-0.000 282 005 969 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 969(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
-0.000 282 005 969(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:

|-0.000 282 005 969| = 0.000 282 005 969


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: 0.000 282 005 969.

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) 0.000 282 005 969 × 2 = 0 + 0.000 564 011 938;
  • 2) 0.000 564 011 938 × 2 = 0 + 0.001 128 023 876;
  • 3) 0.001 128 023 876 × 2 = 0 + 0.002 256 047 752;
  • 4) 0.002 256 047 752 × 2 = 0 + 0.004 512 095 504;
  • 5) 0.004 512 095 504 × 2 = 0 + 0.009 024 191 008;
  • 6) 0.009 024 191 008 × 2 = 0 + 0.018 048 382 016;
  • 7) 0.018 048 382 016 × 2 = 0 + 0.036 096 764 032;
  • 8) 0.036 096 764 032 × 2 = 0 + 0.072 193 528 064;
  • 9) 0.072 193 528 064 × 2 = 0 + 0.144 387 056 128;
  • 10) 0.144 387 056 128 × 2 = 0 + 0.288 774 112 256;
  • 11) 0.288 774 112 256 × 2 = 0 + 0.577 548 224 512;
  • 12) 0.577 548 224 512 × 2 = 1 + 0.155 096 449 024;
  • 13) 0.155 096 449 024 × 2 = 0 + 0.310 192 898 048;
  • 14) 0.310 192 898 048 × 2 = 0 + 0.620 385 796 096;
  • 15) 0.620 385 796 096 × 2 = 1 + 0.240 771 592 192;
  • 16) 0.240 771 592 192 × 2 = 0 + 0.481 543 184 384;
  • 17) 0.481 543 184 384 × 2 = 0 + 0.963 086 368 768;
  • 18) 0.963 086 368 768 × 2 = 1 + 0.926 172 737 536;
  • 19) 0.926 172 737 536 × 2 = 1 + 0.852 345 475 072;
  • 20) 0.852 345 475 072 × 2 = 1 + 0.704 690 950 144;
  • 21) 0.704 690 950 144 × 2 = 1 + 0.409 381 900 288;
  • 22) 0.409 381 900 288 × 2 = 0 + 0.818 763 800 576;
  • 23) 0.818 763 800 576 × 2 = 1 + 0.637 527 601 152;
  • 24) 0.637 527 601 152 × 2 = 1 + 0.275 055 202 304;
  • 25) 0.275 055 202 304 × 2 = 0 + 0.550 110 404 608;
  • 26) 0.550 110 404 608 × 2 = 1 + 0.100 220 809 216;
  • 27) 0.100 220 809 216 × 2 = 0 + 0.200 441 618 432;
  • 28) 0.200 441 618 432 × 2 = 0 + 0.400 883 236 864;
  • 29) 0.400 883 236 864 × 2 = 0 + 0.801 766 473 728;
  • 30) 0.801 766 473 728 × 2 = 1 + 0.603 532 947 456;
  • 31) 0.603 532 947 456 × 2 = 1 + 0.207 065 894 912;
  • 32) 0.207 065 894 912 × 2 = 0 + 0.414 131 789 824;
  • 33) 0.414 131 789 824 × 2 = 0 + 0.828 263 579 648;
  • 34) 0.828 263 579 648 × 2 = 1 + 0.656 527 159 296;
  • 35) 0.656 527 159 296 × 2 = 1 + 0.313 054 318 592;
  • 36) 0.313 054 318 592 × 2 = 0 + 0.626 108 637 184;
  • 37) 0.626 108 637 184 × 2 = 1 + 0.252 217 274 368;
  • 38) 0.252 217 274 368 × 2 = 0 + 0.504 434 548 736;
  • 39) 0.504 434 548 736 × 2 = 1 + 0.008 869 097 472;
  • 40) 0.008 869 097 472 × 2 = 0 + 0.017 738 194 944;
  • 41) 0.017 738 194 944 × 2 = 0 + 0.035 476 389 888;
  • 42) 0.035 476 389 888 × 2 = 0 + 0.070 952 779 776;
  • 43) 0.070 952 779 776 × 2 = 0 + 0.141 905 559 552;
  • 44) 0.141 905 559 552 × 2 = 0 + 0.283 811 119 104;
  • 45) 0.283 811 119 104 × 2 = 0 + 0.567 622 238 208;
  • 46) 0.567 622 238 208 × 2 = 1 + 0.135 244 476 416;
  • 47) 0.135 244 476 416 × 2 = 0 + 0.270 488 952 832;
  • 48) 0.270 488 952 832 × 2 = 0 + 0.540 977 905 664;
  • 49) 0.540 977 905 664 × 2 = 1 + 0.081 955 811 328;
  • 50) 0.081 955 811 328 × 2 = 0 + 0.163 911 622 656;
  • 51) 0.163 911 622 656 × 2 = 0 + 0.327 823 245 312;
  • 52) 0.327 823 245 312 × 2 = 0 + 0.655 646 490 624;
  • 53) 0.655 646 490 624 × 2 = 1 + 0.311 292 981 248;
  • 54) 0.311 292 981 248 × 2 = 0 + 0.622 585 962 496;
  • 55) 0.622 585 962 496 × 2 = 1 + 0.245 171 924 992;
  • 56) 0.245 171 924 992 × 2 = 0 + 0.490 343 849 984;
  • 57) 0.490 343 849 984 × 2 = 0 + 0.980 687 699 968;
  • 58) 0.980 687 699 968 × 2 = 1 + 0.961 375 399 936;
  • 59) 0.961 375 399 936 × 2 = 1 + 0.922 750 799 872;
  • 60) 0.922 750 799 872 × 2 = 1 + 0.845 501 599 744;
  • 61) 0.845 501 599 744 × 2 = 1 + 0.691 003 199 488;
  • 62) 0.691 003 199 488 × 2 = 1 + 0.382 006 398 976;
  • 63) 0.382 006 398 976 × 2 = 0 + 0.764 012 797 952;
  • 64) 0.764 012 797 952 × 2 = 1 + 0.528 025 595 904;

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:


0.000 282 005 969(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101(2)

6. Positive number before normalization:

0.000 282 005 969(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101(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:


0.000 282 005 969(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101(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 0110 1010 0000 0100 1000 1010 0111 1101


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 0110 1010 0000 0100 1000 1010 0111 1101 =

0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101


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 0110 1010 0000 0100 1000 1010 0111 1101


Decimal number -0.000 282 005 969 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0110 1010 0000 0100 1000 1010 0111 1101

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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