-0.016 738 891 601 562 496 530 576 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 576(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
-0.016 738 891 601 562 496 530 576(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:

|-0.016 738 891 601 562 496 530 576| = 0.016 738 891 601 562 496 530 576


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: 0.016 738 891 601 562 496 530 576.

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) 0.016 738 891 601 562 496 530 576 × 2 = 0 + 0.033 477 783 203 124 993 061 152;
  • 2) 0.033 477 783 203 124 993 061 152 × 2 = 0 + 0.066 955 566 406 249 986 122 304;
  • 3) 0.066 955 566 406 249 986 122 304 × 2 = 0 + 0.133 911 132 812 499 972 244 608;
  • 4) 0.133 911 132 812 499 972 244 608 × 2 = 0 + 0.267 822 265 624 999 944 489 216;
  • 5) 0.267 822 265 624 999 944 489 216 × 2 = 0 + 0.535 644 531 249 999 888 978 432;
  • 6) 0.535 644 531 249 999 888 978 432 × 2 = 1 + 0.071 289 062 499 999 777 956 864;
  • 7) 0.071 289 062 499 999 777 956 864 × 2 = 0 + 0.142 578 124 999 999 555 913 728;
  • 8) 0.142 578 124 999 999 555 913 728 × 2 = 0 + 0.285 156 249 999 999 111 827 456;
  • 9) 0.285 156 249 999 999 111 827 456 × 2 = 0 + 0.570 312 499 999 998 223 654 912;
  • 10) 0.570 312 499 999 998 223 654 912 × 2 = 1 + 0.140 624 999 999 996 447 309 824;
  • 11) 0.140 624 999 999 996 447 309 824 × 2 = 0 + 0.281 249 999 999 992 894 619 648;
  • 12) 0.281 249 999 999 992 894 619 648 × 2 = 0 + 0.562 499 999 999 985 789 239 296;
  • 13) 0.562 499 999 999 985 789 239 296 × 2 = 1 + 0.124 999 999 999 971 578 478 592;
  • 14) 0.124 999 999 999 971 578 478 592 × 2 = 0 + 0.249 999 999 999 943 156 957 184;
  • 15) 0.249 999 999 999 943 156 957 184 × 2 = 0 + 0.499 999 999 999 886 313 914 368;
  • 16) 0.499 999 999 999 886 313 914 368 × 2 = 0 + 0.999 999 999 999 772 627 828 736;
  • 17) 0.999 999 999 999 772 627 828 736 × 2 = 1 + 0.999 999 999 999 545 255 657 472;
  • 18) 0.999 999 999 999 545 255 657 472 × 2 = 1 + 0.999 999 999 999 090 511 314 944;
  • 19) 0.999 999 999 999 090 511 314 944 × 2 = 1 + 0.999 999 999 998 181 022 629 888;
  • 20) 0.999 999 999 998 181 022 629 888 × 2 = 1 + 0.999 999 999 996 362 045 259 776;
  • 21) 0.999 999 999 996 362 045 259 776 × 2 = 1 + 0.999 999 999 992 724 090 519 552;
  • 22) 0.999 999 999 992 724 090 519 552 × 2 = 1 + 0.999 999 999 985 448 181 039 104;
  • 23) 0.999 999 999 985 448 181 039 104 × 2 = 1 + 0.999 999 999 970 896 362 078 208;
  • 24) 0.999 999 999 970 896 362 078 208 × 2 = 1 + 0.999 999 999 941 792 724 156 416;
  • 25) 0.999 999 999 941 792 724 156 416 × 2 = 1 + 0.999 999 999 883 585 448 312 832;
  • 26) 0.999 999 999 883 585 448 312 832 × 2 = 1 + 0.999 999 999 767 170 896 625 664;
  • 27) 0.999 999 999 767 170 896 625 664 × 2 = 1 + 0.999 999 999 534 341 793 251 328;
  • 28) 0.999 999 999 534 341 793 251 328 × 2 = 1 + 0.999 999 999 068 683 586 502 656;
  • 29) 0.999 999 999 068 683 586 502 656 × 2 = 1 + 0.999 999 998 137 367 173 005 312;
  • 30) 0.999 999 998 137 367 173 005 312 × 2 = 1 + 0.999 999 996 274 734 346 010 624;
  • 31) 0.999 999 996 274 734 346 010 624 × 2 = 1 + 0.999 999 992 549 468 692 021 248;
  • 32) 0.999 999 992 549 468 692 021 248 × 2 = 1 + 0.999 999 985 098 937 384 042 496;
  • 33) 0.999 999 985 098 937 384 042 496 × 2 = 1 + 0.999 999 970 197 874 768 084 992;
  • 34) 0.999 999 970 197 874 768 084 992 × 2 = 1 + 0.999 999 940 395 749 536 169 984;
  • 35) 0.999 999 940 395 749 536 169 984 × 2 = 1 + 0.999 999 880 791 499 072 339 968;
  • 36) 0.999 999 880 791 499 072 339 968 × 2 = 1 + 0.999 999 761 582 998 144 679 936;
  • 37) 0.999 999 761 582 998 144 679 936 × 2 = 1 + 0.999 999 523 165 996 289 359 872;
  • 38) 0.999 999 523 165 996 289 359 872 × 2 = 1 + 0.999 999 046 331 992 578 719 744;
  • 39) 0.999 999 046 331 992 578 719 744 × 2 = 1 + 0.999 998 092 663 985 157 439 488;
  • 40) 0.999 998 092 663 985 157 439 488 × 2 = 1 + 0.999 996 185 327 970 314 878 976;
  • 41) 0.999 996 185 327 970 314 878 976 × 2 = 1 + 0.999 992 370 655 940 629 757 952;
  • 42) 0.999 992 370 655 940 629 757 952 × 2 = 1 + 0.999 984 741 311 881 259 515 904;
  • 43) 0.999 984 741 311 881 259 515 904 × 2 = 1 + 0.999 969 482 623 762 519 031 808;
  • 44) 0.999 969 482 623 762 519 031 808 × 2 = 1 + 0.999 938 965 247 525 038 063 616;
  • 45) 0.999 938 965 247 525 038 063 616 × 2 = 1 + 0.999 877 930 495 050 076 127 232;
  • 46) 0.999 877 930 495 050 076 127 232 × 2 = 1 + 0.999 755 860 990 100 152 254 464;
  • 47) 0.999 755 860 990 100 152 254 464 × 2 = 1 + 0.999 511 721 980 200 304 508 928;
  • 48) 0.999 511 721 980 200 304 508 928 × 2 = 1 + 0.999 023 443 960 400 609 017 856;
  • 49) 0.999 023 443 960 400 609 017 856 × 2 = 1 + 0.998 046 887 920 801 218 035 712;
  • 50) 0.998 046 887 920 801 218 035 712 × 2 = 1 + 0.996 093 775 841 602 436 071 424;
  • 51) 0.996 093 775 841 602 436 071 424 × 2 = 1 + 0.992 187 551 683 204 872 142 848;
  • 52) 0.992 187 551 683 204 872 142 848 × 2 = 1 + 0.984 375 103 366 409 744 285 696;
  • 53) 0.984 375 103 366 409 744 285 696 × 2 = 1 + 0.968 750 206 732 819 488 571 392;
  • 54) 0.968 750 206 732 819 488 571 392 × 2 = 1 + 0.937 500 413 465 638 977 142 784;
  • 55) 0.937 500 413 465 638 977 142 784 × 2 = 1 + 0.875 000 826 931 277 954 285 568;
  • 56) 0.875 000 826 931 277 954 285 568 × 2 = 1 + 0.750 001 653 862 555 908 571 136;
  • 57) 0.750 001 653 862 555 908 571 136 × 2 = 1 + 0.500 003 307 725 111 817 142 272;
  • 58) 0.500 003 307 725 111 817 142 272 × 2 = 1 + 0.000 006 615 450 223 634 284 544;

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:


0.016 738 891 601 562 496 530 576(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 576(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:


0.016 738 891 601 562 496 530 576(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 -0.016 738 891 601 562 496 530 576 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

Share

» Convert decimal -0.016 738 891 601 562 496 530 623 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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