-0.016 738 891 601 562 497 003 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 003(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
-0.016 738 891 601 562 497 003(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:

|-0.016 738 891 601 562 497 003| = 0.016 738 891 601 562 497 003


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: 0.016 738 891 601 562 497 003.

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) 0.016 738 891 601 562 497 003 × 2 = 0 + 0.033 477 783 203 124 994 006;
  • 2) 0.033 477 783 203 124 994 006 × 2 = 0 + 0.066 955 566 406 249 988 012;
  • 3) 0.066 955 566 406 249 988 012 × 2 = 0 + 0.133 911 132 812 499 976 024;
  • 4) 0.133 911 132 812 499 976 024 × 2 = 0 + 0.267 822 265 624 999 952 048;
  • 5) 0.267 822 265 624 999 952 048 × 2 = 0 + 0.535 644 531 249 999 904 096;
  • 6) 0.535 644 531 249 999 904 096 × 2 = 1 + 0.071 289 062 499 999 808 192;
  • 7) 0.071 289 062 499 999 808 192 × 2 = 0 + 0.142 578 124 999 999 616 384;
  • 8) 0.142 578 124 999 999 616 384 × 2 = 0 + 0.285 156 249 999 999 232 768;
  • 9) 0.285 156 249 999 999 232 768 × 2 = 0 + 0.570 312 499 999 998 465 536;
  • 10) 0.570 312 499 999 998 465 536 × 2 = 1 + 0.140 624 999 999 996 931 072;
  • 11) 0.140 624 999 999 996 931 072 × 2 = 0 + 0.281 249 999 999 993 862 144;
  • 12) 0.281 249 999 999 993 862 144 × 2 = 0 + 0.562 499 999 999 987 724 288;
  • 13) 0.562 499 999 999 987 724 288 × 2 = 1 + 0.124 999 999 999 975 448 576;
  • 14) 0.124 999 999 999 975 448 576 × 2 = 0 + 0.249 999 999 999 950 897 152;
  • 15) 0.249 999 999 999 950 897 152 × 2 = 0 + 0.499 999 999 999 901 794 304;
  • 16) 0.499 999 999 999 901 794 304 × 2 = 0 + 0.999 999 999 999 803 588 608;
  • 17) 0.999 999 999 999 803 588 608 × 2 = 1 + 0.999 999 999 999 607 177 216;
  • 18) 0.999 999 999 999 607 177 216 × 2 = 1 + 0.999 999 999 999 214 354 432;
  • 19) 0.999 999 999 999 214 354 432 × 2 = 1 + 0.999 999 999 998 428 708 864;
  • 20) 0.999 999 999 998 428 708 864 × 2 = 1 + 0.999 999 999 996 857 417 728;
  • 21) 0.999 999 999 996 857 417 728 × 2 = 1 + 0.999 999 999 993 714 835 456;
  • 22) 0.999 999 999 993 714 835 456 × 2 = 1 + 0.999 999 999 987 429 670 912;
  • 23) 0.999 999 999 987 429 670 912 × 2 = 1 + 0.999 999 999 974 859 341 824;
  • 24) 0.999 999 999 974 859 341 824 × 2 = 1 + 0.999 999 999 949 718 683 648;
  • 25) 0.999 999 999 949 718 683 648 × 2 = 1 + 0.999 999 999 899 437 367 296;
  • 26) 0.999 999 999 899 437 367 296 × 2 = 1 + 0.999 999 999 798 874 734 592;
  • 27) 0.999 999 999 798 874 734 592 × 2 = 1 + 0.999 999 999 597 749 469 184;
  • 28) 0.999 999 999 597 749 469 184 × 2 = 1 + 0.999 999 999 195 498 938 368;
  • 29) 0.999 999 999 195 498 938 368 × 2 = 1 + 0.999 999 998 390 997 876 736;
  • 30) 0.999 999 998 390 997 876 736 × 2 = 1 + 0.999 999 996 781 995 753 472;
  • 31) 0.999 999 996 781 995 753 472 × 2 = 1 + 0.999 999 993 563 991 506 944;
  • 32) 0.999 999 993 563 991 506 944 × 2 = 1 + 0.999 999 987 127 983 013 888;
  • 33) 0.999 999 987 127 983 013 888 × 2 = 1 + 0.999 999 974 255 966 027 776;
  • 34) 0.999 999 974 255 966 027 776 × 2 = 1 + 0.999 999 948 511 932 055 552;
  • 35) 0.999 999 948 511 932 055 552 × 2 = 1 + 0.999 999 897 023 864 111 104;
  • 36) 0.999 999 897 023 864 111 104 × 2 = 1 + 0.999 999 794 047 728 222 208;
  • 37) 0.999 999 794 047 728 222 208 × 2 = 1 + 0.999 999 588 095 456 444 416;
  • 38) 0.999 999 588 095 456 444 416 × 2 = 1 + 0.999 999 176 190 912 888 832;
  • 39) 0.999 999 176 190 912 888 832 × 2 = 1 + 0.999 998 352 381 825 777 664;
  • 40) 0.999 998 352 381 825 777 664 × 2 = 1 + 0.999 996 704 763 651 555 328;
  • 41) 0.999 996 704 763 651 555 328 × 2 = 1 + 0.999 993 409 527 303 110 656;
  • 42) 0.999 993 409 527 303 110 656 × 2 = 1 + 0.999 986 819 054 606 221 312;
  • 43) 0.999 986 819 054 606 221 312 × 2 = 1 + 0.999 973 638 109 212 442 624;
  • 44) 0.999 973 638 109 212 442 624 × 2 = 1 + 0.999 947 276 218 424 885 248;
  • 45) 0.999 947 276 218 424 885 248 × 2 = 1 + 0.999 894 552 436 849 770 496;
  • 46) 0.999 894 552 436 849 770 496 × 2 = 1 + 0.999 789 104 873 699 540 992;
  • 47) 0.999 789 104 873 699 540 992 × 2 = 1 + 0.999 578 209 747 399 081 984;
  • 48) 0.999 578 209 747 399 081 984 × 2 = 1 + 0.999 156 419 494 798 163 968;
  • 49) 0.999 156 419 494 798 163 968 × 2 = 1 + 0.998 312 838 989 596 327 936;
  • 50) 0.998 312 838 989 596 327 936 × 2 = 1 + 0.996 625 677 979 192 655 872;
  • 51) 0.996 625 677 979 192 655 872 × 2 = 1 + 0.993 251 355 958 385 311 744;
  • 52) 0.993 251 355 958 385 311 744 × 2 = 1 + 0.986 502 711 916 770 623 488;
  • 53) 0.986 502 711 916 770 623 488 × 2 = 1 + 0.973 005 423 833 541 246 976;
  • 54) 0.973 005 423 833 541 246 976 × 2 = 1 + 0.946 010 847 667 082 493 952;
  • 55) 0.946 010 847 667 082 493 952 × 2 = 1 + 0.892 021 695 334 164 987 904;
  • 56) 0.892 021 695 334 164 987 904 × 2 = 1 + 0.784 043 390 668 329 975 808;
  • 57) 0.784 043 390 668 329 975 808 × 2 = 1 + 0.568 086 781 336 659 951 616;
  • 58) 0.568 086 781 336 659 951 616 × 2 = 1 + 0.136 173 562 673 319 903 232;

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:


0.016 738 891 601 562 497 003(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 497 003(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:


0.016 738 891 601 562 497 003(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 -0.016 738 891 601 562 497 003 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