-0.016 738 891 601 562 496 559 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 559(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
-0.016 738 891 601 562 496 559(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:

|-0.016 738 891 601 562 496 559| = 0.016 738 891 601 562 496 559


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: 0.016 738 891 601 562 496 559.

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) 0.016 738 891 601 562 496 559 × 2 = 0 + 0.033 477 783 203 124 993 118;
  • 2) 0.033 477 783 203 124 993 118 × 2 = 0 + 0.066 955 566 406 249 986 236;
  • 3) 0.066 955 566 406 249 986 236 × 2 = 0 + 0.133 911 132 812 499 972 472;
  • 4) 0.133 911 132 812 499 972 472 × 2 = 0 + 0.267 822 265 624 999 944 944;
  • 5) 0.267 822 265 624 999 944 944 × 2 = 0 + 0.535 644 531 249 999 889 888;
  • 6) 0.535 644 531 249 999 889 888 × 2 = 1 + 0.071 289 062 499 999 779 776;
  • 7) 0.071 289 062 499 999 779 776 × 2 = 0 + 0.142 578 124 999 999 559 552;
  • 8) 0.142 578 124 999 999 559 552 × 2 = 0 + 0.285 156 249 999 999 119 104;
  • 9) 0.285 156 249 999 999 119 104 × 2 = 0 + 0.570 312 499 999 998 238 208;
  • 10) 0.570 312 499 999 998 238 208 × 2 = 1 + 0.140 624 999 999 996 476 416;
  • 11) 0.140 624 999 999 996 476 416 × 2 = 0 + 0.281 249 999 999 992 952 832;
  • 12) 0.281 249 999 999 992 952 832 × 2 = 0 + 0.562 499 999 999 985 905 664;
  • 13) 0.562 499 999 999 985 905 664 × 2 = 1 + 0.124 999 999 999 971 811 328;
  • 14) 0.124 999 999 999 971 811 328 × 2 = 0 + 0.249 999 999 999 943 622 656;
  • 15) 0.249 999 999 999 943 622 656 × 2 = 0 + 0.499 999 999 999 887 245 312;
  • 16) 0.499 999 999 999 887 245 312 × 2 = 0 + 0.999 999 999 999 774 490 624;
  • 17) 0.999 999 999 999 774 490 624 × 2 = 1 + 0.999 999 999 999 548 981 248;
  • 18) 0.999 999 999 999 548 981 248 × 2 = 1 + 0.999 999 999 999 097 962 496;
  • 19) 0.999 999 999 999 097 962 496 × 2 = 1 + 0.999 999 999 998 195 924 992;
  • 20) 0.999 999 999 998 195 924 992 × 2 = 1 + 0.999 999 999 996 391 849 984;
  • 21) 0.999 999 999 996 391 849 984 × 2 = 1 + 0.999 999 999 992 783 699 968;
  • 22) 0.999 999 999 992 783 699 968 × 2 = 1 + 0.999 999 999 985 567 399 936;
  • 23) 0.999 999 999 985 567 399 936 × 2 = 1 + 0.999 999 999 971 134 799 872;
  • 24) 0.999 999 999 971 134 799 872 × 2 = 1 + 0.999 999 999 942 269 599 744;
  • 25) 0.999 999 999 942 269 599 744 × 2 = 1 + 0.999 999 999 884 539 199 488;
  • 26) 0.999 999 999 884 539 199 488 × 2 = 1 + 0.999 999 999 769 078 398 976;
  • 27) 0.999 999 999 769 078 398 976 × 2 = 1 + 0.999 999 999 538 156 797 952;
  • 28) 0.999 999 999 538 156 797 952 × 2 = 1 + 0.999 999 999 076 313 595 904;
  • 29) 0.999 999 999 076 313 595 904 × 2 = 1 + 0.999 999 998 152 627 191 808;
  • 30) 0.999 999 998 152 627 191 808 × 2 = 1 + 0.999 999 996 305 254 383 616;
  • 31) 0.999 999 996 305 254 383 616 × 2 = 1 + 0.999 999 992 610 508 767 232;
  • 32) 0.999 999 992 610 508 767 232 × 2 = 1 + 0.999 999 985 221 017 534 464;
  • 33) 0.999 999 985 221 017 534 464 × 2 = 1 + 0.999 999 970 442 035 068 928;
  • 34) 0.999 999 970 442 035 068 928 × 2 = 1 + 0.999 999 940 884 070 137 856;
  • 35) 0.999 999 940 884 070 137 856 × 2 = 1 + 0.999 999 881 768 140 275 712;
  • 36) 0.999 999 881 768 140 275 712 × 2 = 1 + 0.999 999 763 536 280 551 424;
  • 37) 0.999 999 763 536 280 551 424 × 2 = 1 + 0.999 999 527 072 561 102 848;
  • 38) 0.999 999 527 072 561 102 848 × 2 = 1 + 0.999 999 054 145 122 205 696;
  • 39) 0.999 999 054 145 122 205 696 × 2 = 1 + 0.999 998 108 290 244 411 392;
  • 40) 0.999 998 108 290 244 411 392 × 2 = 1 + 0.999 996 216 580 488 822 784;
  • 41) 0.999 996 216 580 488 822 784 × 2 = 1 + 0.999 992 433 160 977 645 568;
  • 42) 0.999 992 433 160 977 645 568 × 2 = 1 + 0.999 984 866 321 955 291 136;
  • 43) 0.999 984 866 321 955 291 136 × 2 = 1 + 0.999 969 732 643 910 582 272;
  • 44) 0.999 969 732 643 910 582 272 × 2 = 1 + 0.999 939 465 287 821 164 544;
  • 45) 0.999 939 465 287 821 164 544 × 2 = 1 + 0.999 878 930 575 642 329 088;
  • 46) 0.999 878 930 575 642 329 088 × 2 = 1 + 0.999 757 861 151 284 658 176;
  • 47) 0.999 757 861 151 284 658 176 × 2 = 1 + 0.999 515 722 302 569 316 352;
  • 48) 0.999 515 722 302 569 316 352 × 2 = 1 + 0.999 031 444 605 138 632 704;
  • 49) 0.999 031 444 605 138 632 704 × 2 = 1 + 0.998 062 889 210 277 265 408;
  • 50) 0.998 062 889 210 277 265 408 × 2 = 1 + 0.996 125 778 420 554 530 816;
  • 51) 0.996 125 778 420 554 530 816 × 2 = 1 + 0.992 251 556 841 109 061 632;
  • 52) 0.992 251 556 841 109 061 632 × 2 = 1 + 0.984 503 113 682 218 123 264;
  • 53) 0.984 503 113 682 218 123 264 × 2 = 1 + 0.969 006 227 364 436 246 528;
  • 54) 0.969 006 227 364 436 246 528 × 2 = 1 + 0.938 012 454 728 872 493 056;
  • 55) 0.938 012 454 728 872 493 056 × 2 = 1 + 0.876 024 909 457 744 986 112;
  • 56) 0.876 024 909 457 744 986 112 × 2 = 1 + 0.752 049 818 915 489 972 224;
  • 57) 0.752 049 818 915 489 972 224 × 2 = 1 + 0.504 099 637 830 979 944 448;
  • 58) 0.504 099 637 830 979 944 448 × 2 = 1 + 0.008 199 275 661 959 888 896;

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:


0.016 738 891 601 562 496 559(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 559(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:


0.016 738 891 601 562 496 559(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 -0.016 738 891 601 562 496 559 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