-0.000 000 017 188 905 742 258 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 017 188 905 742 258 91(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
-0.000 000 017 188 905 742 258 91(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:

|-0.000 000 017 188 905 742 258 91| = 0.000 000 017 188 905 742 258 91


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: 0.000 000 017 188 905 742 258 91.

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) 0.000 000 017 188 905 742 258 91 × 2 = 0 + 0.000 000 034 377 811 484 517 82;
  • 2) 0.000 000 034 377 811 484 517 82 × 2 = 0 + 0.000 000 068 755 622 969 035 64;
  • 3) 0.000 000 068 755 622 969 035 64 × 2 = 0 + 0.000 000 137 511 245 938 071 28;
  • 4) 0.000 000 137 511 245 938 071 28 × 2 = 0 + 0.000 000 275 022 491 876 142 56;
  • 5) 0.000 000 275 022 491 876 142 56 × 2 = 0 + 0.000 000 550 044 983 752 285 12;
  • 6) 0.000 000 550 044 983 752 285 12 × 2 = 0 + 0.000 001 100 089 967 504 570 24;
  • 7) 0.000 001 100 089 967 504 570 24 × 2 = 0 + 0.000 002 200 179 935 009 140 48;
  • 8) 0.000 002 200 179 935 009 140 48 × 2 = 0 + 0.000 004 400 359 870 018 280 96;
  • 9) 0.000 004 400 359 870 018 280 96 × 2 = 0 + 0.000 008 800 719 740 036 561 92;
  • 10) 0.000 008 800 719 740 036 561 92 × 2 = 0 + 0.000 017 601 439 480 073 123 84;
  • 11) 0.000 017 601 439 480 073 123 84 × 2 = 0 + 0.000 035 202 878 960 146 247 68;
  • 12) 0.000 035 202 878 960 146 247 68 × 2 = 0 + 0.000 070 405 757 920 292 495 36;
  • 13) 0.000 070 405 757 920 292 495 36 × 2 = 0 + 0.000 140 811 515 840 584 990 72;
  • 14) 0.000 140 811 515 840 584 990 72 × 2 = 0 + 0.000 281 623 031 681 169 981 44;
  • 15) 0.000 281 623 031 681 169 981 44 × 2 = 0 + 0.000 563 246 063 362 339 962 88;
  • 16) 0.000 563 246 063 362 339 962 88 × 2 = 0 + 0.001 126 492 126 724 679 925 76;
  • 17) 0.001 126 492 126 724 679 925 76 × 2 = 0 + 0.002 252 984 253 449 359 851 52;
  • 18) 0.002 252 984 253 449 359 851 52 × 2 = 0 + 0.004 505 968 506 898 719 703 04;
  • 19) 0.004 505 968 506 898 719 703 04 × 2 = 0 + 0.009 011 937 013 797 439 406 08;
  • 20) 0.009 011 937 013 797 439 406 08 × 2 = 0 + 0.018 023 874 027 594 878 812 16;
  • 21) 0.018 023 874 027 594 878 812 16 × 2 = 0 + 0.036 047 748 055 189 757 624 32;
  • 22) 0.036 047 748 055 189 757 624 32 × 2 = 0 + 0.072 095 496 110 379 515 248 64;
  • 23) 0.072 095 496 110 379 515 248 64 × 2 = 0 + 0.144 190 992 220 759 030 497 28;
  • 24) 0.144 190 992 220 759 030 497 28 × 2 = 0 + 0.288 381 984 441 518 060 994 56;
  • 25) 0.288 381 984 441 518 060 994 56 × 2 = 0 + 0.576 763 968 883 036 121 989 12;
  • 26) 0.576 763 968 883 036 121 989 12 × 2 = 1 + 0.153 527 937 766 072 243 978 24;
  • 27) 0.153 527 937 766 072 243 978 24 × 2 = 0 + 0.307 055 875 532 144 487 956 48;
  • 28) 0.307 055 875 532 144 487 956 48 × 2 = 0 + 0.614 111 751 064 288 975 912 96;
  • 29) 0.614 111 751 064 288 975 912 96 × 2 = 1 + 0.228 223 502 128 577 951 825 92;
  • 30) 0.228 223 502 128 577 951 825 92 × 2 = 0 + 0.456 447 004 257 155 903 651 84;
  • 31) 0.456 447 004 257 155 903 651 84 × 2 = 0 + 0.912 894 008 514 311 807 303 68;
  • 32) 0.912 894 008 514 311 807 303 68 × 2 = 1 + 0.825 788 017 028 623 614 607 36;
  • 33) 0.825 788 017 028 623 614 607 36 × 2 = 1 + 0.651 576 034 057 247 229 214 72;
  • 34) 0.651 576 034 057 247 229 214 72 × 2 = 1 + 0.303 152 068 114 494 458 429 44;
  • 35) 0.303 152 068 114 494 458 429 44 × 2 = 0 + 0.606 304 136 228 988 916 858 88;
  • 36) 0.606 304 136 228 988 916 858 88 × 2 = 1 + 0.212 608 272 457 977 833 717 76;
  • 37) 0.212 608 272 457 977 833 717 76 × 2 = 0 + 0.425 216 544 915 955 667 435 52;
  • 38) 0.425 216 544 915 955 667 435 52 × 2 = 0 + 0.850 433 089 831 911 334 871 04;
  • 39) 0.850 433 089 831 911 334 871 04 × 2 = 1 + 0.700 866 179 663 822 669 742 08;
  • 40) 0.700 866 179 663 822 669 742 08 × 2 = 1 + 0.401 732 359 327 645 339 484 16;
  • 41) 0.401 732 359 327 645 339 484 16 × 2 = 0 + 0.803 464 718 655 290 678 968 32;
  • 42) 0.803 464 718 655 290 678 968 32 × 2 = 1 + 0.606 929 437 310 581 357 936 64;
  • 43) 0.606 929 437 310 581 357 936 64 × 2 = 1 + 0.213 858 874 621 162 715 873 28;
  • 44) 0.213 858 874 621 162 715 873 28 × 2 = 0 + 0.427 717 749 242 325 431 746 56;
  • 45) 0.427 717 749 242 325 431 746 56 × 2 = 0 + 0.855 435 498 484 650 863 493 12;
  • 46) 0.855 435 498 484 650 863 493 12 × 2 = 1 + 0.710 870 996 969 301 726 986 24;
  • 47) 0.710 870 996 969 301 726 986 24 × 2 = 1 + 0.421 741 993 938 603 453 972 48;
  • 48) 0.421 741 993 938 603 453 972 48 × 2 = 0 + 0.843 483 987 877 206 907 944 96;
  • 49) 0.843 483 987 877 206 907 944 96 × 2 = 1 + 0.686 967 975 754 413 815 889 92;
  • 50) 0.686 967 975 754 413 815 889 92 × 2 = 1 + 0.373 935 951 508 827 631 779 84;
  • 51) 0.373 935 951 508 827 631 779 84 × 2 = 0 + 0.747 871 903 017 655 263 559 68;
  • 52) 0.747 871 903 017 655 263 559 68 × 2 = 1 + 0.495 743 806 035 310 527 119 36;
  • 53) 0.495 743 806 035 310 527 119 36 × 2 = 0 + 0.991 487 612 070 621 054 238 72;
  • 54) 0.991 487 612 070 621 054 238 72 × 2 = 1 + 0.982 975 224 141 242 108 477 44;
  • 55) 0.982 975 224 141 242 108 477 44 × 2 = 1 + 0.965 950 448 282 484 216 954 88;
  • 56) 0.965 950 448 282 484 216 954 88 × 2 = 1 + 0.931 900 896 564 968 433 909 76;
  • 57) 0.931 900 896 564 968 433 909 76 × 2 = 1 + 0.863 801 793 129 936 867 819 52;
  • 58) 0.863 801 793 129 936 867 819 52 × 2 = 1 + 0.727 603 586 259 873 735 639 04;
  • 59) 0.727 603 586 259 873 735 639 04 × 2 = 1 + 0.455 207 172 519 747 471 278 08;
  • 60) 0.455 207 172 519 747 471 278 08 × 2 = 0 + 0.910 414 345 039 494 942 556 16;
  • 61) 0.910 414 345 039 494 942 556 16 × 2 = 1 + 0.820 828 690 078 989 885 112 32;
  • 62) 0.820 828 690 078 989 885 112 32 × 2 = 1 + 0.641 657 380 157 979 770 224 64;
  • 63) 0.641 657 380 157 979 770 224 64 × 2 = 1 + 0.283 314 760 315 959 540 449 28;
  • 64) 0.283 314 760 315 959 540 449 28 × 2 = 0 + 0.566 629 520 631 919 080 898 56;
  • 65) 0.566 629 520 631 919 080 898 56 × 2 = 1 + 0.133 259 041 263 838 161 797 12;
  • 66) 0.133 259 041 263 838 161 797 12 × 2 = 0 + 0.266 518 082 527 676 323 594 24;
  • 67) 0.266 518 082 527 676 323 594 24 × 2 = 0 + 0.533 036 165 055 352 647 188 48;
  • 68) 0.533 036 165 055 352 647 188 48 × 2 = 1 + 0.066 072 330 110 705 294 376 96;
  • 69) 0.066 072 330 110 705 294 376 96 × 2 = 0 + 0.132 144 660 221 410 588 753 92;
  • 70) 0.132 144 660 221 410 588 753 92 × 2 = 0 + 0.264 289 320 442 821 177 507 84;
  • 71) 0.264 289 320 442 821 177 507 84 × 2 = 0 + 0.528 578 640 885 642 355 015 68;
  • 72) 0.528 578 640 885 642 355 015 68 × 2 = 1 + 0.057 157 281 771 284 710 031 36;
  • 73) 0.057 157 281 771 284 710 031 36 × 2 = 0 + 0.114 314 563 542 569 420 062 72;
  • 74) 0.114 314 563 542 569 420 062 72 × 2 = 0 + 0.228 629 127 085 138 840 125 44;
  • 75) 0.228 629 127 085 138 840 125 44 × 2 = 0 + 0.457 258 254 170 277 680 250 88;
  • 76) 0.457 258 254 170 277 680 250 88 × 2 = 0 + 0.914 516 508 340 555 360 501 76;
  • 77) 0.914 516 508 340 555 360 501 76 × 2 = 1 + 0.829 033 016 681 110 721 003 52;
  • 78) 0.829 033 016 681 110 721 003 52 × 2 = 1 + 0.658 066 033 362 221 442 007 04;

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:


0.000 000 017 188 905 742 258 91(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0001 0000 11(2)

6. Positive number before normalization:

0.000 000 017 188 905 742 258 91(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0001 0000 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 017 188 905 742 258 91(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0001 0000 11(2) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0001 0000 11(2) × 20 =

1.0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011(2) × 2-26


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): -26

Mantissa (not normalized):
1.0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

997(10) =

011 1110 0101(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. 0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011 =

0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011


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 1110 0101

Mantissa (52 bits) =
0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011


Decimal number -0.000 000 017 188 905 742 258 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0101 - 0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 0100 0011

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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