-0.016 738 891 601 562 496 418 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 418(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
-0.016 738 891 601 562 496 418(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:

|-0.016 738 891 601 562 496 418| = 0.016 738 891 601 562 496 418


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: 0.016 738 891 601 562 496 418.

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) 0.016 738 891 601 562 496 418 × 2 = 0 + 0.033 477 783 203 124 992 836;
  • 2) 0.033 477 783 203 124 992 836 × 2 = 0 + 0.066 955 566 406 249 985 672;
  • 3) 0.066 955 566 406 249 985 672 × 2 = 0 + 0.133 911 132 812 499 971 344;
  • 4) 0.133 911 132 812 499 971 344 × 2 = 0 + 0.267 822 265 624 999 942 688;
  • 5) 0.267 822 265 624 999 942 688 × 2 = 0 + 0.535 644 531 249 999 885 376;
  • 6) 0.535 644 531 249 999 885 376 × 2 = 1 + 0.071 289 062 499 999 770 752;
  • 7) 0.071 289 062 499 999 770 752 × 2 = 0 + 0.142 578 124 999 999 541 504;
  • 8) 0.142 578 124 999 999 541 504 × 2 = 0 + 0.285 156 249 999 999 083 008;
  • 9) 0.285 156 249 999 999 083 008 × 2 = 0 + 0.570 312 499 999 998 166 016;
  • 10) 0.570 312 499 999 998 166 016 × 2 = 1 + 0.140 624 999 999 996 332 032;
  • 11) 0.140 624 999 999 996 332 032 × 2 = 0 + 0.281 249 999 999 992 664 064;
  • 12) 0.281 249 999 999 992 664 064 × 2 = 0 + 0.562 499 999 999 985 328 128;
  • 13) 0.562 499 999 999 985 328 128 × 2 = 1 + 0.124 999 999 999 970 656 256;
  • 14) 0.124 999 999 999 970 656 256 × 2 = 0 + 0.249 999 999 999 941 312 512;
  • 15) 0.249 999 999 999 941 312 512 × 2 = 0 + 0.499 999 999 999 882 625 024;
  • 16) 0.499 999 999 999 882 625 024 × 2 = 0 + 0.999 999 999 999 765 250 048;
  • 17) 0.999 999 999 999 765 250 048 × 2 = 1 + 0.999 999 999 999 530 500 096;
  • 18) 0.999 999 999 999 530 500 096 × 2 = 1 + 0.999 999 999 999 061 000 192;
  • 19) 0.999 999 999 999 061 000 192 × 2 = 1 + 0.999 999 999 998 122 000 384;
  • 20) 0.999 999 999 998 122 000 384 × 2 = 1 + 0.999 999 999 996 244 000 768;
  • 21) 0.999 999 999 996 244 000 768 × 2 = 1 + 0.999 999 999 992 488 001 536;
  • 22) 0.999 999 999 992 488 001 536 × 2 = 1 + 0.999 999 999 984 976 003 072;
  • 23) 0.999 999 999 984 976 003 072 × 2 = 1 + 0.999 999 999 969 952 006 144;
  • 24) 0.999 999 999 969 952 006 144 × 2 = 1 + 0.999 999 999 939 904 012 288;
  • 25) 0.999 999 999 939 904 012 288 × 2 = 1 + 0.999 999 999 879 808 024 576;
  • 26) 0.999 999 999 879 808 024 576 × 2 = 1 + 0.999 999 999 759 616 049 152;
  • 27) 0.999 999 999 759 616 049 152 × 2 = 1 + 0.999 999 999 519 232 098 304;
  • 28) 0.999 999 999 519 232 098 304 × 2 = 1 + 0.999 999 999 038 464 196 608;
  • 29) 0.999 999 999 038 464 196 608 × 2 = 1 + 0.999 999 998 076 928 393 216;
  • 30) 0.999 999 998 076 928 393 216 × 2 = 1 + 0.999 999 996 153 856 786 432;
  • 31) 0.999 999 996 153 856 786 432 × 2 = 1 + 0.999 999 992 307 713 572 864;
  • 32) 0.999 999 992 307 713 572 864 × 2 = 1 + 0.999 999 984 615 427 145 728;
  • 33) 0.999 999 984 615 427 145 728 × 2 = 1 + 0.999 999 969 230 854 291 456;
  • 34) 0.999 999 969 230 854 291 456 × 2 = 1 + 0.999 999 938 461 708 582 912;
  • 35) 0.999 999 938 461 708 582 912 × 2 = 1 + 0.999 999 876 923 417 165 824;
  • 36) 0.999 999 876 923 417 165 824 × 2 = 1 + 0.999 999 753 846 834 331 648;
  • 37) 0.999 999 753 846 834 331 648 × 2 = 1 + 0.999 999 507 693 668 663 296;
  • 38) 0.999 999 507 693 668 663 296 × 2 = 1 + 0.999 999 015 387 337 326 592;
  • 39) 0.999 999 015 387 337 326 592 × 2 = 1 + 0.999 998 030 774 674 653 184;
  • 40) 0.999 998 030 774 674 653 184 × 2 = 1 + 0.999 996 061 549 349 306 368;
  • 41) 0.999 996 061 549 349 306 368 × 2 = 1 + 0.999 992 123 098 698 612 736;
  • 42) 0.999 992 123 098 698 612 736 × 2 = 1 + 0.999 984 246 197 397 225 472;
  • 43) 0.999 984 246 197 397 225 472 × 2 = 1 + 0.999 968 492 394 794 450 944;
  • 44) 0.999 968 492 394 794 450 944 × 2 = 1 + 0.999 936 984 789 588 901 888;
  • 45) 0.999 936 984 789 588 901 888 × 2 = 1 + 0.999 873 969 579 177 803 776;
  • 46) 0.999 873 969 579 177 803 776 × 2 = 1 + 0.999 747 939 158 355 607 552;
  • 47) 0.999 747 939 158 355 607 552 × 2 = 1 + 0.999 495 878 316 711 215 104;
  • 48) 0.999 495 878 316 711 215 104 × 2 = 1 + 0.998 991 756 633 422 430 208;
  • 49) 0.998 991 756 633 422 430 208 × 2 = 1 + 0.997 983 513 266 844 860 416;
  • 50) 0.997 983 513 266 844 860 416 × 2 = 1 + 0.995 967 026 533 689 720 832;
  • 51) 0.995 967 026 533 689 720 832 × 2 = 1 + 0.991 934 053 067 379 441 664;
  • 52) 0.991 934 053 067 379 441 664 × 2 = 1 + 0.983 868 106 134 758 883 328;
  • 53) 0.983 868 106 134 758 883 328 × 2 = 1 + 0.967 736 212 269 517 766 656;
  • 54) 0.967 736 212 269 517 766 656 × 2 = 1 + 0.935 472 424 539 035 533 312;
  • 55) 0.935 472 424 539 035 533 312 × 2 = 1 + 0.870 944 849 078 071 066 624;
  • 56) 0.870 944 849 078 071 066 624 × 2 = 1 + 0.741 889 698 156 142 133 248;
  • 57) 0.741 889 698 156 142 133 248 × 2 = 1 + 0.483 779 396 312 284 266 496;
  • 58) 0.483 779 396 312 284 266 496 × 2 = 0 + 0.967 558 792 624 568 532 992;

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:


0.016 738 891 601 562 496 418(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 418(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 418(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 418 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 1110

Share

» Convert decimal -0.016 738 891 601 562 496 391 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