-0.016 738 891 601 562 496 530 553 066 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 066 3(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
-0.016 738 891 601 562 496 530 553 066 3(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:

|-0.016 738 891 601 562 496 530 553 066 3| = 0.016 738 891 601 562 496 530 553 066 3


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: 0.016 738 891 601 562 496 530 553 066 3.

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) 0.016 738 891 601 562 496 530 553 066 3 × 2 = 0 + 0.033 477 783 203 124 993 061 106 132 6;
  • 2) 0.033 477 783 203 124 993 061 106 132 6 × 2 = 0 + 0.066 955 566 406 249 986 122 212 265 2;
  • 3) 0.066 955 566 406 249 986 122 212 265 2 × 2 = 0 + 0.133 911 132 812 499 972 244 424 530 4;
  • 4) 0.133 911 132 812 499 972 244 424 530 4 × 2 = 0 + 0.267 822 265 624 999 944 488 849 060 8;
  • 5) 0.267 822 265 624 999 944 488 849 060 8 × 2 = 0 + 0.535 644 531 249 999 888 977 698 121 6;
  • 6) 0.535 644 531 249 999 888 977 698 121 6 × 2 = 1 + 0.071 289 062 499 999 777 955 396 243 2;
  • 7) 0.071 289 062 499 999 777 955 396 243 2 × 2 = 0 + 0.142 578 124 999 999 555 910 792 486 4;
  • 8) 0.142 578 124 999 999 555 910 792 486 4 × 2 = 0 + 0.285 156 249 999 999 111 821 584 972 8;
  • 9) 0.285 156 249 999 999 111 821 584 972 8 × 2 = 0 + 0.570 312 499 999 998 223 643 169 945 6;
  • 10) 0.570 312 499 999 998 223 643 169 945 6 × 2 = 1 + 0.140 624 999 999 996 447 286 339 891 2;
  • 11) 0.140 624 999 999 996 447 286 339 891 2 × 2 = 0 + 0.281 249 999 999 992 894 572 679 782 4;
  • 12) 0.281 249 999 999 992 894 572 679 782 4 × 2 = 0 + 0.562 499 999 999 985 789 145 359 564 8;
  • 13) 0.562 499 999 999 985 789 145 359 564 8 × 2 = 1 + 0.124 999 999 999 971 578 290 719 129 6;
  • 14) 0.124 999 999 999 971 578 290 719 129 6 × 2 = 0 + 0.249 999 999 999 943 156 581 438 259 2;
  • 15) 0.249 999 999 999 943 156 581 438 259 2 × 2 = 0 + 0.499 999 999 999 886 313 162 876 518 4;
  • 16) 0.499 999 999 999 886 313 162 876 518 4 × 2 = 0 + 0.999 999 999 999 772 626 325 753 036 8;
  • 17) 0.999 999 999 999 772 626 325 753 036 8 × 2 = 1 + 0.999 999 999 999 545 252 651 506 073 6;
  • 18) 0.999 999 999 999 545 252 651 506 073 6 × 2 = 1 + 0.999 999 999 999 090 505 303 012 147 2;
  • 19) 0.999 999 999 999 090 505 303 012 147 2 × 2 = 1 + 0.999 999 999 998 181 010 606 024 294 4;
  • 20) 0.999 999 999 998 181 010 606 024 294 4 × 2 = 1 + 0.999 999 999 996 362 021 212 048 588 8;
  • 21) 0.999 999 999 996 362 021 212 048 588 8 × 2 = 1 + 0.999 999 999 992 724 042 424 097 177 6;
  • 22) 0.999 999 999 992 724 042 424 097 177 6 × 2 = 1 + 0.999 999 999 985 448 084 848 194 355 2;
  • 23) 0.999 999 999 985 448 084 848 194 355 2 × 2 = 1 + 0.999 999 999 970 896 169 696 388 710 4;
  • 24) 0.999 999 999 970 896 169 696 388 710 4 × 2 = 1 + 0.999 999 999 941 792 339 392 777 420 8;
  • 25) 0.999 999 999 941 792 339 392 777 420 8 × 2 = 1 + 0.999 999 999 883 584 678 785 554 841 6;
  • 26) 0.999 999 999 883 584 678 785 554 841 6 × 2 = 1 + 0.999 999 999 767 169 357 571 109 683 2;
  • 27) 0.999 999 999 767 169 357 571 109 683 2 × 2 = 1 + 0.999 999 999 534 338 715 142 219 366 4;
  • 28) 0.999 999 999 534 338 715 142 219 366 4 × 2 = 1 + 0.999 999 999 068 677 430 284 438 732 8;
  • 29) 0.999 999 999 068 677 430 284 438 732 8 × 2 = 1 + 0.999 999 998 137 354 860 568 877 465 6;
  • 30) 0.999 999 998 137 354 860 568 877 465 6 × 2 = 1 + 0.999 999 996 274 709 721 137 754 931 2;
  • 31) 0.999 999 996 274 709 721 137 754 931 2 × 2 = 1 + 0.999 999 992 549 419 442 275 509 862 4;
  • 32) 0.999 999 992 549 419 442 275 509 862 4 × 2 = 1 + 0.999 999 985 098 838 884 551 019 724 8;
  • 33) 0.999 999 985 098 838 884 551 019 724 8 × 2 = 1 + 0.999 999 970 197 677 769 102 039 449 6;
  • 34) 0.999 999 970 197 677 769 102 039 449 6 × 2 = 1 + 0.999 999 940 395 355 538 204 078 899 2;
  • 35) 0.999 999 940 395 355 538 204 078 899 2 × 2 = 1 + 0.999 999 880 790 711 076 408 157 798 4;
  • 36) 0.999 999 880 790 711 076 408 157 798 4 × 2 = 1 + 0.999 999 761 581 422 152 816 315 596 8;
  • 37) 0.999 999 761 581 422 152 816 315 596 8 × 2 = 1 + 0.999 999 523 162 844 305 632 631 193 6;
  • 38) 0.999 999 523 162 844 305 632 631 193 6 × 2 = 1 + 0.999 999 046 325 688 611 265 262 387 2;
  • 39) 0.999 999 046 325 688 611 265 262 387 2 × 2 = 1 + 0.999 998 092 651 377 222 530 524 774 4;
  • 40) 0.999 998 092 651 377 222 530 524 774 4 × 2 = 1 + 0.999 996 185 302 754 445 061 049 548 8;
  • 41) 0.999 996 185 302 754 445 061 049 548 8 × 2 = 1 + 0.999 992 370 605 508 890 122 099 097 6;
  • 42) 0.999 992 370 605 508 890 122 099 097 6 × 2 = 1 + 0.999 984 741 211 017 780 244 198 195 2;
  • 43) 0.999 984 741 211 017 780 244 198 195 2 × 2 = 1 + 0.999 969 482 422 035 560 488 396 390 4;
  • 44) 0.999 969 482 422 035 560 488 396 390 4 × 2 = 1 + 0.999 938 964 844 071 120 976 792 780 8;
  • 45) 0.999 938 964 844 071 120 976 792 780 8 × 2 = 1 + 0.999 877 929 688 142 241 953 585 561 6;
  • 46) 0.999 877 929 688 142 241 953 585 561 6 × 2 = 1 + 0.999 755 859 376 284 483 907 171 123 2;
  • 47) 0.999 755 859 376 284 483 907 171 123 2 × 2 = 1 + 0.999 511 718 752 568 967 814 342 246 4;
  • 48) 0.999 511 718 752 568 967 814 342 246 4 × 2 = 1 + 0.999 023 437 505 137 935 628 684 492 8;
  • 49) 0.999 023 437 505 137 935 628 684 492 8 × 2 = 1 + 0.998 046 875 010 275 871 257 368 985 6;
  • 50) 0.998 046 875 010 275 871 257 368 985 6 × 2 = 1 + 0.996 093 750 020 551 742 514 737 971 2;
  • 51) 0.996 093 750 020 551 742 514 737 971 2 × 2 = 1 + 0.992 187 500 041 103 485 029 475 942 4;
  • 52) 0.992 187 500 041 103 485 029 475 942 4 × 2 = 1 + 0.984 375 000 082 206 970 058 951 884 8;
  • 53) 0.984 375 000 082 206 970 058 951 884 8 × 2 = 1 + 0.968 750 000 164 413 940 117 903 769 6;
  • 54) 0.968 750 000 164 413 940 117 903 769 6 × 2 = 1 + 0.937 500 000 328 827 880 235 807 539 2;
  • 55) 0.937 500 000 328 827 880 235 807 539 2 × 2 = 1 + 0.875 000 000 657 655 760 471 615 078 4;
  • 56) 0.875 000 000 657 655 760 471 615 078 4 × 2 = 1 + 0.750 000 001 315 311 520 943 230 156 8;
  • 57) 0.750 000 001 315 311 520 943 230 156 8 × 2 = 1 + 0.500 000 002 630 623 041 886 460 313 6;
  • 58) 0.500 000 002 630 623 041 886 460 313 6 × 2 = 1 + 0.000 000 005 261 246 083 772 920 627 2;

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:


0.016 738 891 601 562 496 530 553 066 3(10) =


0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 066 3(10) =


0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 530 553 066 3(10) =


0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =


0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =


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


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 1111 =


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


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 1111


Decimal number -0.016 738 891 601 562 496 530 553 066 3 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 1111


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