-0.016 738 891 601 562 496 530 249 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 249(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
-0.016 738 891 601 562 496 530 249(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:

|-0.016 738 891 601 562 496 530 249| = 0.016 738 891 601 562 496 530 249


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: 0.016 738 891 601 562 496 530 249.

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) 0.016 738 891 601 562 496 530 249 × 2 = 0 + 0.033 477 783 203 124 993 060 498;
  • 2) 0.033 477 783 203 124 993 060 498 × 2 = 0 + 0.066 955 566 406 249 986 120 996;
  • 3) 0.066 955 566 406 249 986 120 996 × 2 = 0 + 0.133 911 132 812 499 972 241 992;
  • 4) 0.133 911 132 812 499 972 241 992 × 2 = 0 + 0.267 822 265 624 999 944 483 984;
  • 5) 0.267 822 265 624 999 944 483 984 × 2 = 0 + 0.535 644 531 249 999 888 967 968;
  • 6) 0.535 644 531 249 999 888 967 968 × 2 = 1 + 0.071 289 062 499 999 777 935 936;
  • 7) 0.071 289 062 499 999 777 935 936 × 2 = 0 + 0.142 578 124 999 999 555 871 872;
  • 8) 0.142 578 124 999 999 555 871 872 × 2 = 0 + 0.285 156 249 999 999 111 743 744;
  • 9) 0.285 156 249 999 999 111 743 744 × 2 = 0 + 0.570 312 499 999 998 223 487 488;
  • 10) 0.570 312 499 999 998 223 487 488 × 2 = 1 + 0.140 624 999 999 996 446 974 976;
  • 11) 0.140 624 999 999 996 446 974 976 × 2 = 0 + 0.281 249 999 999 992 893 949 952;
  • 12) 0.281 249 999 999 992 893 949 952 × 2 = 0 + 0.562 499 999 999 985 787 899 904;
  • 13) 0.562 499 999 999 985 787 899 904 × 2 = 1 + 0.124 999 999 999 971 575 799 808;
  • 14) 0.124 999 999 999 971 575 799 808 × 2 = 0 + 0.249 999 999 999 943 151 599 616;
  • 15) 0.249 999 999 999 943 151 599 616 × 2 = 0 + 0.499 999 999 999 886 303 199 232;
  • 16) 0.499 999 999 999 886 303 199 232 × 2 = 0 + 0.999 999 999 999 772 606 398 464;
  • 17) 0.999 999 999 999 772 606 398 464 × 2 = 1 + 0.999 999 999 999 545 212 796 928;
  • 18) 0.999 999 999 999 545 212 796 928 × 2 = 1 + 0.999 999 999 999 090 425 593 856;
  • 19) 0.999 999 999 999 090 425 593 856 × 2 = 1 + 0.999 999 999 998 180 851 187 712;
  • 20) 0.999 999 999 998 180 851 187 712 × 2 = 1 + 0.999 999 999 996 361 702 375 424;
  • 21) 0.999 999 999 996 361 702 375 424 × 2 = 1 + 0.999 999 999 992 723 404 750 848;
  • 22) 0.999 999 999 992 723 404 750 848 × 2 = 1 + 0.999 999 999 985 446 809 501 696;
  • 23) 0.999 999 999 985 446 809 501 696 × 2 = 1 + 0.999 999 999 970 893 619 003 392;
  • 24) 0.999 999 999 970 893 619 003 392 × 2 = 1 + 0.999 999 999 941 787 238 006 784;
  • 25) 0.999 999 999 941 787 238 006 784 × 2 = 1 + 0.999 999 999 883 574 476 013 568;
  • 26) 0.999 999 999 883 574 476 013 568 × 2 = 1 + 0.999 999 999 767 148 952 027 136;
  • 27) 0.999 999 999 767 148 952 027 136 × 2 = 1 + 0.999 999 999 534 297 904 054 272;
  • 28) 0.999 999 999 534 297 904 054 272 × 2 = 1 + 0.999 999 999 068 595 808 108 544;
  • 29) 0.999 999 999 068 595 808 108 544 × 2 = 1 + 0.999 999 998 137 191 616 217 088;
  • 30) 0.999 999 998 137 191 616 217 088 × 2 = 1 + 0.999 999 996 274 383 232 434 176;
  • 31) 0.999 999 996 274 383 232 434 176 × 2 = 1 + 0.999 999 992 548 766 464 868 352;
  • 32) 0.999 999 992 548 766 464 868 352 × 2 = 1 + 0.999 999 985 097 532 929 736 704;
  • 33) 0.999 999 985 097 532 929 736 704 × 2 = 1 + 0.999 999 970 195 065 859 473 408;
  • 34) 0.999 999 970 195 065 859 473 408 × 2 = 1 + 0.999 999 940 390 131 718 946 816;
  • 35) 0.999 999 940 390 131 718 946 816 × 2 = 1 + 0.999 999 880 780 263 437 893 632;
  • 36) 0.999 999 880 780 263 437 893 632 × 2 = 1 + 0.999 999 761 560 526 875 787 264;
  • 37) 0.999 999 761 560 526 875 787 264 × 2 = 1 + 0.999 999 523 121 053 751 574 528;
  • 38) 0.999 999 523 121 053 751 574 528 × 2 = 1 + 0.999 999 046 242 107 503 149 056;
  • 39) 0.999 999 046 242 107 503 149 056 × 2 = 1 + 0.999 998 092 484 215 006 298 112;
  • 40) 0.999 998 092 484 215 006 298 112 × 2 = 1 + 0.999 996 184 968 430 012 596 224;
  • 41) 0.999 996 184 968 430 012 596 224 × 2 = 1 + 0.999 992 369 936 860 025 192 448;
  • 42) 0.999 992 369 936 860 025 192 448 × 2 = 1 + 0.999 984 739 873 720 050 384 896;
  • 43) 0.999 984 739 873 720 050 384 896 × 2 = 1 + 0.999 969 479 747 440 100 769 792;
  • 44) 0.999 969 479 747 440 100 769 792 × 2 = 1 + 0.999 938 959 494 880 201 539 584;
  • 45) 0.999 938 959 494 880 201 539 584 × 2 = 1 + 0.999 877 918 989 760 403 079 168;
  • 46) 0.999 877 918 989 760 403 079 168 × 2 = 1 + 0.999 755 837 979 520 806 158 336;
  • 47) 0.999 755 837 979 520 806 158 336 × 2 = 1 + 0.999 511 675 959 041 612 316 672;
  • 48) 0.999 511 675 959 041 612 316 672 × 2 = 1 + 0.999 023 351 918 083 224 633 344;
  • 49) 0.999 023 351 918 083 224 633 344 × 2 = 1 + 0.998 046 703 836 166 449 266 688;
  • 50) 0.998 046 703 836 166 449 266 688 × 2 = 1 + 0.996 093 407 672 332 898 533 376;
  • 51) 0.996 093 407 672 332 898 533 376 × 2 = 1 + 0.992 186 815 344 665 797 066 752;
  • 52) 0.992 186 815 344 665 797 066 752 × 2 = 1 + 0.984 373 630 689 331 594 133 504;
  • 53) 0.984 373 630 689 331 594 133 504 × 2 = 1 + 0.968 747 261 378 663 188 267 008;
  • 54) 0.968 747 261 378 663 188 267 008 × 2 = 1 + 0.937 494 522 757 326 376 534 016;
  • 55) 0.937 494 522 757 326 376 534 016 × 2 = 1 + 0.874 989 045 514 652 753 068 032;
  • 56) 0.874 989 045 514 652 753 068 032 × 2 = 1 + 0.749 978 091 029 305 506 136 064;
  • 57) 0.749 978 091 029 305 506 136 064 × 2 = 1 + 0.499 956 182 058 611 012 272 128;
  • 58) 0.499 956 182 058 611 012 272 128 × 2 = 0 + 0.999 912 364 117 222 024 544 256;

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:


0.016 738 891 601 562 496 530 249(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 249(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 249(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 249 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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