-0.016 738 891 601 562 496 516 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 516 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
-0.016 738 891 601 562 496 516 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:

|-0.016 738 891 601 562 496 516 3| = 0.016 738 891 601 562 496 516 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: 0.016 738 891 601 562 496 516 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) 0.016 738 891 601 562 496 516 3 × 2 = 0 + 0.033 477 783 203 124 993 032 6;
  • 2) 0.033 477 783 203 124 993 032 6 × 2 = 0 + 0.066 955 566 406 249 986 065 2;
  • 3) 0.066 955 566 406 249 986 065 2 × 2 = 0 + 0.133 911 132 812 499 972 130 4;
  • 4) 0.133 911 132 812 499 972 130 4 × 2 = 0 + 0.267 822 265 624 999 944 260 8;
  • 5) 0.267 822 265 624 999 944 260 8 × 2 = 0 + 0.535 644 531 249 999 888 521 6;
  • 6) 0.535 644 531 249 999 888 521 6 × 2 = 1 + 0.071 289 062 499 999 777 043 2;
  • 7) 0.071 289 062 499 999 777 043 2 × 2 = 0 + 0.142 578 124 999 999 554 086 4;
  • 8) 0.142 578 124 999 999 554 086 4 × 2 = 0 + 0.285 156 249 999 999 108 172 8;
  • 9) 0.285 156 249 999 999 108 172 8 × 2 = 0 + 0.570 312 499 999 998 216 345 6;
  • 10) 0.570 312 499 999 998 216 345 6 × 2 = 1 + 0.140 624 999 999 996 432 691 2;
  • 11) 0.140 624 999 999 996 432 691 2 × 2 = 0 + 0.281 249 999 999 992 865 382 4;
  • 12) 0.281 249 999 999 992 865 382 4 × 2 = 0 + 0.562 499 999 999 985 730 764 8;
  • 13) 0.562 499 999 999 985 730 764 8 × 2 = 1 + 0.124 999 999 999 971 461 529 6;
  • 14) 0.124 999 999 999 971 461 529 6 × 2 = 0 + 0.249 999 999 999 942 923 059 2;
  • 15) 0.249 999 999 999 942 923 059 2 × 2 = 0 + 0.499 999 999 999 885 846 118 4;
  • 16) 0.499 999 999 999 885 846 118 4 × 2 = 0 + 0.999 999 999 999 771 692 236 8;
  • 17) 0.999 999 999 999 771 692 236 8 × 2 = 1 + 0.999 999 999 999 543 384 473 6;
  • 18) 0.999 999 999 999 543 384 473 6 × 2 = 1 + 0.999 999 999 999 086 768 947 2;
  • 19) 0.999 999 999 999 086 768 947 2 × 2 = 1 + 0.999 999 999 998 173 537 894 4;
  • 20) 0.999 999 999 998 173 537 894 4 × 2 = 1 + 0.999 999 999 996 347 075 788 8;
  • 21) 0.999 999 999 996 347 075 788 8 × 2 = 1 + 0.999 999 999 992 694 151 577 6;
  • 22) 0.999 999 999 992 694 151 577 6 × 2 = 1 + 0.999 999 999 985 388 303 155 2;
  • 23) 0.999 999 999 985 388 303 155 2 × 2 = 1 + 0.999 999 999 970 776 606 310 4;
  • 24) 0.999 999 999 970 776 606 310 4 × 2 = 1 + 0.999 999 999 941 553 212 620 8;
  • 25) 0.999 999 999 941 553 212 620 8 × 2 = 1 + 0.999 999 999 883 106 425 241 6;
  • 26) 0.999 999 999 883 106 425 241 6 × 2 = 1 + 0.999 999 999 766 212 850 483 2;
  • 27) 0.999 999 999 766 212 850 483 2 × 2 = 1 + 0.999 999 999 532 425 700 966 4;
  • 28) 0.999 999 999 532 425 700 966 4 × 2 = 1 + 0.999 999 999 064 851 401 932 8;
  • 29) 0.999 999 999 064 851 401 932 8 × 2 = 1 + 0.999 999 998 129 702 803 865 6;
  • 30) 0.999 999 998 129 702 803 865 6 × 2 = 1 + 0.999 999 996 259 405 607 731 2;
  • 31) 0.999 999 996 259 405 607 731 2 × 2 = 1 + 0.999 999 992 518 811 215 462 4;
  • 32) 0.999 999 992 518 811 215 462 4 × 2 = 1 + 0.999 999 985 037 622 430 924 8;
  • 33) 0.999 999 985 037 622 430 924 8 × 2 = 1 + 0.999 999 970 075 244 861 849 6;
  • 34) 0.999 999 970 075 244 861 849 6 × 2 = 1 + 0.999 999 940 150 489 723 699 2;
  • 35) 0.999 999 940 150 489 723 699 2 × 2 = 1 + 0.999 999 880 300 979 447 398 4;
  • 36) 0.999 999 880 300 979 447 398 4 × 2 = 1 + 0.999 999 760 601 958 894 796 8;
  • 37) 0.999 999 760 601 958 894 796 8 × 2 = 1 + 0.999 999 521 203 917 789 593 6;
  • 38) 0.999 999 521 203 917 789 593 6 × 2 = 1 + 0.999 999 042 407 835 579 187 2;
  • 39) 0.999 999 042 407 835 579 187 2 × 2 = 1 + 0.999 998 084 815 671 158 374 4;
  • 40) 0.999 998 084 815 671 158 374 4 × 2 = 1 + 0.999 996 169 631 342 316 748 8;
  • 41) 0.999 996 169 631 342 316 748 8 × 2 = 1 + 0.999 992 339 262 684 633 497 6;
  • 42) 0.999 992 339 262 684 633 497 6 × 2 = 1 + 0.999 984 678 525 369 266 995 2;
  • 43) 0.999 984 678 525 369 266 995 2 × 2 = 1 + 0.999 969 357 050 738 533 990 4;
  • 44) 0.999 969 357 050 738 533 990 4 × 2 = 1 + 0.999 938 714 101 477 067 980 8;
  • 45) 0.999 938 714 101 477 067 980 8 × 2 = 1 + 0.999 877 428 202 954 135 961 6;
  • 46) 0.999 877 428 202 954 135 961 6 × 2 = 1 + 0.999 754 856 405 908 271 923 2;
  • 47) 0.999 754 856 405 908 271 923 2 × 2 = 1 + 0.999 509 712 811 816 543 846 4;
  • 48) 0.999 509 712 811 816 543 846 4 × 2 = 1 + 0.999 019 425 623 633 087 692 8;
  • 49) 0.999 019 425 623 633 087 692 8 × 2 = 1 + 0.998 038 851 247 266 175 385 6;
  • 50) 0.998 038 851 247 266 175 385 6 × 2 = 1 + 0.996 077 702 494 532 350 771 2;
  • 51) 0.996 077 702 494 532 350 771 2 × 2 = 1 + 0.992 155 404 989 064 701 542 4;
  • 52) 0.992 155 404 989 064 701 542 4 × 2 = 1 + 0.984 310 809 978 129 403 084 8;
  • 53) 0.984 310 809 978 129 403 084 8 × 2 = 1 + 0.968 621 619 956 258 806 169 6;
  • 54) 0.968 621 619 956 258 806 169 6 × 2 = 1 + 0.937 243 239 912 517 612 339 2;
  • 55) 0.937 243 239 912 517 612 339 2 × 2 = 1 + 0.874 486 479 825 035 224 678 4;
  • 56) 0.874 486 479 825 035 224 678 4 × 2 = 1 + 0.748 972 959 650 070 449 356 8;
  • 57) 0.748 972 959 650 070 449 356 8 × 2 = 1 + 0.497 945 919 300 140 898 713 6;
  • 58) 0.497 945 919 300 140 898 713 6 × 2 = 0 + 0.995 891 838 600 281 797 427 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:


0.016 738 891 601 562 496 516 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 516 3(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:


0.016 738 891 601 562 496 516 3(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 -0.016 738 891 601 562 496 516 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 1110

Share

» Convert decimal -0.016 738 891 601 562 496 519 9 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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