 976 158 142 089 843 750 000 000 211 353 6 × 2 = 0 + 0.799 999 952 316 284 179 687 500 000 000 422 707 2;
- 20) 0.799 999 952 316 284 179 687 500 000 000 422 707 2 × 2 = 1 + 0.599 999 904 632 568 359 375 000 000 000 845 414 4;
- 21) 0.599 999 904 632 568 359 375 000 000 000 845 414 4 × 2 = 1 + 0.199 999 809 265 136 718 750 000 000 001 690 828 8;
- 22) 0.199 999 809 265 136 718 750 000 000 001 690 828 8 × 2 = 0 + 0.399 999 618 530 273 437 500 000 000 003 381 657 6;
- 23) 0.399 999 618 530 273 437 500 000 000 003 381 657 6 × 2 = 0 + 0.799 999 237 060 546 875 000 000 000 006 763 315 2;
- 24) 0.799 999 237 060 546 875 000 000 000 006 763 315 2 × 2 = 1 + 0.599 998 474 121 093 750 000 000 000 013 526 630 4;
- 25) 0.599 998 474 121 093 750 000 000 000 013 526 630 4 × 2 = 1 + 0.199 996 948 242 187 500 000 000 000 027 053 260 8;
- 26) 0.199 996 948 242 187 500 000 000 000 027 053 260 8 × 2 = 0 + 0.399 993 896 484 375 000 000 000 000 054 106 521 6;
- 27) 0.399 993 896 484 375 000 000 000 000 054 106 521 6 × 2 = 0 + 0.799 987 792 968 750 000 000 000 000 108 213 043 2;
- 28) 0.799 987 792 968 750 000 000 000 000 108 213 043 2 × 2 = 1 + 0.599 975 585 937 500 000 000 000 000 216 426 086 4;
- 29) 0.599 975 585 937 500 000 000 000 000 216 426 086 4 × 2 = 1 + 0.199 951 171 875 000 000 000 000 000 432 852 172 8;
- 30) 0.199 951 171 875 000 000 000 000 000 432 852 172 8 × 2 = 0 + 0.399 902 343 750 000 000 000 000 000 865 704 345 6;
- 31) 0.399 902 343 750 000 000 000 000 000 865 704 345 6 × 2 = 0 + 0.799 804 687 500 000 000 000 000 001 731 408 691 2;
- 32) 0.799 804 687 500 000 000 000 000 001 731 408 691 2 × 2 = 1 + 0.599 609 375 000 000 000 000 000 003 462 817 382 4;
- 33) 0.599 609 375 000 000 000 000 000 003 462 817 382 4 × 2 = 1 + 0.199 218 750 000 000 000 000 000 006 925 634 764 8;
- 34) 0.199 218 750 000 000 000 000 000 006 925 634 764 8 × 2 = 0 + 0.398 437 500 000 000 000 000 000 013 851 269 529 6;
- 35) 0.398 437 500 000 000 000 000 000 013 851 269 529 6 × 2 = 0 + 0.796 875 000 000 000 000 000 000 027 702 539 059 2;
- 36) 0.796 875 000 000 000 000 000 000 027 702 539 059 2 × 2 = 1 + 0.593 750 000 000 000 000 000 000 055 405 078 118 4;
- 37) 0.593 750 000 000 000 000 000 000 055 405 078 118 4 × 2 = 1 + 0.187 500 000 000 000 000 000 000 110 810 156 236 8;
- 38) 0.187 500 000 000 000 000 000 000 110 810 156 236 8 × 2 = 0 + 0.375 000 000 000 000 000 000 000 221 620 312 473 6;
- 39) 0.375 000 000 000 000 000 000 000 221 620 312 473 6 × 2 = 0 + 0.750 000 000 000 000 000 000 000 443 240 624 947 2;
- 40) 0.750 000 000 000 000 000 000 000 443 240 624 947 2 × 2 = 1 + 0.500 000 000 000 000 000 000 000 886 481 249 894 4;
- 41) 0.500 000 000 000 000 000 000 000 886 481 249 894 4 × 2 = 1 + 0.000 000 000 000 000 000 000 001 772 962 499 788 8;
- 42) 0.000 000 000 000 000 000 000 001 772 962 499 788 8 × 2 = 0 + 0.000 000 000 000 000 000 000 003 545 924 999 577 6;
- 43) 0.000 000 000 000 000 000 000 003 545 924 999 577 6 × 2 = 0 + 0.000 000 000 000 000 000 000 007 091 849 999 155 2;
- 44) 0.000 000 000 000 000 000 000 007 091 849 999 155 2 × 2 = 0 + 0.000 000 000 000 000 000 000 014 183 699 998 310 4;
- 45) 0.000 000 000 000 000 000 000 014 183 699 998 310 4 × 2 = 0 + 0.000 000 000 000 000 000 000 028 367 399 996 620 8;
- 46) 0.000 000 000 000 000 000 000 028 367 399 996 620 8 × 2 = 0 + 0.000 000 000 000 000 000 000 056 734 799 993 241 6;
- 47) 0.000 000 000 000 000 000 000 056 734 799 993 241 6 × 2 = 0 + 0.000 000 000 000 000 000 000 113 469 599 986 483 2;
- 48) 0.000 000 000 000 000 000 000 113 469 599 986 483 2 × 2 = 0 + 0.000 000 000 000 000 000 000 226 939 199 972 966 4;
- 49) 0.000 000 000 000 000 000 000 226 939 199 972 966 4 × 2 = 0 + 0.000 000 000 000 000 000 000 453 878 399 945 932 8;
- 50) 0.000 000 000 000 000 000 000 453 878 399 945 932 8 × 2 = 0 + 0.000 000 000 000 000 000 000 907 756 799 891 865 6;
- 51) 0.000 000 000 000 000 000 000 907 756 799 891 865 6 × 2 = 0 + 0.000 000 000 000 000 000 001 815 513 599 783 731 2;
- 52) 0.000 000 000 000 000 000 001 815 513 599 783 731 2 × 2 = 0 + 0.000 000 000 000 000 000 003 631 027 199 567 462 4;
- 53) 0.000 000 000 000 000 000 003 631 027 199 567 462 4 × 2 = 0 + 0.000 000 000 000 000 000 007 262 054 399 134 924 8;
We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)
5. Construct the base 2 representation of the fractional part of the number.
Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:
0.599 999 999 999 909 050 529 822 707 176 208 496 9(10) =
0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)
6. Positive number before normalization:
1 880.599 999 999 999 909 050 529 822 707 176 208 496 9(10) =
111 0101 1000.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)
The last steps we'll go through to make the conversion:
Normalize the binary representation of the number.
Adjust the exponent.
Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Normalize the mantissa.
7. Normalize the binary representation of the number.
Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:
1 880.599 999 999 999 909 050 529 822 707 176 208 496 9(10) =
111 0101 1000.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) =
111 0101 1000.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) × 20 =
1.1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0000 0000 000(2) × 210
8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign 1 (a negative number)
Exponent (unadjusted): 10
Mantissa (not normalized):
1.1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0000 0000 000
9. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
10 + 2(11-1) - 1 =
(10 + 1 023)(10) =
1 033(10)
10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.
Use the same technique of repeatedly dividing by 2:
- division = quotient + remainder;
- 1 033 ÷ 2 = 516 + 1;
- 516 ÷ 2 = 258 + 0;
- 258 ÷ 2 = 129 + 0;
- 129 ÷ 2 = 64 + 1;
- 64 ÷ 2 = 32 + 0;
- 32 ÷ 2 = 16 + 0;
- 16 ÷ 2 = 8 + 0;
- 8 ÷ 2 = 4 + 0;
- 4 ÷ 2 = 2 + 0;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
11. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above.
Exponent (adjusted) =
1033(10) =
100 0000 1001(2)
12. Normalize the mantissa.
a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.
b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).
Mantissa (normalized) =
1. 1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 000 0000 0000 =
1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110
13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:
Sign (1 bit) =
1 (a negative number)
Exponent (11 bits) =
100 0000 1001
Mantissa (52 bits) =
1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110
The base ten decimal number -1 880.599 999 999 999 909 050 529 822 707 176 208 496 9 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 1001 - 1101 0110 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110
Sign (1 bit):
Exponent (11 bits):
1
620
610
600
590
580
570
561
550
540
531
52
Mantissa (52 bits):
1
511
500
491
480
471
461
450
440
430
421
410
400
391
381
370
360
351
341
330
320
311
301
290
280
271
261
250
240
231
221
210
200
191
181
170
160
151
141
130
120
111
101
90
80
71
61
50
40
31
21
10
0
Convert to 64 bit double precision IEEE 754 binary floating point representation standard
A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)
The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation
How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard
Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:
- 1. If the number to be converted is negative, start with its the positive version.
- 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
- 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
- 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
- 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
- 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 - 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
- 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.
Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:
- 1. Start with the positive version of the number:
|-31.640 215| = 31.640 215
- 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
- division = quotient + remainder;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
- We have encountered a quotient that is ZERO => FULL STOP
- 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:
31(10) = 1 1111(2)
- 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
- #) multiplying = integer + fractional part;
- 1) 0.640 215 × 2 = 1 + 0.280 43;
- 2) 0.280 43 × 2 = 0 + 0.560 86;
- 3) 0.560 86 × 2 = 1 + 0.121 72;
- 4) 0.121 72 × 2 = 0 + 0.243 44;
- 5) 0.243 44 × 2 = 0 + 0.486 88;
- 6) 0.486 88 × 2 = 0 + 0.973 76;
- 7) 0.973 76 × 2 = 1 + 0.947 52;
- 8) 0.947 52 × 2 = 1 + 0.895 04;
- 9) 0.895 04 × 2 = 1 + 0.790 08;
- 10) 0.790 08 × 2 = 1 + 0.580 16;
- 11) 0.580 16 × 2 = 1 + 0.160 32;
- 12) 0.160 32 × 2 = 0 + 0.320 64;
- 13) 0.320 64 × 2 = 0 + 0.641 28;
- 14) 0.641 28 × 2 = 1 + 0.282 56;
- 15) 0.282 56 × 2 = 0 + 0.565 12;
- 16) 0.565 12 × 2 = 1 + 0.130 24;
- 17) 0.130 24 × 2 = 0 + 0.260 48;
- 18) 0.260 48 × 2 = 0 + 0.520 96;
- 19) 0.520 96 × 2 = 1 + 0.041 92;
- 20) 0.041 92 × 2 = 0 + 0.083 84;
- 21) 0.083 84 × 2 = 0 + 0.167 68;
- 22) 0.167 68 × 2 = 0 + 0.335 36;
- 23) 0.335 36 × 2 = 0 + 0.670 72;
- 24) 0.670 72 × 2 = 1 + 0.341 44;
- 25) 0.341 44 × 2 = 0 + 0.682 88;
- 26) 0.682 88 × 2 = 1 + 0.365 76;
- 27) 0.365 76 × 2 = 0 + 0.731 52;
- 28) 0.731 52 × 2 = 1 + 0.463 04;
- 29) 0.463 04 × 2 = 0 + 0.926 08;
- 30) 0.926 08 × 2 = 1 + 0.852 16;
- 31) 0.852 16 × 2 = 1 + 0.704 32;
- 32) 0.704 32 × 2 = 1 + 0.408 64;
- 33) 0.408 64 × 2 = 0 + 0.817 28;
- 34) 0.817 28 × 2 = 1 + 0.634 56;
- 35) 0.634 56 × 2 = 1 + 0.269 12;
- 36) 0.269 12 × 2 = 0 + 0.538 24;
- 37) 0.538 24 × 2 = 1 + 0.076 48;
- 38) 0.076 48 × 2 = 0 + 0.152 96;
- 39) 0.152 96 × 2 = 0 + 0.305 92;
- 40) 0.305 92 × 2 = 0 + 0.611 84;
- 41) 0.611 84 × 2 = 1 + 0.223 68;
- 42) 0.223 68 × 2 = 0 + 0.447 36;
- 43) 0.447 36 × 2 = 0 + 0.894 72;
- 44) 0.894 72 × 2 = 1 + 0.789 44;
- 45) 0.789 44 × 2 = 1 + 0.578 88;
- 46) 0.578 88 × 2 = 1 + 0.157 76;
- 47) 0.157 76 × 2 = 0 + 0.315 52;
- 48) 0.315 52 × 2 = 0 + 0.631 04;
- 49) 0.631 04 × 2 = 1 + 0.262 08;
- 50) 0.262 08 × 2 = 0 + 0.524 16;
- 51) 0.524 16 × 2 = 1 + 0.048 32;
- 52) 0.048 32 × 2 = 0 + 0.096 64;
- 53) 0.096 64 × 2 = 0 + 0.193 28;
- We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:
0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 6. Summarizing - the positive number before normalization:
31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:
31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24
- 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign: 1 (a negative number)
Exponent (unadjusted): 4
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
- 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)
- 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
- Conclusion:
Sign (1 bit) = 1 (a negative number)
Exponent (8 bits) = 100 0000 0011
Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Available Base Conversions Between Decimal and Binary Systems
Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):
1. Integer -> Binary
2. Decimal -> Binary
3. Binary -> Integer
4. Binary -> Decimal