-0.016 738 891 601 562 496 799 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 799(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
-0.016 738 891 601 562 496 799(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:

|-0.016 738 891 601 562 496 799| = 0.016 738 891 601 562 496 799


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: 0.016 738 891 601 562 496 799.

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) 0.016 738 891 601 562 496 799 × 2 = 0 + 0.033 477 783 203 124 993 598;
  • 2) 0.033 477 783 203 124 993 598 × 2 = 0 + 0.066 955 566 406 249 987 196;
  • 3) 0.066 955 566 406 249 987 196 × 2 = 0 + 0.133 911 132 812 499 974 392;
  • 4) 0.133 911 132 812 499 974 392 × 2 = 0 + 0.267 822 265 624 999 948 784;
  • 5) 0.267 822 265 624 999 948 784 × 2 = 0 + 0.535 644 531 249 999 897 568;
  • 6) 0.535 644 531 249 999 897 568 × 2 = 1 + 0.071 289 062 499 999 795 136;
  • 7) 0.071 289 062 499 999 795 136 × 2 = 0 + 0.142 578 124 999 999 590 272;
  • 8) 0.142 578 124 999 999 590 272 × 2 = 0 + 0.285 156 249 999 999 180 544;
  • 9) 0.285 156 249 999 999 180 544 × 2 = 0 + 0.570 312 499 999 998 361 088;
  • 10) 0.570 312 499 999 998 361 088 × 2 = 1 + 0.140 624 999 999 996 722 176;
  • 11) 0.140 624 999 999 996 722 176 × 2 = 0 + 0.281 249 999 999 993 444 352;
  • 12) 0.281 249 999 999 993 444 352 × 2 = 0 + 0.562 499 999 999 986 888 704;
  • 13) 0.562 499 999 999 986 888 704 × 2 = 1 + 0.124 999 999 999 973 777 408;
  • 14) 0.124 999 999 999 973 777 408 × 2 = 0 + 0.249 999 999 999 947 554 816;
  • 15) 0.249 999 999 999 947 554 816 × 2 = 0 + 0.499 999 999 999 895 109 632;
  • 16) 0.499 999 999 999 895 109 632 × 2 = 0 + 0.999 999 999 999 790 219 264;
  • 17) 0.999 999 999 999 790 219 264 × 2 = 1 + 0.999 999 999 999 580 438 528;
  • 18) 0.999 999 999 999 580 438 528 × 2 = 1 + 0.999 999 999 999 160 877 056;
  • 19) 0.999 999 999 999 160 877 056 × 2 = 1 + 0.999 999 999 998 321 754 112;
  • 20) 0.999 999 999 998 321 754 112 × 2 = 1 + 0.999 999 999 996 643 508 224;
  • 21) 0.999 999 999 996 643 508 224 × 2 = 1 + 0.999 999 999 993 287 016 448;
  • 22) 0.999 999 999 993 287 016 448 × 2 = 1 + 0.999 999 999 986 574 032 896;
  • 23) 0.999 999 999 986 574 032 896 × 2 = 1 + 0.999 999 999 973 148 065 792;
  • 24) 0.999 999 999 973 148 065 792 × 2 = 1 + 0.999 999 999 946 296 131 584;
  • 25) 0.999 999 999 946 296 131 584 × 2 = 1 + 0.999 999 999 892 592 263 168;
  • 26) 0.999 999 999 892 592 263 168 × 2 = 1 + 0.999 999 999 785 184 526 336;
  • 27) 0.999 999 999 785 184 526 336 × 2 = 1 + 0.999 999 999 570 369 052 672;
  • 28) 0.999 999 999 570 369 052 672 × 2 = 1 + 0.999 999 999 140 738 105 344;
  • 29) 0.999 999 999 140 738 105 344 × 2 = 1 + 0.999 999 998 281 476 210 688;
  • 30) 0.999 999 998 281 476 210 688 × 2 = 1 + 0.999 999 996 562 952 421 376;
  • 31) 0.999 999 996 562 952 421 376 × 2 = 1 + 0.999 999 993 125 904 842 752;
  • 32) 0.999 999 993 125 904 842 752 × 2 = 1 + 0.999 999 986 251 809 685 504;
  • 33) 0.999 999 986 251 809 685 504 × 2 = 1 + 0.999 999 972 503 619 371 008;
  • 34) 0.999 999 972 503 619 371 008 × 2 = 1 + 0.999 999 945 007 238 742 016;
  • 35) 0.999 999 945 007 238 742 016 × 2 = 1 + 0.999 999 890 014 477 484 032;
  • 36) 0.999 999 890 014 477 484 032 × 2 = 1 + 0.999 999 780 028 954 968 064;
  • 37) 0.999 999 780 028 954 968 064 × 2 = 1 + 0.999 999 560 057 909 936 128;
  • 38) 0.999 999 560 057 909 936 128 × 2 = 1 + 0.999 999 120 115 819 872 256;
  • 39) 0.999 999 120 115 819 872 256 × 2 = 1 + 0.999 998 240 231 639 744 512;
  • 40) 0.999 998 240 231 639 744 512 × 2 = 1 + 0.999 996 480 463 279 489 024;
  • 41) 0.999 996 480 463 279 489 024 × 2 = 1 + 0.999 992 960 926 558 978 048;
  • 42) 0.999 992 960 926 558 978 048 × 2 = 1 + 0.999 985 921 853 117 956 096;
  • 43) 0.999 985 921 853 117 956 096 × 2 = 1 + 0.999 971 843 706 235 912 192;
  • 44) 0.999 971 843 706 235 912 192 × 2 = 1 + 0.999 943 687 412 471 824 384;
  • 45) 0.999 943 687 412 471 824 384 × 2 = 1 + 0.999 887 374 824 943 648 768;
  • 46) 0.999 887 374 824 943 648 768 × 2 = 1 + 0.999 774 749 649 887 297 536;
  • 47) 0.999 774 749 649 887 297 536 × 2 = 1 + 0.999 549 499 299 774 595 072;
  • 48) 0.999 549 499 299 774 595 072 × 2 = 1 + 0.999 098 998 599 549 190 144;
  • 49) 0.999 098 998 599 549 190 144 × 2 = 1 + 0.998 197 997 199 098 380 288;
  • 50) 0.998 197 997 199 098 380 288 × 2 = 1 + 0.996 395 994 398 196 760 576;
  • 51) 0.996 395 994 398 196 760 576 × 2 = 1 + 0.992 791 988 796 393 521 152;
  • 52) 0.992 791 988 796 393 521 152 × 2 = 1 + 0.985 583 977 592 787 042 304;
  • 53) 0.985 583 977 592 787 042 304 × 2 = 1 + 0.971 167 955 185 574 084 608;
  • 54) 0.971 167 955 185 574 084 608 × 2 = 1 + 0.942 335 910 371 148 169 216;
  • 55) 0.942 335 910 371 148 169 216 × 2 = 1 + 0.884 671 820 742 296 338 432;
  • 56) 0.884 671 820 742 296 338 432 × 2 = 1 + 0.769 343 641 484 592 676 864;
  • 57) 0.769 343 641 484 592 676 864 × 2 = 1 + 0.538 687 282 969 185 353 728;
  • 58) 0.538 687 282 969 185 353 728 × 2 = 1 + 0.077 374 565 938 370 707 456;

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:


0.016 738 891 601 562 496 799(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 799(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:


0.016 738 891 601 562 496 799(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 -0.016 738 891 601 562 496 799 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