-0.016 738 891 601 562 531 086 244 689 504 383 017 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 531 086 244 689 504 383 017(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.016 738 891 601 562 531 086 244 689 504 383 017(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.016 738 891 601 562 531 086 244 689 504 383 017| = 0.016 738 891 601 562 531 086 244 689 504 383 017


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 531 086 244 689 504 383 017.

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) 0.016 738 891 601 562 531 086 244 689 504 383 017 × 2 = 0 + 0.033 477 783 203 125 062 172 489 379 008 766 034;
  • 2) 0.033 477 783 203 125 062 172 489 379 008 766 034 × 2 = 0 + 0.066 955 566 406 250 124 344 978 758 017 532 068;
  • 3) 0.066 955 566 406 250 124 344 978 758 017 532 068 × 2 = 0 + 0.133 911 132 812 500 248 689 957 516 035 064 136;
  • 4) 0.133 911 132 812 500 248 689 957 516 035 064 136 × 2 = 0 + 0.267 822 265 625 000 497 379 915 032 070 128 272;
  • 5) 0.267 822 265 625 000 497 379 915 032 070 128 272 × 2 = 0 + 0.535 644 531 250 000 994 759 830 064 140 256 544;
  • 6) 0.535 644 531 250 000 994 759 830 064 140 256 544 × 2 = 1 + 0.071 289 062 500 001 989 519 660 128 280 513 088;
  • 7) 0.071 289 062 500 001 989 519 660 128 280 513 088 × 2 = 0 + 0.142 578 125 000 003 979 039 320 256 561 026 176;
  • 8) 0.142 578 125 000 003 979 039 320 256 561 026 176 × 2 = 0 + 0.285 156 250 000 007 958 078 640 513 122 052 352;
  • 9) 0.285 156 250 000 007 958 078 640 513 122 052 352 × 2 = 0 + 0.570 312 500 000 015 916 157 281 026 244 104 704;
  • 10) 0.570 312 500 000 015 916 157 281 026 244 104 704 × 2 = 1 + 0.140 625 000 000 031 832 314 562 052 488 209 408;
  • 11) 0.140 625 000 000 031 832 314 562 052 488 209 408 × 2 = 0 + 0.281 250 000 000 063 664 629 124 104 976 418 816;
  • 12) 0.281 250 000 000 063 664 629 124 104 976 418 816 × 2 = 0 + 0.562 500 000 000 127 329 258 248 209 952 837 632;
  • 13) 0.562 500 000 000 127 329 258 248 209 952 837 632 × 2 = 1 + 0.125 000 000 000 254 658 516 496 419 905 675 264;
  • 14) 0.125 000 000 000 254 658 516 496 419 905 675 264 × 2 = 0 + 0.250 000 000 000 509 317 032 992 839 811 350 528;
  • 15) 0.250 000 000 000 509 317 032 992 839 811 350 528 × 2 = 0 + 0.500 000 000 001 018 634 065 985 679 622 701 056;
  • 16) 0.500 000 000 001 018 634 065 985 679 622 701 056 × 2 = 1 + 0.000 000 000 002 037 268 131 971 359 245 402 112;
  • 17) 0.000 000 000 002 037 268 131 971 359 245 402 112 × 2 = 0 + 0.000 000 000 004 074 536 263 942 718 490 804 224;
  • 18) 0.000 000 000 004 074 536 263 942 718 490 804 224 × 2 = 0 + 0.000 000 000 008 149 072 527 885 436 981 608 448;
  • 19) 0.000 000 000 008 149 072 527 885 436 981 608 448 × 2 = 0 + 0.000 000 000 016 298 145 055 770 873 963 216 896;
  • 20) 0.000 000 000 016 298 145 055 770 873 963 216 896 × 2 = 0 + 0.000 000 000 032 596 290 111 541 747 926 433 792;
  • 21) 0.000 000 000 032 596 290 111 541 747 926 433 792 × 2 = 0 + 0.000 000 000 065 192 580 223 083 495 852 867 584;
  • 22) 0.000 000 000 065 192 580 223 083 495 852 867 584 × 2 = 0 + 0.000 000 000 130 385 160 446 166 991 705 735 168;
  • 23) 0.000 000 000 130 385 160 446 166 991 705 735 168 × 2 = 0 + 0.000 000 000 260 770 320 892 333 983 411 470 336;
  • 24) 0.000 000 000 260 770 320 892 333 983 411 470 336 × 2 = 0 + 0.000 000 000 521 540 641 784 667 966 822 940 672;
  • 25) 0.000 000 000 521 540 641 784 667 966 822 940 672 × 2 = 0 + 0.000 000 001 043 081 283 569 335 933 645 881 344;
  • 26) 0.000 000 001 043 081 283 569 335 933 645 881 344 × 2 = 0 + 0.000 000 002 086 162 567 138 671 867 291 762 688;
  • 27) 0.000 000 002 086 162 567 138 671 867 291 762 688 × 2 = 0 + 0.000 000 004 172 325 134 277 343 734 583 525 376;
  • 28) 0.000 000 004 172 325 134 277 343 734 583 525 376 × 2 = 0 + 0.000 000 008 344 650 268 554 687 469 167 050 752;
  • 29) 0.000 000 008 344 650 268 554 687 469 167 050 752 × 2 = 0 + 0.000 000 016 689 300 537 109 374 938 334 101 504;
  • 30) 0.000 000 016 689 300 537 109 374 938 334 101 504 × 2 = 0 + 0.000 000 033 378 601 074 218 749 876 668 203 008;
  • 31) 0.000 000 033 378 601 074 218 749 876 668 203 008 × 2 = 0 + 0.000 000 066 757 202 148 437 499 753 336 406 016;
  • 32) 0.000 000 066 757 202 148 437 499 753 336 406 016 × 2 = 0 + 0.000 000 133 514 404 296 874 999 506 672 812 032;
  • 33) 0.000 000 133 514 404 296 874 999 506 672 812 032 × 2 = 0 + 0.000 000 267 028 808 593 749 999 013 345 624 064;
  • 34) 0.000 000 267 028 808 593 749 999 013 345 624 064 × 2 = 0 + 0.000 000 534 057 617 187 499 998 026 691 248 128;
  • 35) 0.000 000 534 057 617 187 499 998 026 691 248 128 × 2 = 0 + 0.000 001 068 115 234 374 999 996 053 382 496 256;
  • 36) 0.000 001 068 115 234 374 999 996 053 382 496 256 × 2 = 0 + 0.000 002 136 230 468 749 999 992 106 764 992 512;
  • 37) 0.000 002 136 230 468 749 999 992 106 764 992 512 × 2 = 0 + 0.000 004 272 460 937 499 999 984 213 529 985 024;
  • 38) 0.000 004 272 460 937 499 999 984 213 529 985 024 × 2 = 0 + 0.000 008 544 921 874 999 999 968 427 059 970 048;
  • 39) 0.000 008 544 921 874 999 999 968 427 059 970 048 × 2 = 0 + 0.000 017 089 843 749 999 999 936 854 119 940 096;
  • 40) 0.000 017 089 843 749 999 999 936 854 119 940 096 × 2 = 0 + 0.000 034 179 687 499 999 999 873 708 239 880 192;
  • 41) 0.000 034 179 687 499 999 999 873 708 239 880 192 × 2 = 0 + 0.000 068 359 374 999 999 999 747 416 479 760 384;
  • 42) 0.000 068 359 374 999 999 999 747 416 479 760 384 × 2 = 0 + 0.000 136 718 749 999 999 999 494 832 959 520 768;
  • 43) 0.000 136 718 749 999 999 999 494 832 959 520 768 × 2 = 0 + 0.000 273 437 499 999 999 998 989 665 919 041 536;
  • 44) 0.000 273 437 499 999 999 998 989 665 919 041 536 × 2 = 0 + 0.000 546 874 999 999 999 997 979 331 838 083 072;
  • 45) 0.000 546 874 999 999 999 997 979 331 838 083 072 × 2 = 0 + 0.001 093 749 999 999 999 995 958 663 676 166 144;
  • 46) 0.001 093 749 999 999 999 995 958 663 676 166 144 × 2 = 0 + 0.002 187 499 999 999 999 991 917 327 352 332 288;
  • 47) 0.002 187 499 999 999 999 991 917 327 352 332 288 × 2 = 0 + 0.004 374 999 999 999 999 983 834 654 704 664 576;
  • 48) 0.004 374 999 999 999 999 983 834 654 704 664 576 × 2 = 0 + 0.008 749 999 999 999 999 967 669 309 409 329 152;
  • 49) 0.008 749 999 999 999 999 967 669 309 409 329 152 × 2 = 0 + 0.017 499 999 999 999 999 935 338 618 818 658 304;
  • 50) 0.017 499 999 999 999 999 935 338 618 818 658 304 × 2 = 0 + 0.034 999 999 999 999 999 870 677 237 637 316 608;
  • 51) 0.034 999 999 999 999 999 870 677 237 637 316 608 × 2 = 0 + 0.069 999 999 999 999 999 741 354 475 274 633 216;
  • 52) 0.069 999 999 999 999 999 741 354 475 274 633 216 × 2 = 0 + 0.139 999 999 999 999 999 482 708 950 549 266 432;
  • 53) 0.139 999 999 999 999 999 482 708 950 549 266 432 × 2 = 0 + 0.279 999 999 999 999 998 965 417 901 098 532 864;
  • 54) 0.279 999 999 999 999 998 965 417 901 098 532 864 × 2 = 0 + 0.559 999 999 999 999 997 930 835 802 197 065 728;
  • 55) 0.559 999 999 999 999 997 930 835 802 197 065 728 × 2 = 1 + 0.119 999 999 999 999 995 861 671 604 394 131 456;
  • 56) 0.119 999 999 999 999 995 861 671 604 394 131 456 × 2 = 0 + 0.239 999 999 999 999 991 723 343 208 788 262 912;
  • 57) 0.239 999 999 999 999 991 723 343 208 788 262 912 × 2 = 0 + 0.479 999 999 999 999 983 446 686 417 576 525 824;
  • 58) 0.479 999 999 999 999 983 446 686 417 576 525 824 × 2 = 0 + 0.959 999 999 999 999 966 893 372 835 153 051 648;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.016 738 891 601 562 531 086 244 689 504 383 017(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

6. Positive number before normalization:

0.016 738 891 601 562 531 086 244 689 504 383 017(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 531 086 244 689 504 383 017(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1017(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


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) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


Decimal number -0.016 738 891 601 562 531 086 244 689 504 383 017 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000

» Convert decimal -0.016 738 891 601 562 531 086 244 689 504 382 926 to 64 bit double precision IEEE 754 binary floating point representation standard

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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