-0.016 738 891 601 562 496 547 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 547 2(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
-0.016 738 891 601 562 496 547 2(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:

|-0.016 738 891 601 562 496 547 2| = 0.016 738 891 601 562 496 547 2


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: 0.016 738 891 601 562 496 547 2.

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) 0.016 738 891 601 562 496 547 2 × 2 = 0 + 0.033 477 783 203 124 993 094 4;
  • 2) 0.033 477 783 203 124 993 094 4 × 2 = 0 + 0.066 955 566 406 249 986 188 8;
  • 3) 0.066 955 566 406 249 986 188 8 × 2 = 0 + 0.133 911 132 812 499 972 377 6;
  • 4) 0.133 911 132 812 499 972 377 6 × 2 = 0 + 0.267 822 265 624 999 944 755 2;
  • 5) 0.267 822 265 624 999 944 755 2 × 2 = 0 + 0.535 644 531 249 999 889 510 4;
  • 6) 0.535 644 531 249 999 889 510 4 × 2 = 1 + 0.071 289 062 499 999 779 020 8;
  • 7) 0.071 289 062 499 999 779 020 8 × 2 = 0 + 0.142 578 124 999 999 558 041 6;
  • 8) 0.142 578 124 999 999 558 041 6 × 2 = 0 + 0.285 156 249 999 999 116 083 2;
  • 9) 0.285 156 249 999 999 116 083 2 × 2 = 0 + 0.570 312 499 999 998 232 166 4;
  • 10) 0.570 312 499 999 998 232 166 4 × 2 = 1 + 0.140 624 999 999 996 464 332 8;
  • 11) 0.140 624 999 999 996 464 332 8 × 2 = 0 + 0.281 249 999 999 992 928 665 6;
  • 12) 0.281 249 999 999 992 928 665 6 × 2 = 0 + 0.562 499 999 999 985 857 331 2;
  • 13) 0.562 499 999 999 985 857 331 2 × 2 = 1 + 0.124 999 999 999 971 714 662 4;
  • 14) 0.124 999 999 999 971 714 662 4 × 2 = 0 + 0.249 999 999 999 943 429 324 8;
  • 15) 0.249 999 999 999 943 429 324 8 × 2 = 0 + 0.499 999 999 999 886 858 649 6;
  • 16) 0.499 999 999 999 886 858 649 6 × 2 = 0 + 0.999 999 999 999 773 717 299 2;
  • 17) 0.999 999 999 999 773 717 299 2 × 2 = 1 + 0.999 999 999 999 547 434 598 4;
  • 18) 0.999 999 999 999 547 434 598 4 × 2 = 1 + 0.999 999 999 999 094 869 196 8;
  • 19) 0.999 999 999 999 094 869 196 8 × 2 = 1 + 0.999 999 999 998 189 738 393 6;
  • 20) 0.999 999 999 998 189 738 393 6 × 2 = 1 + 0.999 999 999 996 379 476 787 2;
  • 21) 0.999 999 999 996 379 476 787 2 × 2 = 1 + 0.999 999 999 992 758 953 574 4;
  • 22) 0.999 999 999 992 758 953 574 4 × 2 = 1 + 0.999 999 999 985 517 907 148 8;
  • 23) 0.999 999 999 985 517 907 148 8 × 2 = 1 + 0.999 999 999 971 035 814 297 6;
  • 24) 0.999 999 999 971 035 814 297 6 × 2 = 1 + 0.999 999 999 942 071 628 595 2;
  • 25) 0.999 999 999 942 071 628 595 2 × 2 = 1 + 0.999 999 999 884 143 257 190 4;
  • 26) 0.999 999 999 884 143 257 190 4 × 2 = 1 + 0.999 999 999 768 286 514 380 8;
  • 27) 0.999 999 999 768 286 514 380 8 × 2 = 1 + 0.999 999 999 536 573 028 761 6;
  • 28) 0.999 999 999 536 573 028 761 6 × 2 = 1 + 0.999 999 999 073 146 057 523 2;
  • 29) 0.999 999 999 073 146 057 523 2 × 2 = 1 + 0.999 999 998 146 292 115 046 4;
  • 30) 0.999 999 998 146 292 115 046 4 × 2 = 1 + 0.999 999 996 292 584 230 092 8;
  • 31) 0.999 999 996 292 584 230 092 8 × 2 = 1 + 0.999 999 992 585 168 460 185 6;
  • 32) 0.999 999 992 585 168 460 185 6 × 2 = 1 + 0.999 999 985 170 336 920 371 2;
  • 33) 0.999 999 985 170 336 920 371 2 × 2 = 1 + 0.999 999 970 340 673 840 742 4;
  • 34) 0.999 999 970 340 673 840 742 4 × 2 = 1 + 0.999 999 940 681 347 681 484 8;
  • 35) 0.999 999 940 681 347 681 484 8 × 2 = 1 + 0.999 999 881 362 695 362 969 6;
  • 36) 0.999 999 881 362 695 362 969 6 × 2 = 1 + 0.999 999 762 725 390 725 939 2;
  • 37) 0.999 999 762 725 390 725 939 2 × 2 = 1 + 0.999 999 525 450 781 451 878 4;
  • 38) 0.999 999 525 450 781 451 878 4 × 2 = 1 + 0.999 999 050 901 562 903 756 8;
  • 39) 0.999 999 050 901 562 903 756 8 × 2 = 1 + 0.999 998 101 803 125 807 513 6;
  • 40) 0.999 998 101 803 125 807 513 6 × 2 = 1 + 0.999 996 203 606 251 615 027 2;
  • 41) 0.999 996 203 606 251 615 027 2 × 2 = 1 + 0.999 992 407 212 503 230 054 4;
  • 42) 0.999 992 407 212 503 230 054 4 × 2 = 1 + 0.999 984 814 425 006 460 108 8;
  • 43) 0.999 984 814 425 006 460 108 8 × 2 = 1 + 0.999 969 628 850 012 920 217 6;
  • 44) 0.999 969 628 850 012 920 217 6 × 2 = 1 + 0.999 939 257 700 025 840 435 2;
  • 45) 0.999 939 257 700 025 840 435 2 × 2 = 1 + 0.999 878 515 400 051 680 870 4;
  • 46) 0.999 878 515 400 051 680 870 4 × 2 = 1 + 0.999 757 030 800 103 361 740 8;
  • 47) 0.999 757 030 800 103 361 740 8 × 2 = 1 + 0.999 514 061 600 206 723 481 6;
  • 48) 0.999 514 061 600 206 723 481 6 × 2 = 1 + 0.999 028 123 200 413 446 963 2;
  • 49) 0.999 028 123 200 413 446 963 2 × 2 = 1 + 0.998 056 246 400 826 893 926 4;
  • 50) 0.998 056 246 400 826 893 926 4 × 2 = 1 + 0.996 112 492 801 653 787 852 8;
  • 51) 0.996 112 492 801 653 787 852 8 × 2 = 1 + 0.992 224 985 603 307 575 705 6;
  • 52) 0.992 224 985 603 307 575 705 6 × 2 = 1 + 0.984 449 971 206 615 151 411 2;
  • 53) 0.984 449 971 206 615 151 411 2 × 2 = 1 + 0.968 899 942 413 230 302 822 4;
  • 54) 0.968 899 942 413 230 302 822 4 × 2 = 1 + 0.937 799 884 826 460 605 644 8;
  • 55) 0.937 799 884 826 460 605 644 8 × 2 = 1 + 0.875 599 769 652 921 211 289 6;
  • 56) 0.875 599 769 652 921 211 289 6 × 2 = 1 + 0.751 199 539 305 842 422 579 2;
  • 57) 0.751 199 539 305 842 422 579 2 × 2 = 1 + 0.502 399 078 611 684 845 158 4;
  • 58) 0.502 399 078 611 684 845 158 4 × 2 = 1 + 0.004 798 157 223 369 690 316 8;

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:


0.016 738 891 601 562 496 547 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 547 2(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:


0.016 738 891 601 562 496 547 2(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 -0.016 738 891 601 562 496 547 2 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

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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