-0.016 738 891 601 562 496 530 571 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 571(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
-0.016 738 891 601 562 496 530 571(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:

|-0.016 738 891 601 562 496 530 571| = 0.016 738 891 601 562 496 530 571


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: 0.016 738 891 601 562 496 530 571.

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) 0.016 738 891 601 562 496 530 571 × 2 = 0 + 0.033 477 783 203 124 993 061 142;
  • 2) 0.033 477 783 203 124 993 061 142 × 2 = 0 + 0.066 955 566 406 249 986 122 284;
  • 3) 0.066 955 566 406 249 986 122 284 × 2 = 0 + 0.133 911 132 812 499 972 244 568;
  • 4) 0.133 911 132 812 499 972 244 568 × 2 = 0 + 0.267 822 265 624 999 944 489 136;
  • 5) 0.267 822 265 624 999 944 489 136 × 2 = 0 + 0.535 644 531 249 999 888 978 272;
  • 6) 0.535 644 531 249 999 888 978 272 × 2 = 1 + 0.071 289 062 499 999 777 956 544;
  • 7) 0.071 289 062 499 999 777 956 544 × 2 = 0 + 0.142 578 124 999 999 555 913 088;
  • 8) 0.142 578 124 999 999 555 913 088 × 2 = 0 + 0.285 156 249 999 999 111 826 176;
  • 9) 0.285 156 249 999 999 111 826 176 × 2 = 0 + 0.570 312 499 999 998 223 652 352;
  • 10) 0.570 312 499 999 998 223 652 352 × 2 = 1 + 0.140 624 999 999 996 447 304 704;
  • 11) 0.140 624 999 999 996 447 304 704 × 2 = 0 + 0.281 249 999 999 992 894 609 408;
  • 12) 0.281 249 999 999 992 894 609 408 × 2 = 0 + 0.562 499 999 999 985 789 218 816;
  • 13) 0.562 499 999 999 985 789 218 816 × 2 = 1 + 0.124 999 999 999 971 578 437 632;
  • 14) 0.124 999 999 999 971 578 437 632 × 2 = 0 + 0.249 999 999 999 943 156 875 264;
  • 15) 0.249 999 999 999 943 156 875 264 × 2 = 0 + 0.499 999 999 999 886 313 750 528;
  • 16) 0.499 999 999 999 886 313 750 528 × 2 = 0 + 0.999 999 999 999 772 627 501 056;
  • 17) 0.999 999 999 999 772 627 501 056 × 2 = 1 + 0.999 999 999 999 545 255 002 112;
  • 18) 0.999 999 999 999 545 255 002 112 × 2 = 1 + 0.999 999 999 999 090 510 004 224;
  • 19) 0.999 999 999 999 090 510 004 224 × 2 = 1 + 0.999 999 999 998 181 020 008 448;
  • 20) 0.999 999 999 998 181 020 008 448 × 2 = 1 + 0.999 999 999 996 362 040 016 896;
  • 21) 0.999 999 999 996 362 040 016 896 × 2 = 1 + 0.999 999 999 992 724 080 033 792;
  • 22) 0.999 999 999 992 724 080 033 792 × 2 = 1 + 0.999 999 999 985 448 160 067 584;
  • 23) 0.999 999 999 985 448 160 067 584 × 2 = 1 + 0.999 999 999 970 896 320 135 168;
  • 24) 0.999 999 999 970 896 320 135 168 × 2 = 1 + 0.999 999 999 941 792 640 270 336;
  • 25) 0.999 999 999 941 792 640 270 336 × 2 = 1 + 0.999 999 999 883 585 280 540 672;
  • 26) 0.999 999 999 883 585 280 540 672 × 2 = 1 + 0.999 999 999 767 170 561 081 344;
  • 27) 0.999 999 999 767 170 561 081 344 × 2 = 1 + 0.999 999 999 534 341 122 162 688;
  • 28) 0.999 999 999 534 341 122 162 688 × 2 = 1 + 0.999 999 999 068 682 244 325 376;
  • 29) 0.999 999 999 068 682 244 325 376 × 2 = 1 + 0.999 999 998 137 364 488 650 752;
  • 30) 0.999 999 998 137 364 488 650 752 × 2 = 1 + 0.999 999 996 274 728 977 301 504;
  • 31) 0.999 999 996 274 728 977 301 504 × 2 = 1 + 0.999 999 992 549 457 954 603 008;
  • 32) 0.999 999 992 549 457 954 603 008 × 2 = 1 + 0.999 999 985 098 915 909 206 016;
  • 33) 0.999 999 985 098 915 909 206 016 × 2 = 1 + 0.999 999 970 197 831 818 412 032;
  • 34) 0.999 999 970 197 831 818 412 032 × 2 = 1 + 0.999 999 940 395 663 636 824 064;
  • 35) 0.999 999 940 395 663 636 824 064 × 2 = 1 + 0.999 999 880 791 327 273 648 128;
  • 36) 0.999 999 880 791 327 273 648 128 × 2 = 1 + 0.999 999 761 582 654 547 296 256;
  • 37) 0.999 999 761 582 654 547 296 256 × 2 = 1 + 0.999 999 523 165 309 094 592 512;
  • 38) 0.999 999 523 165 309 094 592 512 × 2 = 1 + 0.999 999 046 330 618 189 185 024;
  • 39) 0.999 999 046 330 618 189 185 024 × 2 = 1 + 0.999 998 092 661 236 378 370 048;
  • 40) 0.999 998 092 661 236 378 370 048 × 2 = 1 + 0.999 996 185 322 472 756 740 096;
  • 41) 0.999 996 185 322 472 756 740 096 × 2 = 1 + 0.999 992 370 644 945 513 480 192;
  • 42) 0.999 992 370 644 945 513 480 192 × 2 = 1 + 0.999 984 741 289 891 026 960 384;
  • 43) 0.999 984 741 289 891 026 960 384 × 2 = 1 + 0.999 969 482 579 782 053 920 768;
  • 44) 0.999 969 482 579 782 053 920 768 × 2 = 1 + 0.999 938 965 159 564 107 841 536;
  • 45) 0.999 938 965 159 564 107 841 536 × 2 = 1 + 0.999 877 930 319 128 215 683 072;
  • 46) 0.999 877 930 319 128 215 683 072 × 2 = 1 + 0.999 755 860 638 256 431 366 144;
  • 47) 0.999 755 860 638 256 431 366 144 × 2 = 1 + 0.999 511 721 276 512 862 732 288;
  • 48) 0.999 511 721 276 512 862 732 288 × 2 = 1 + 0.999 023 442 553 025 725 464 576;
  • 49) 0.999 023 442 553 025 725 464 576 × 2 = 1 + 0.998 046 885 106 051 450 929 152;
  • 50) 0.998 046 885 106 051 450 929 152 × 2 = 1 + 0.996 093 770 212 102 901 858 304;
  • 51) 0.996 093 770 212 102 901 858 304 × 2 = 1 + 0.992 187 540 424 205 803 716 608;
  • 52) 0.992 187 540 424 205 803 716 608 × 2 = 1 + 0.984 375 080 848 411 607 433 216;
  • 53) 0.984 375 080 848 411 607 433 216 × 2 = 1 + 0.968 750 161 696 823 214 866 432;
  • 54) 0.968 750 161 696 823 214 866 432 × 2 = 1 + 0.937 500 323 393 646 429 732 864;
  • 55) 0.937 500 323 393 646 429 732 864 × 2 = 1 + 0.875 000 646 787 292 859 465 728;
  • 56) 0.875 000 646 787 292 859 465 728 × 2 = 1 + 0.750 001 293 574 585 718 931 456;
  • 57) 0.750 001 293 574 585 718 931 456 × 2 = 1 + 0.500 002 587 149 171 437 862 912;
  • 58) 0.500 002 587 149 171 437 862 912 × 2 = 1 + 0.000 005 174 298 342 875 725 824;

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:


0.016 738 891 601 562 496 530 571(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 571(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:


0.016 738 891 601 562 496 530 571(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 -0.016 738 891 601 562 496 530 571 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 -0.016 738 891 601 562 496 530 668 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