-0.016 738 891 601 562 496 753 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 753(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
-0.016 738 891 601 562 496 753(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:

|-0.016 738 891 601 562 496 753| = 0.016 738 891 601 562 496 753


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: 0.016 738 891 601 562 496 753.

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) 0.016 738 891 601 562 496 753 × 2 = 0 + 0.033 477 783 203 124 993 506;
  • 2) 0.033 477 783 203 124 993 506 × 2 = 0 + 0.066 955 566 406 249 987 012;
  • 3) 0.066 955 566 406 249 987 012 × 2 = 0 + 0.133 911 132 812 499 974 024;
  • 4) 0.133 911 132 812 499 974 024 × 2 = 0 + 0.267 822 265 624 999 948 048;
  • 5) 0.267 822 265 624 999 948 048 × 2 = 0 + 0.535 644 531 249 999 896 096;
  • 6) 0.535 644 531 249 999 896 096 × 2 = 1 + 0.071 289 062 499 999 792 192;
  • 7) 0.071 289 062 499 999 792 192 × 2 = 0 + 0.142 578 124 999 999 584 384;
  • 8) 0.142 578 124 999 999 584 384 × 2 = 0 + 0.285 156 249 999 999 168 768;
  • 9) 0.285 156 249 999 999 168 768 × 2 = 0 + 0.570 312 499 999 998 337 536;
  • 10) 0.570 312 499 999 998 337 536 × 2 = 1 + 0.140 624 999 999 996 675 072;
  • 11) 0.140 624 999 999 996 675 072 × 2 = 0 + 0.281 249 999 999 993 350 144;
  • 12) 0.281 249 999 999 993 350 144 × 2 = 0 + 0.562 499 999 999 986 700 288;
  • 13) 0.562 499 999 999 986 700 288 × 2 = 1 + 0.124 999 999 999 973 400 576;
  • 14) 0.124 999 999 999 973 400 576 × 2 = 0 + 0.249 999 999 999 946 801 152;
  • 15) 0.249 999 999 999 946 801 152 × 2 = 0 + 0.499 999 999 999 893 602 304;
  • 16) 0.499 999 999 999 893 602 304 × 2 = 0 + 0.999 999 999 999 787 204 608;
  • 17) 0.999 999 999 999 787 204 608 × 2 = 1 + 0.999 999 999 999 574 409 216;
  • 18) 0.999 999 999 999 574 409 216 × 2 = 1 + 0.999 999 999 999 148 818 432;
  • 19) 0.999 999 999 999 148 818 432 × 2 = 1 + 0.999 999 999 998 297 636 864;
  • 20) 0.999 999 999 998 297 636 864 × 2 = 1 + 0.999 999 999 996 595 273 728;
  • 21) 0.999 999 999 996 595 273 728 × 2 = 1 + 0.999 999 999 993 190 547 456;
  • 22) 0.999 999 999 993 190 547 456 × 2 = 1 + 0.999 999 999 986 381 094 912;
  • 23) 0.999 999 999 986 381 094 912 × 2 = 1 + 0.999 999 999 972 762 189 824;
  • 24) 0.999 999 999 972 762 189 824 × 2 = 1 + 0.999 999 999 945 524 379 648;
  • 25) 0.999 999 999 945 524 379 648 × 2 = 1 + 0.999 999 999 891 048 759 296;
  • 26) 0.999 999 999 891 048 759 296 × 2 = 1 + 0.999 999 999 782 097 518 592;
  • 27) 0.999 999 999 782 097 518 592 × 2 = 1 + 0.999 999 999 564 195 037 184;
  • 28) 0.999 999 999 564 195 037 184 × 2 = 1 + 0.999 999 999 128 390 074 368;
  • 29) 0.999 999 999 128 390 074 368 × 2 = 1 + 0.999 999 998 256 780 148 736;
  • 30) 0.999 999 998 256 780 148 736 × 2 = 1 + 0.999 999 996 513 560 297 472;
  • 31) 0.999 999 996 513 560 297 472 × 2 = 1 + 0.999 999 993 027 120 594 944;
  • 32) 0.999 999 993 027 120 594 944 × 2 = 1 + 0.999 999 986 054 241 189 888;
  • 33) 0.999 999 986 054 241 189 888 × 2 = 1 + 0.999 999 972 108 482 379 776;
  • 34) 0.999 999 972 108 482 379 776 × 2 = 1 + 0.999 999 944 216 964 759 552;
  • 35) 0.999 999 944 216 964 759 552 × 2 = 1 + 0.999 999 888 433 929 519 104;
  • 36) 0.999 999 888 433 929 519 104 × 2 = 1 + 0.999 999 776 867 859 038 208;
  • 37) 0.999 999 776 867 859 038 208 × 2 = 1 + 0.999 999 553 735 718 076 416;
  • 38) 0.999 999 553 735 718 076 416 × 2 = 1 + 0.999 999 107 471 436 152 832;
  • 39) 0.999 999 107 471 436 152 832 × 2 = 1 + 0.999 998 214 942 872 305 664;
  • 40) 0.999 998 214 942 872 305 664 × 2 = 1 + 0.999 996 429 885 744 611 328;
  • 41) 0.999 996 429 885 744 611 328 × 2 = 1 + 0.999 992 859 771 489 222 656;
  • 42) 0.999 992 859 771 489 222 656 × 2 = 1 + 0.999 985 719 542 978 445 312;
  • 43) 0.999 985 719 542 978 445 312 × 2 = 1 + 0.999 971 439 085 956 890 624;
  • 44) 0.999 971 439 085 956 890 624 × 2 = 1 + 0.999 942 878 171 913 781 248;
  • 45) 0.999 942 878 171 913 781 248 × 2 = 1 + 0.999 885 756 343 827 562 496;
  • 46) 0.999 885 756 343 827 562 496 × 2 = 1 + 0.999 771 512 687 655 124 992;
  • 47) 0.999 771 512 687 655 124 992 × 2 = 1 + 0.999 543 025 375 310 249 984;
  • 48) 0.999 543 025 375 310 249 984 × 2 = 1 + 0.999 086 050 750 620 499 968;
  • 49) 0.999 086 050 750 620 499 968 × 2 = 1 + 0.998 172 101 501 240 999 936;
  • 50) 0.998 172 101 501 240 999 936 × 2 = 1 + 0.996 344 203 002 481 999 872;
  • 51) 0.996 344 203 002 481 999 872 × 2 = 1 + 0.992 688 406 004 963 999 744;
  • 52) 0.992 688 406 004 963 999 744 × 2 = 1 + 0.985 376 812 009 927 999 488;
  • 53) 0.985 376 812 009 927 999 488 × 2 = 1 + 0.970 753 624 019 855 998 976;
  • 54) 0.970 753 624 019 855 998 976 × 2 = 1 + 0.941 507 248 039 711 997 952;
  • 55) 0.941 507 248 039 711 997 952 × 2 = 1 + 0.883 014 496 079 423 995 904;
  • 56) 0.883 014 496 079 423 995 904 × 2 = 1 + 0.766 028 992 158 847 991 808;
  • 57) 0.766 028 992 158 847 991 808 × 2 = 1 + 0.532 057 984 317 695 983 616;
  • 58) 0.532 057 984 317 695 983 616 × 2 = 1 + 0.064 115 968 635 391 967 232;

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:


0.016 738 891 601 562 496 753(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 753(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:


0.016 738 891 601 562 496 753(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 -0.016 738 891 601 562 496 753 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