-0.000 282 005 959 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 959 5(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
-0.000 282 005 959 5(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:

|-0.000 282 005 959 5| = 0.000 282 005 959 5


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: 0.000 282 005 959 5.

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) 0.000 282 005 959 5 × 2 = 0 + 0.000 564 011 919;
  • 2) 0.000 564 011 919 × 2 = 0 + 0.001 128 023 838;
  • 3) 0.001 128 023 838 × 2 = 0 + 0.002 256 047 676;
  • 4) 0.002 256 047 676 × 2 = 0 + 0.004 512 095 352;
  • 5) 0.004 512 095 352 × 2 = 0 + 0.009 024 190 704;
  • 6) 0.009 024 190 704 × 2 = 0 + 0.018 048 381 408;
  • 7) 0.018 048 381 408 × 2 = 0 + 0.036 096 762 816;
  • 8) 0.036 096 762 816 × 2 = 0 + 0.072 193 525 632;
  • 9) 0.072 193 525 632 × 2 = 0 + 0.144 387 051 264;
  • 10) 0.144 387 051 264 × 2 = 0 + 0.288 774 102 528;
  • 11) 0.288 774 102 528 × 2 = 0 + 0.577 548 205 056;
  • 12) 0.577 548 205 056 × 2 = 1 + 0.155 096 410 112;
  • 13) 0.155 096 410 112 × 2 = 0 + 0.310 192 820 224;
  • 14) 0.310 192 820 224 × 2 = 0 + 0.620 385 640 448;
  • 15) 0.620 385 640 448 × 2 = 1 + 0.240 771 280 896;
  • 16) 0.240 771 280 896 × 2 = 0 + 0.481 542 561 792;
  • 17) 0.481 542 561 792 × 2 = 0 + 0.963 085 123 584;
  • 18) 0.963 085 123 584 × 2 = 1 + 0.926 170 247 168;
  • 19) 0.926 170 247 168 × 2 = 1 + 0.852 340 494 336;
  • 20) 0.852 340 494 336 × 2 = 1 + 0.704 680 988 672;
  • 21) 0.704 680 988 672 × 2 = 1 + 0.409 361 977 344;
  • 22) 0.409 361 977 344 × 2 = 0 + 0.818 723 954 688;
  • 23) 0.818 723 954 688 × 2 = 1 + 0.637 447 909 376;
  • 24) 0.637 447 909 376 × 2 = 1 + 0.274 895 818 752;
  • 25) 0.274 895 818 752 × 2 = 0 + 0.549 791 637 504;
  • 26) 0.549 791 637 504 × 2 = 1 + 0.099 583 275 008;
  • 27) 0.099 583 275 008 × 2 = 0 + 0.199 166 550 016;
  • 28) 0.199 166 550 016 × 2 = 0 + 0.398 333 100 032;
  • 29) 0.398 333 100 032 × 2 = 0 + 0.796 666 200 064;
  • 30) 0.796 666 200 064 × 2 = 1 + 0.593 332 400 128;
  • 31) 0.593 332 400 128 × 2 = 1 + 0.186 664 800 256;
  • 32) 0.186 664 800 256 × 2 = 0 + 0.373 329 600 512;
  • 33) 0.373 329 600 512 × 2 = 0 + 0.746 659 201 024;
  • 34) 0.746 659 201 024 × 2 = 1 + 0.493 318 402 048;
  • 35) 0.493 318 402 048 × 2 = 0 + 0.986 636 804 096;
  • 36) 0.986 636 804 096 × 2 = 1 + 0.973 273 608 192;
  • 37) 0.973 273 608 192 × 2 = 1 + 0.946 547 216 384;
  • 38) 0.946 547 216 384 × 2 = 1 + 0.893 094 432 768;
  • 39) 0.893 094 432 768 × 2 = 1 + 0.786 188 865 536;
  • 40) 0.786 188 865 536 × 2 = 1 + 0.572 377 731 072;
  • 41) 0.572 377 731 072 × 2 = 1 + 0.144 755 462 144;
  • 42) 0.144 755 462 144 × 2 = 0 + 0.289 510 924 288;
  • 43) 0.289 510 924 288 × 2 = 0 + 0.579 021 848 576;
  • 44) 0.579 021 848 576 × 2 = 1 + 0.158 043 697 152;
  • 45) 0.158 043 697 152 × 2 = 0 + 0.316 087 394 304;
  • 46) 0.316 087 394 304 × 2 = 0 + 0.632 174 788 608;
  • 47) 0.632 174 788 608 × 2 = 1 + 0.264 349 577 216;
  • 48) 0.264 349 577 216 × 2 = 0 + 0.528 699 154 432;
  • 49) 0.528 699 154 432 × 2 = 1 + 0.057 398 308 864;
  • 50) 0.057 398 308 864 × 2 = 0 + 0.114 796 617 728;
  • 51) 0.114 796 617 728 × 2 = 0 + 0.229 593 235 456;
  • 52) 0.229 593 235 456 × 2 = 0 + 0.459 186 470 912;
  • 53) 0.459 186 470 912 × 2 = 0 + 0.918 372 941 824;
  • 54) 0.918 372 941 824 × 2 = 1 + 0.836 745 883 648;
  • 55) 0.836 745 883 648 × 2 = 1 + 0.673 491 767 296;
  • 56) 0.673 491 767 296 × 2 = 1 + 0.346 983 534 592;
  • 57) 0.346 983 534 592 × 2 = 0 + 0.693 967 069 184;
  • 58) 0.693 967 069 184 × 2 = 1 + 0.387 934 138 368;
  • 59) 0.387 934 138 368 × 2 = 0 + 0.775 868 276 736;
  • 60) 0.775 868 276 736 × 2 = 1 + 0.551 736 553 472;
  • 61) 0.551 736 553 472 × 2 = 1 + 0.103 473 106 944;
  • 62) 0.103 473 106 944 × 2 = 0 + 0.206 946 213 888;
  • 63) 0.206 946 213 888 × 2 = 0 + 0.413 892 427 776;
  • 64) 0.413 892 427 776 × 2 = 0 + 0.827 784 855 552;

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:


0.000 282 005 959 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000(2)

6. Positive number before normalization:

0.000 282 005 959 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 959 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000 =

0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000


Decimal number -0.000 282 005 959 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1111 1001 0010 1000 0111 0101 1000

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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