-0.016 738 891 601 562 496 711 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 711(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
-0.016 738 891 601 562 496 711(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:

|-0.016 738 891 601 562 496 711| = 0.016 738 891 601 562 496 711


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: 0.016 738 891 601 562 496 711.

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) 0.016 738 891 601 562 496 711 × 2 = 0 + 0.033 477 783 203 124 993 422;
  • 2) 0.033 477 783 203 124 993 422 × 2 = 0 + 0.066 955 566 406 249 986 844;
  • 3) 0.066 955 566 406 249 986 844 × 2 = 0 + 0.133 911 132 812 499 973 688;
  • 4) 0.133 911 132 812 499 973 688 × 2 = 0 + 0.267 822 265 624 999 947 376;
  • 5) 0.267 822 265 624 999 947 376 × 2 = 0 + 0.535 644 531 249 999 894 752;
  • 6) 0.535 644 531 249 999 894 752 × 2 = 1 + 0.071 289 062 499 999 789 504;
  • 7) 0.071 289 062 499 999 789 504 × 2 = 0 + 0.142 578 124 999 999 579 008;
  • 8) 0.142 578 124 999 999 579 008 × 2 = 0 + 0.285 156 249 999 999 158 016;
  • 9) 0.285 156 249 999 999 158 016 × 2 = 0 + 0.570 312 499 999 998 316 032;
  • 10) 0.570 312 499 999 998 316 032 × 2 = 1 + 0.140 624 999 999 996 632 064;
  • 11) 0.140 624 999 999 996 632 064 × 2 = 0 + 0.281 249 999 999 993 264 128;
  • 12) 0.281 249 999 999 993 264 128 × 2 = 0 + 0.562 499 999 999 986 528 256;
  • 13) 0.562 499 999 999 986 528 256 × 2 = 1 + 0.124 999 999 999 973 056 512;
  • 14) 0.124 999 999 999 973 056 512 × 2 = 0 + 0.249 999 999 999 946 113 024;
  • 15) 0.249 999 999 999 946 113 024 × 2 = 0 + 0.499 999 999 999 892 226 048;
  • 16) 0.499 999 999 999 892 226 048 × 2 = 0 + 0.999 999 999 999 784 452 096;
  • 17) 0.999 999 999 999 784 452 096 × 2 = 1 + 0.999 999 999 999 568 904 192;
  • 18) 0.999 999 999 999 568 904 192 × 2 = 1 + 0.999 999 999 999 137 808 384;
  • 19) 0.999 999 999 999 137 808 384 × 2 = 1 + 0.999 999 999 998 275 616 768;
  • 20) 0.999 999 999 998 275 616 768 × 2 = 1 + 0.999 999 999 996 551 233 536;
  • 21) 0.999 999 999 996 551 233 536 × 2 = 1 + 0.999 999 999 993 102 467 072;
  • 22) 0.999 999 999 993 102 467 072 × 2 = 1 + 0.999 999 999 986 204 934 144;
  • 23) 0.999 999 999 986 204 934 144 × 2 = 1 + 0.999 999 999 972 409 868 288;
  • 24) 0.999 999 999 972 409 868 288 × 2 = 1 + 0.999 999 999 944 819 736 576;
  • 25) 0.999 999 999 944 819 736 576 × 2 = 1 + 0.999 999 999 889 639 473 152;
  • 26) 0.999 999 999 889 639 473 152 × 2 = 1 + 0.999 999 999 779 278 946 304;
  • 27) 0.999 999 999 779 278 946 304 × 2 = 1 + 0.999 999 999 558 557 892 608;
  • 28) 0.999 999 999 558 557 892 608 × 2 = 1 + 0.999 999 999 117 115 785 216;
  • 29) 0.999 999 999 117 115 785 216 × 2 = 1 + 0.999 999 998 234 231 570 432;
  • 30) 0.999 999 998 234 231 570 432 × 2 = 1 + 0.999 999 996 468 463 140 864;
  • 31) 0.999 999 996 468 463 140 864 × 2 = 1 + 0.999 999 992 936 926 281 728;
  • 32) 0.999 999 992 936 926 281 728 × 2 = 1 + 0.999 999 985 873 852 563 456;
  • 33) 0.999 999 985 873 852 563 456 × 2 = 1 + 0.999 999 971 747 705 126 912;
  • 34) 0.999 999 971 747 705 126 912 × 2 = 1 + 0.999 999 943 495 410 253 824;
  • 35) 0.999 999 943 495 410 253 824 × 2 = 1 + 0.999 999 886 990 820 507 648;
  • 36) 0.999 999 886 990 820 507 648 × 2 = 1 + 0.999 999 773 981 641 015 296;
  • 37) 0.999 999 773 981 641 015 296 × 2 = 1 + 0.999 999 547 963 282 030 592;
  • 38) 0.999 999 547 963 282 030 592 × 2 = 1 + 0.999 999 095 926 564 061 184;
  • 39) 0.999 999 095 926 564 061 184 × 2 = 1 + 0.999 998 191 853 128 122 368;
  • 40) 0.999 998 191 853 128 122 368 × 2 = 1 + 0.999 996 383 706 256 244 736;
  • 41) 0.999 996 383 706 256 244 736 × 2 = 1 + 0.999 992 767 412 512 489 472;
  • 42) 0.999 992 767 412 512 489 472 × 2 = 1 + 0.999 985 534 825 024 978 944;
  • 43) 0.999 985 534 825 024 978 944 × 2 = 1 + 0.999 971 069 650 049 957 888;
  • 44) 0.999 971 069 650 049 957 888 × 2 = 1 + 0.999 942 139 300 099 915 776;
  • 45) 0.999 942 139 300 099 915 776 × 2 = 1 + 0.999 884 278 600 199 831 552;
  • 46) 0.999 884 278 600 199 831 552 × 2 = 1 + 0.999 768 557 200 399 663 104;
  • 47) 0.999 768 557 200 399 663 104 × 2 = 1 + 0.999 537 114 400 799 326 208;
  • 48) 0.999 537 114 400 799 326 208 × 2 = 1 + 0.999 074 228 801 598 652 416;
  • 49) 0.999 074 228 801 598 652 416 × 2 = 1 + 0.998 148 457 603 197 304 832;
  • 50) 0.998 148 457 603 197 304 832 × 2 = 1 + 0.996 296 915 206 394 609 664;
  • 51) 0.996 296 915 206 394 609 664 × 2 = 1 + 0.992 593 830 412 789 219 328;
  • 52) 0.992 593 830 412 789 219 328 × 2 = 1 + 0.985 187 660 825 578 438 656;
  • 53) 0.985 187 660 825 578 438 656 × 2 = 1 + 0.970 375 321 651 156 877 312;
  • 54) 0.970 375 321 651 156 877 312 × 2 = 1 + 0.940 750 643 302 313 754 624;
  • 55) 0.940 750 643 302 313 754 624 × 2 = 1 + 0.881 501 286 604 627 509 248;
  • 56) 0.881 501 286 604 627 509 248 × 2 = 1 + 0.763 002 573 209 255 018 496;
  • 57) 0.763 002 573 209 255 018 496 × 2 = 1 + 0.526 005 146 418 510 036 992;
  • 58) 0.526 005 146 418 510 036 992 × 2 = 1 + 0.052 010 292 837 020 073 984;

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:


0.016 738 891 601 562 496 711(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 711(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:


0.016 738 891 601 562 496 711(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 -0.016 738 891 601 562 496 711 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 -0.016 738 891 601 562 496 739 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: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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