-0.000 282 005 900 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 900 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
-0.000 282 005 900 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:

|-0.000 282 005 900 5| = 0.000 282 005 900 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: 0.000 282 005 900 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) 0.000 282 005 900 5 × 2 = 0 + 0.000 564 011 801;
  • 2) 0.000 564 011 801 × 2 = 0 + 0.001 128 023 602;
  • 3) 0.001 128 023 602 × 2 = 0 + 0.002 256 047 204;
  • 4) 0.002 256 047 204 × 2 = 0 + 0.004 512 094 408;
  • 5) 0.004 512 094 408 × 2 = 0 + 0.009 024 188 816;
  • 6) 0.009 024 188 816 × 2 = 0 + 0.018 048 377 632;
  • 7) 0.018 048 377 632 × 2 = 0 + 0.036 096 755 264;
  • 8) 0.036 096 755 264 × 2 = 0 + 0.072 193 510 528;
  • 9) 0.072 193 510 528 × 2 = 0 + 0.144 387 021 056;
  • 10) 0.144 387 021 056 × 2 = 0 + 0.288 774 042 112;
  • 11) 0.288 774 042 112 × 2 = 0 + 0.577 548 084 224;
  • 12) 0.577 548 084 224 × 2 = 1 + 0.155 096 168 448;
  • 13) 0.155 096 168 448 × 2 = 0 + 0.310 192 336 896;
  • 14) 0.310 192 336 896 × 2 = 0 + 0.620 384 673 792;
  • 15) 0.620 384 673 792 × 2 = 1 + 0.240 769 347 584;
  • 16) 0.240 769 347 584 × 2 = 0 + 0.481 538 695 168;
  • 17) 0.481 538 695 168 × 2 = 0 + 0.963 077 390 336;
  • 18) 0.963 077 390 336 × 2 = 1 + 0.926 154 780 672;
  • 19) 0.926 154 780 672 × 2 = 1 + 0.852 309 561 344;
  • 20) 0.852 309 561 344 × 2 = 1 + 0.704 619 122 688;
  • 21) 0.704 619 122 688 × 2 = 1 + 0.409 238 245 376;
  • 22) 0.409 238 245 376 × 2 = 0 + 0.818 476 490 752;
  • 23) 0.818 476 490 752 × 2 = 1 + 0.636 952 981 504;
  • 24) 0.636 952 981 504 × 2 = 1 + 0.273 905 963 008;
  • 25) 0.273 905 963 008 × 2 = 0 + 0.547 811 926 016;
  • 26) 0.547 811 926 016 × 2 = 1 + 0.095 623 852 032;
  • 27) 0.095 623 852 032 × 2 = 0 + 0.191 247 704 064;
  • 28) 0.191 247 704 064 × 2 = 0 + 0.382 495 408 128;
  • 29) 0.382 495 408 128 × 2 = 0 + 0.764 990 816 256;
  • 30) 0.764 990 816 256 × 2 = 1 + 0.529 981 632 512;
  • 31) 0.529 981 632 512 × 2 = 1 + 0.059 963 265 024;
  • 32) 0.059 963 265 024 × 2 = 0 + 0.119 926 530 048;
  • 33) 0.119 926 530 048 × 2 = 0 + 0.239 853 060 096;
  • 34) 0.239 853 060 096 × 2 = 0 + 0.479 706 120 192;
  • 35) 0.479 706 120 192 × 2 = 0 + 0.959 412 240 384;
  • 36) 0.959 412 240 384 × 2 = 1 + 0.918 824 480 768;
  • 37) 0.918 824 480 768 × 2 = 1 + 0.837 648 961 536;
  • 38) 0.837 648 961 536 × 2 = 1 + 0.675 297 923 072;
  • 39) 0.675 297 923 072 × 2 = 1 + 0.350 595 846 144;
  • 40) 0.350 595 846 144 × 2 = 0 + 0.701 191 692 288;
  • 41) 0.701 191 692 288 × 2 = 1 + 0.402 383 384 576;
  • 42) 0.402 383 384 576 × 2 = 0 + 0.804 766 769 152;
  • 43) 0.804 766 769 152 × 2 = 1 + 0.609 533 538 304;
  • 44) 0.609 533 538 304 × 2 = 1 + 0.219 067 076 608;
  • 45) 0.219 067 076 608 × 2 = 0 + 0.438 134 153 216;
  • 46) 0.438 134 153 216 × 2 = 0 + 0.876 268 306 432;
  • 47) 0.876 268 306 432 × 2 = 1 + 0.752 536 612 864;
  • 48) 0.752 536 612 864 × 2 = 1 + 0.505 073 225 728;
  • 49) 0.505 073 225 728 × 2 = 1 + 0.010 146 451 456;
  • 50) 0.010 146 451 456 × 2 = 0 + 0.020 292 902 912;
  • 51) 0.020 292 902 912 × 2 = 0 + 0.040 585 805 824;
  • 52) 0.040 585 805 824 × 2 = 0 + 0.081 171 611 648;
  • 53) 0.081 171 611 648 × 2 = 0 + 0.162 343 223 296;
  • 54) 0.162 343 223 296 × 2 = 0 + 0.324 686 446 592;
  • 55) 0.324 686 446 592 × 2 = 0 + 0.649 372 893 184;
  • 56) 0.649 372 893 184 × 2 = 1 + 0.298 745 786 368;
  • 57) 0.298 745 786 368 × 2 = 0 + 0.597 491 572 736;
  • 58) 0.597 491 572 736 × 2 = 1 + 0.194 983 145 472;
  • 59) 0.194 983 145 472 × 2 = 0 + 0.389 966 290 944;
  • 60) 0.389 966 290 944 × 2 = 0 + 0.779 932 581 888;
  • 61) 0.779 932 581 888 × 2 = 1 + 0.559 865 163 776;
  • 62) 0.559 865 163 776 × 2 = 1 + 0.119 730 327 552;
  • 63) 0.119 730 327 552 × 2 = 0 + 0.239 460 655 104;
  • 64) 0.239 460 655 104 × 2 = 0 + 0.478 921 310 208;

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:


0.000 282 005 900 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100(2)

6. Positive number before normalization:

0.000 282 005 900 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100(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:


0.000 282 005 900 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100(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 0001 1110 1011 0011 1000 0001 0100 1100


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 0001 1110 1011 0011 1000 0001 0100 1100 =

0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100


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 0001 1110 1011 0011 1000 0001 0100 1100


Decimal number -0.000 282 005 900 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1110 1011 0011 1000 0001 0100 1100

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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