-0.016 738 891 601 562 496 555 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 555 7(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
-0.016 738 891 601 562 496 555 7(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:

|-0.016 738 891 601 562 496 555 7| = 0.016 738 891 601 562 496 555 7


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: 0.016 738 891 601 562 496 555 7.

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) 0.016 738 891 601 562 496 555 7 × 2 = 0 + 0.033 477 783 203 124 993 111 4;
  • 2) 0.033 477 783 203 124 993 111 4 × 2 = 0 + 0.066 955 566 406 249 986 222 8;
  • 3) 0.066 955 566 406 249 986 222 8 × 2 = 0 + 0.133 911 132 812 499 972 445 6;
  • 4) 0.133 911 132 812 499 972 445 6 × 2 = 0 + 0.267 822 265 624 999 944 891 2;
  • 5) 0.267 822 265 624 999 944 891 2 × 2 = 0 + 0.535 644 531 249 999 889 782 4;
  • 6) 0.535 644 531 249 999 889 782 4 × 2 = 1 + 0.071 289 062 499 999 779 564 8;
  • 7) 0.071 289 062 499 999 779 564 8 × 2 = 0 + 0.142 578 124 999 999 559 129 6;
  • 8) 0.142 578 124 999 999 559 129 6 × 2 = 0 + 0.285 156 249 999 999 118 259 2;
  • 9) 0.285 156 249 999 999 118 259 2 × 2 = 0 + 0.570 312 499 999 998 236 518 4;
  • 10) 0.570 312 499 999 998 236 518 4 × 2 = 1 + 0.140 624 999 999 996 473 036 8;
  • 11) 0.140 624 999 999 996 473 036 8 × 2 = 0 + 0.281 249 999 999 992 946 073 6;
  • 12) 0.281 249 999 999 992 946 073 6 × 2 = 0 + 0.562 499 999 999 985 892 147 2;
  • 13) 0.562 499 999 999 985 892 147 2 × 2 = 1 + 0.124 999 999 999 971 784 294 4;
  • 14) 0.124 999 999 999 971 784 294 4 × 2 = 0 + 0.249 999 999 999 943 568 588 8;
  • 15) 0.249 999 999 999 943 568 588 8 × 2 = 0 + 0.499 999 999 999 887 137 177 6;
  • 16) 0.499 999 999 999 887 137 177 6 × 2 = 0 + 0.999 999 999 999 774 274 355 2;
  • 17) 0.999 999 999 999 774 274 355 2 × 2 = 1 + 0.999 999 999 999 548 548 710 4;
  • 18) 0.999 999 999 999 548 548 710 4 × 2 = 1 + 0.999 999 999 999 097 097 420 8;
  • 19) 0.999 999 999 999 097 097 420 8 × 2 = 1 + 0.999 999 999 998 194 194 841 6;
  • 20) 0.999 999 999 998 194 194 841 6 × 2 = 1 + 0.999 999 999 996 388 389 683 2;
  • 21) 0.999 999 999 996 388 389 683 2 × 2 = 1 + 0.999 999 999 992 776 779 366 4;
  • 22) 0.999 999 999 992 776 779 366 4 × 2 = 1 + 0.999 999 999 985 553 558 732 8;
  • 23) 0.999 999 999 985 553 558 732 8 × 2 = 1 + 0.999 999 999 971 107 117 465 6;
  • 24) 0.999 999 999 971 107 117 465 6 × 2 = 1 + 0.999 999 999 942 214 234 931 2;
  • 25) 0.999 999 999 942 214 234 931 2 × 2 = 1 + 0.999 999 999 884 428 469 862 4;
  • 26) 0.999 999 999 884 428 469 862 4 × 2 = 1 + 0.999 999 999 768 856 939 724 8;
  • 27) 0.999 999 999 768 856 939 724 8 × 2 = 1 + 0.999 999 999 537 713 879 449 6;
  • 28) 0.999 999 999 537 713 879 449 6 × 2 = 1 + 0.999 999 999 075 427 758 899 2;
  • 29) 0.999 999 999 075 427 758 899 2 × 2 = 1 + 0.999 999 998 150 855 517 798 4;
  • 30) 0.999 999 998 150 855 517 798 4 × 2 = 1 + 0.999 999 996 301 711 035 596 8;
  • 31) 0.999 999 996 301 711 035 596 8 × 2 = 1 + 0.999 999 992 603 422 071 193 6;
  • 32) 0.999 999 992 603 422 071 193 6 × 2 = 1 + 0.999 999 985 206 844 142 387 2;
  • 33) 0.999 999 985 206 844 142 387 2 × 2 = 1 + 0.999 999 970 413 688 284 774 4;
  • 34) 0.999 999 970 413 688 284 774 4 × 2 = 1 + 0.999 999 940 827 376 569 548 8;
  • 35) 0.999 999 940 827 376 569 548 8 × 2 = 1 + 0.999 999 881 654 753 139 097 6;
  • 36) 0.999 999 881 654 753 139 097 6 × 2 = 1 + 0.999 999 763 309 506 278 195 2;
  • 37) 0.999 999 763 309 506 278 195 2 × 2 = 1 + 0.999 999 526 619 012 556 390 4;
  • 38) 0.999 999 526 619 012 556 390 4 × 2 = 1 + 0.999 999 053 238 025 112 780 8;
  • 39) 0.999 999 053 238 025 112 780 8 × 2 = 1 + 0.999 998 106 476 050 225 561 6;
  • 40) 0.999 998 106 476 050 225 561 6 × 2 = 1 + 0.999 996 212 952 100 451 123 2;
  • 41) 0.999 996 212 952 100 451 123 2 × 2 = 1 + 0.999 992 425 904 200 902 246 4;
  • 42) 0.999 992 425 904 200 902 246 4 × 2 = 1 + 0.999 984 851 808 401 804 492 8;
  • 43) 0.999 984 851 808 401 804 492 8 × 2 = 1 + 0.999 969 703 616 803 608 985 6;
  • 44) 0.999 969 703 616 803 608 985 6 × 2 = 1 + 0.999 939 407 233 607 217 971 2;
  • 45) 0.999 939 407 233 607 217 971 2 × 2 = 1 + 0.999 878 814 467 214 435 942 4;
  • 46) 0.999 878 814 467 214 435 942 4 × 2 = 1 + 0.999 757 628 934 428 871 884 8;
  • 47) 0.999 757 628 934 428 871 884 8 × 2 = 1 + 0.999 515 257 868 857 743 769 6;
  • 48) 0.999 515 257 868 857 743 769 6 × 2 = 1 + 0.999 030 515 737 715 487 539 2;
  • 49) 0.999 030 515 737 715 487 539 2 × 2 = 1 + 0.998 061 031 475 430 975 078 4;
  • 50) 0.998 061 031 475 430 975 078 4 × 2 = 1 + 0.996 122 062 950 861 950 156 8;
  • 51) 0.996 122 062 950 861 950 156 8 × 2 = 1 + 0.992 244 125 901 723 900 313 6;
  • 52) 0.992 244 125 901 723 900 313 6 × 2 = 1 + 0.984 488 251 803 447 800 627 2;
  • 53) 0.984 488 251 803 447 800 627 2 × 2 = 1 + 0.968 976 503 606 895 601 254 4;
  • 54) 0.968 976 503 606 895 601 254 4 × 2 = 1 + 0.937 953 007 213 791 202 508 8;
  • 55) 0.937 953 007 213 791 202 508 8 × 2 = 1 + 0.875 906 014 427 582 405 017 6;
  • 56) 0.875 906 014 427 582 405 017 6 × 2 = 1 + 0.751 812 028 855 164 810 035 2;
  • 57) 0.751 812 028 855 164 810 035 2 × 2 = 1 + 0.503 624 057 710 329 620 070 4;
  • 58) 0.503 624 057 710 329 620 070 4 × 2 = 1 + 0.007 248 115 420 659 240 140 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:


0.016 738 891 601 562 496 555 7(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 555 7(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:


0.016 738 891 601 562 496 555 7(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 -0.016 738 891 601 562 496 555 7 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