-0.016 738 891 601 562 496 883 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 883(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
-0.016 738 891 601 562 496 883(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:

|-0.016 738 891 601 562 496 883| = 0.016 738 891 601 562 496 883


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: 0.016 738 891 601 562 496 883.

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) 0.016 738 891 601 562 496 883 × 2 = 0 + 0.033 477 783 203 124 993 766;
  • 2) 0.033 477 783 203 124 993 766 × 2 = 0 + 0.066 955 566 406 249 987 532;
  • 3) 0.066 955 566 406 249 987 532 × 2 = 0 + 0.133 911 132 812 499 975 064;
  • 4) 0.133 911 132 812 499 975 064 × 2 = 0 + 0.267 822 265 624 999 950 128;
  • 5) 0.267 822 265 624 999 950 128 × 2 = 0 + 0.535 644 531 249 999 900 256;
  • 6) 0.535 644 531 249 999 900 256 × 2 = 1 + 0.071 289 062 499 999 800 512;
  • 7) 0.071 289 062 499 999 800 512 × 2 = 0 + 0.142 578 124 999 999 601 024;
  • 8) 0.142 578 124 999 999 601 024 × 2 = 0 + 0.285 156 249 999 999 202 048;
  • 9) 0.285 156 249 999 999 202 048 × 2 = 0 + 0.570 312 499 999 998 404 096;
  • 10) 0.570 312 499 999 998 404 096 × 2 = 1 + 0.140 624 999 999 996 808 192;
  • 11) 0.140 624 999 999 996 808 192 × 2 = 0 + 0.281 249 999 999 993 616 384;
  • 12) 0.281 249 999 999 993 616 384 × 2 = 0 + 0.562 499 999 999 987 232 768;
  • 13) 0.562 499 999 999 987 232 768 × 2 = 1 + 0.124 999 999 999 974 465 536;
  • 14) 0.124 999 999 999 974 465 536 × 2 = 0 + 0.249 999 999 999 948 931 072;
  • 15) 0.249 999 999 999 948 931 072 × 2 = 0 + 0.499 999 999 999 897 862 144;
  • 16) 0.499 999 999 999 897 862 144 × 2 = 0 + 0.999 999 999 999 795 724 288;
  • 17) 0.999 999 999 999 795 724 288 × 2 = 1 + 0.999 999 999 999 591 448 576;
  • 18) 0.999 999 999 999 591 448 576 × 2 = 1 + 0.999 999 999 999 182 897 152;
  • 19) 0.999 999 999 999 182 897 152 × 2 = 1 + 0.999 999 999 998 365 794 304;
  • 20) 0.999 999 999 998 365 794 304 × 2 = 1 + 0.999 999 999 996 731 588 608;
  • 21) 0.999 999 999 996 731 588 608 × 2 = 1 + 0.999 999 999 993 463 177 216;
  • 22) 0.999 999 999 993 463 177 216 × 2 = 1 + 0.999 999 999 986 926 354 432;
  • 23) 0.999 999 999 986 926 354 432 × 2 = 1 + 0.999 999 999 973 852 708 864;
  • 24) 0.999 999 999 973 852 708 864 × 2 = 1 + 0.999 999 999 947 705 417 728;
  • 25) 0.999 999 999 947 705 417 728 × 2 = 1 + 0.999 999 999 895 410 835 456;
  • 26) 0.999 999 999 895 410 835 456 × 2 = 1 + 0.999 999 999 790 821 670 912;
  • 27) 0.999 999 999 790 821 670 912 × 2 = 1 + 0.999 999 999 581 643 341 824;
  • 28) 0.999 999 999 581 643 341 824 × 2 = 1 + 0.999 999 999 163 286 683 648;
  • 29) 0.999 999 999 163 286 683 648 × 2 = 1 + 0.999 999 998 326 573 367 296;
  • 30) 0.999 999 998 326 573 367 296 × 2 = 1 + 0.999 999 996 653 146 734 592;
  • 31) 0.999 999 996 653 146 734 592 × 2 = 1 + 0.999 999 993 306 293 469 184;
  • 32) 0.999 999 993 306 293 469 184 × 2 = 1 + 0.999 999 986 612 586 938 368;
  • 33) 0.999 999 986 612 586 938 368 × 2 = 1 + 0.999 999 973 225 173 876 736;
  • 34) 0.999 999 973 225 173 876 736 × 2 = 1 + 0.999 999 946 450 347 753 472;
  • 35) 0.999 999 946 450 347 753 472 × 2 = 1 + 0.999 999 892 900 695 506 944;
  • 36) 0.999 999 892 900 695 506 944 × 2 = 1 + 0.999 999 785 801 391 013 888;
  • 37) 0.999 999 785 801 391 013 888 × 2 = 1 + 0.999 999 571 602 782 027 776;
  • 38) 0.999 999 571 602 782 027 776 × 2 = 1 + 0.999 999 143 205 564 055 552;
  • 39) 0.999 999 143 205 564 055 552 × 2 = 1 + 0.999 998 286 411 128 111 104;
  • 40) 0.999 998 286 411 128 111 104 × 2 = 1 + 0.999 996 572 822 256 222 208;
  • 41) 0.999 996 572 822 256 222 208 × 2 = 1 + 0.999 993 145 644 512 444 416;
  • 42) 0.999 993 145 644 512 444 416 × 2 = 1 + 0.999 986 291 289 024 888 832;
  • 43) 0.999 986 291 289 024 888 832 × 2 = 1 + 0.999 972 582 578 049 777 664;
  • 44) 0.999 972 582 578 049 777 664 × 2 = 1 + 0.999 945 165 156 099 555 328;
  • 45) 0.999 945 165 156 099 555 328 × 2 = 1 + 0.999 890 330 312 199 110 656;
  • 46) 0.999 890 330 312 199 110 656 × 2 = 1 + 0.999 780 660 624 398 221 312;
  • 47) 0.999 780 660 624 398 221 312 × 2 = 1 + 0.999 561 321 248 796 442 624;
  • 48) 0.999 561 321 248 796 442 624 × 2 = 1 + 0.999 122 642 497 592 885 248;
  • 49) 0.999 122 642 497 592 885 248 × 2 = 1 + 0.998 245 284 995 185 770 496;
  • 50) 0.998 245 284 995 185 770 496 × 2 = 1 + 0.996 490 569 990 371 540 992;
  • 51) 0.996 490 569 990 371 540 992 × 2 = 1 + 0.992 981 139 980 743 081 984;
  • 52) 0.992 981 139 980 743 081 984 × 2 = 1 + 0.985 962 279 961 486 163 968;
  • 53) 0.985 962 279 961 486 163 968 × 2 = 1 + 0.971 924 559 922 972 327 936;
  • 54) 0.971 924 559 922 972 327 936 × 2 = 1 + 0.943 849 119 845 944 655 872;
  • 55) 0.943 849 119 845 944 655 872 × 2 = 1 + 0.887 698 239 691 889 311 744;
  • 56) 0.887 698 239 691 889 311 744 × 2 = 1 + 0.775 396 479 383 778 623 488;
  • 57) 0.775 396 479 383 778 623 488 × 2 = 1 + 0.550 792 958 767 557 246 976;
  • 58) 0.550 792 958 767 557 246 976 × 2 = 1 + 0.101 585 917 535 114 493 952;

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:


0.016 738 891 601 562 496 883(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 883(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:


0.016 738 891 601 562 496 883(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 -0.016 738 891 601 562 496 883 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