-0.016 738 891 601 562 496 453 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 453(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
-0.016 738 891 601 562 496 453(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:

|-0.016 738 891 601 562 496 453| = 0.016 738 891 601 562 496 453


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: 0.016 738 891 601 562 496 453.

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) 0.016 738 891 601 562 496 453 × 2 = 0 + 0.033 477 783 203 124 992 906;
  • 2) 0.033 477 783 203 124 992 906 × 2 = 0 + 0.066 955 566 406 249 985 812;
  • 3) 0.066 955 566 406 249 985 812 × 2 = 0 + 0.133 911 132 812 499 971 624;
  • 4) 0.133 911 132 812 499 971 624 × 2 = 0 + 0.267 822 265 624 999 943 248;
  • 5) 0.267 822 265 624 999 943 248 × 2 = 0 + 0.535 644 531 249 999 886 496;
  • 6) 0.535 644 531 249 999 886 496 × 2 = 1 + 0.071 289 062 499 999 772 992;
  • 7) 0.071 289 062 499 999 772 992 × 2 = 0 + 0.142 578 124 999 999 545 984;
  • 8) 0.142 578 124 999 999 545 984 × 2 = 0 + 0.285 156 249 999 999 091 968;
  • 9) 0.285 156 249 999 999 091 968 × 2 = 0 + 0.570 312 499 999 998 183 936;
  • 10) 0.570 312 499 999 998 183 936 × 2 = 1 + 0.140 624 999 999 996 367 872;
  • 11) 0.140 624 999 999 996 367 872 × 2 = 0 + 0.281 249 999 999 992 735 744;
  • 12) 0.281 249 999 999 992 735 744 × 2 = 0 + 0.562 499 999 999 985 471 488;
  • 13) 0.562 499 999 999 985 471 488 × 2 = 1 + 0.124 999 999 999 970 942 976;
  • 14) 0.124 999 999 999 970 942 976 × 2 = 0 + 0.249 999 999 999 941 885 952;
  • 15) 0.249 999 999 999 941 885 952 × 2 = 0 + 0.499 999 999 999 883 771 904;
  • 16) 0.499 999 999 999 883 771 904 × 2 = 0 + 0.999 999 999 999 767 543 808;
  • 17) 0.999 999 999 999 767 543 808 × 2 = 1 + 0.999 999 999 999 535 087 616;
  • 18) 0.999 999 999 999 535 087 616 × 2 = 1 + 0.999 999 999 999 070 175 232;
  • 19) 0.999 999 999 999 070 175 232 × 2 = 1 + 0.999 999 999 998 140 350 464;
  • 20) 0.999 999 999 998 140 350 464 × 2 = 1 + 0.999 999 999 996 280 700 928;
  • 21) 0.999 999 999 996 280 700 928 × 2 = 1 + 0.999 999 999 992 561 401 856;
  • 22) 0.999 999 999 992 561 401 856 × 2 = 1 + 0.999 999 999 985 122 803 712;
  • 23) 0.999 999 999 985 122 803 712 × 2 = 1 + 0.999 999 999 970 245 607 424;
  • 24) 0.999 999 999 970 245 607 424 × 2 = 1 + 0.999 999 999 940 491 214 848;
  • 25) 0.999 999 999 940 491 214 848 × 2 = 1 + 0.999 999 999 880 982 429 696;
  • 26) 0.999 999 999 880 982 429 696 × 2 = 1 + 0.999 999 999 761 964 859 392;
  • 27) 0.999 999 999 761 964 859 392 × 2 = 1 + 0.999 999 999 523 929 718 784;
  • 28) 0.999 999 999 523 929 718 784 × 2 = 1 + 0.999 999 999 047 859 437 568;
  • 29) 0.999 999 999 047 859 437 568 × 2 = 1 + 0.999 999 998 095 718 875 136;
  • 30) 0.999 999 998 095 718 875 136 × 2 = 1 + 0.999 999 996 191 437 750 272;
  • 31) 0.999 999 996 191 437 750 272 × 2 = 1 + 0.999 999 992 382 875 500 544;
  • 32) 0.999 999 992 382 875 500 544 × 2 = 1 + 0.999 999 984 765 751 001 088;
  • 33) 0.999 999 984 765 751 001 088 × 2 = 1 + 0.999 999 969 531 502 002 176;
  • 34) 0.999 999 969 531 502 002 176 × 2 = 1 + 0.999 999 939 063 004 004 352;
  • 35) 0.999 999 939 063 004 004 352 × 2 = 1 + 0.999 999 878 126 008 008 704;
  • 36) 0.999 999 878 126 008 008 704 × 2 = 1 + 0.999 999 756 252 016 017 408;
  • 37) 0.999 999 756 252 016 017 408 × 2 = 1 + 0.999 999 512 504 032 034 816;
  • 38) 0.999 999 512 504 032 034 816 × 2 = 1 + 0.999 999 025 008 064 069 632;
  • 39) 0.999 999 025 008 064 069 632 × 2 = 1 + 0.999 998 050 016 128 139 264;
  • 40) 0.999 998 050 016 128 139 264 × 2 = 1 + 0.999 996 100 032 256 278 528;
  • 41) 0.999 996 100 032 256 278 528 × 2 = 1 + 0.999 992 200 064 512 557 056;
  • 42) 0.999 992 200 064 512 557 056 × 2 = 1 + 0.999 984 400 129 025 114 112;
  • 43) 0.999 984 400 129 025 114 112 × 2 = 1 + 0.999 968 800 258 050 228 224;
  • 44) 0.999 968 800 258 050 228 224 × 2 = 1 + 0.999 937 600 516 100 456 448;
  • 45) 0.999 937 600 516 100 456 448 × 2 = 1 + 0.999 875 201 032 200 912 896;
  • 46) 0.999 875 201 032 200 912 896 × 2 = 1 + 0.999 750 402 064 401 825 792;
  • 47) 0.999 750 402 064 401 825 792 × 2 = 1 + 0.999 500 804 128 803 651 584;
  • 48) 0.999 500 804 128 803 651 584 × 2 = 1 + 0.999 001 608 257 607 303 168;
  • 49) 0.999 001 608 257 607 303 168 × 2 = 1 + 0.998 003 216 515 214 606 336;
  • 50) 0.998 003 216 515 214 606 336 × 2 = 1 + 0.996 006 433 030 429 212 672;
  • 51) 0.996 006 433 030 429 212 672 × 2 = 1 + 0.992 012 866 060 858 425 344;
  • 52) 0.992 012 866 060 858 425 344 × 2 = 1 + 0.984 025 732 121 716 850 688;
  • 53) 0.984 025 732 121 716 850 688 × 2 = 1 + 0.968 051 464 243 433 701 376;
  • 54) 0.968 051 464 243 433 701 376 × 2 = 1 + 0.936 102 928 486 867 402 752;
  • 55) 0.936 102 928 486 867 402 752 × 2 = 1 + 0.872 205 856 973 734 805 504;
  • 56) 0.872 205 856 973 734 805 504 × 2 = 1 + 0.744 411 713 947 469 611 008;
  • 57) 0.744 411 713 947 469 611 008 × 2 = 1 + 0.488 823 427 894 939 222 016;
  • 58) 0.488 823 427 894 939 222 016 × 2 = 0 + 0.977 646 855 789 878 444 032;

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:


0.016 738 891 601 562 496 453(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 453(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 453(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 453 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 1110

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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