-0.000 282 005 908 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 908(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
-0.000 282 005 908(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:

|-0.000 282 005 908| = 0.000 282 005 908


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: 0.000 282 005 908.

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) 0.000 282 005 908 × 2 = 0 + 0.000 564 011 816;
  • 2) 0.000 564 011 816 × 2 = 0 + 0.001 128 023 632;
  • 3) 0.001 128 023 632 × 2 = 0 + 0.002 256 047 264;
  • 4) 0.002 256 047 264 × 2 = 0 + 0.004 512 094 528;
  • 5) 0.004 512 094 528 × 2 = 0 + 0.009 024 189 056;
  • 6) 0.009 024 189 056 × 2 = 0 + 0.018 048 378 112;
  • 7) 0.018 048 378 112 × 2 = 0 + 0.036 096 756 224;
  • 8) 0.036 096 756 224 × 2 = 0 + 0.072 193 512 448;
  • 9) 0.072 193 512 448 × 2 = 0 + 0.144 387 024 896;
  • 10) 0.144 387 024 896 × 2 = 0 + 0.288 774 049 792;
  • 11) 0.288 774 049 792 × 2 = 0 + 0.577 548 099 584;
  • 12) 0.577 548 099 584 × 2 = 1 + 0.155 096 199 168;
  • 13) 0.155 096 199 168 × 2 = 0 + 0.310 192 398 336;
  • 14) 0.310 192 398 336 × 2 = 0 + 0.620 384 796 672;
  • 15) 0.620 384 796 672 × 2 = 1 + 0.240 769 593 344;
  • 16) 0.240 769 593 344 × 2 = 0 + 0.481 539 186 688;
  • 17) 0.481 539 186 688 × 2 = 0 + 0.963 078 373 376;
  • 18) 0.963 078 373 376 × 2 = 1 + 0.926 156 746 752;
  • 19) 0.926 156 746 752 × 2 = 1 + 0.852 313 493 504;
  • 20) 0.852 313 493 504 × 2 = 1 + 0.704 626 987 008;
  • 21) 0.704 626 987 008 × 2 = 1 + 0.409 253 974 016;
  • 22) 0.409 253 974 016 × 2 = 0 + 0.818 507 948 032;
  • 23) 0.818 507 948 032 × 2 = 1 + 0.637 015 896 064;
  • 24) 0.637 015 896 064 × 2 = 1 + 0.274 031 792 128;
  • 25) 0.274 031 792 128 × 2 = 0 + 0.548 063 584 256;
  • 26) 0.548 063 584 256 × 2 = 1 + 0.096 127 168 512;
  • 27) 0.096 127 168 512 × 2 = 0 + 0.192 254 337 024;
  • 28) 0.192 254 337 024 × 2 = 0 + 0.384 508 674 048;
  • 29) 0.384 508 674 048 × 2 = 0 + 0.769 017 348 096;
  • 30) 0.769 017 348 096 × 2 = 1 + 0.538 034 696 192;
  • 31) 0.538 034 696 192 × 2 = 1 + 0.076 069 392 384;
  • 32) 0.076 069 392 384 × 2 = 0 + 0.152 138 784 768;
  • 33) 0.152 138 784 768 × 2 = 0 + 0.304 277 569 536;
  • 34) 0.304 277 569 536 × 2 = 0 + 0.608 555 139 072;
  • 35) 0.608 555 139 072 × 2 = 1 + 0.217 110 278 144;
  • 36) 0.217 110 278 144 × 2 = 0 + 0.434 220 556 288;
  • 37) 0.434 220 556 288 × 2 = 0 + 0.868 441 112 576;
  • 38) 0.868 441 112 576 × 2 = 1 + 0.736 882 225 152;
  • 39) 0.736 882 225 152 × 2 = 1 + 0.473 764 450 304;
  • 40) 0.473 764 450 304 × 2 = 0 + 0.947 528 900 608;
  • 41) 0.947 528 900 608 × 2 = 1 + 0.895 057 801 216;
  • 42) 0.895 057 801 216 × 2 = 1 + 0.790 115 602 432;
  • 43) 0.790 115 602 432 × 2 = 1 + 0.580 231 204 864;
  • 44) 0.580 231 204 864 × 2 = 1 + 0.160 462 409 728;
  • 45) 0.160 462 409 728 × 2 = 0 + 0.320 924 819 456;
  • 46) 0.320 924 819 456 × 2 = 0 + 0.641 849 638 912;
  • 47) 0.641 849 638 912 × 2 = 1 + 0.283 699 277 824;
  • 48) 0.283 699 277 824 × 2 = 0 + 0.567 398 555 648;
  • 49) 0.567 398 555 648 × 2 = 1 + 0.134 797 111 296;
  • 50) 0.134 797 111 296 × 2 = 0 + 0.269 594 222 592;
  • 51) 0.269 594 222 592 × 2 = 0 + 0.539 188 445 184;
  • 52) 0.539 188 445 184 × 2 = 1 + 0.078 376 890 368;
  • 53) 0.078 376 890 368 × 2 = 0 + 0.156 753 780 736;
  • 54) 0.156 753 780 736 × 2 = 0 + 0.313 507 561 472;
  • 55) 0.313 507 561 472 × 2 = 0 + 0.627 015 122 944;
  • 56) 0.627 015 122 944 × 2 = 1 + 0.254 030 245 888;
  • 57) 0.254 030 245 888 × 2 = 0 + 0.508 060 491 776;
  • 58) 0.508 060 491 776 × 2 = 1 + 0.016 120 983 552;
  • 59) 0.016 120 983 552 × 2 = 0 + 0.032 241 967 104;
  • 60) 0.032 241 967 104 × 2 = 0 + 0.064 483 934 208;
  • 61) 0.064 483 934 208 × 2 = 0 + 0.128 967 868 416;
  • 62) 0.128 967 868 416 × 2 = 0 + 0.257 935 736 832;
  • 63) 0.257 935 736 832 × 2 = 0 + 0.515 871 473 664;
  • 64) 0.515 871 473 664 × 2 = 1 + 0.031 742 947 328;

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:


0.000 282 005 908(10) =

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

6. Positive number before normalization:

0.000 282 005 908(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0110 1111 0010 1001 0001 0100 0001(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:


0.000 282 005 908(10) =

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

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

1.0010 0111 1011 0100 0110 0010 0110 1111 0010 1001 0001 0100 0001(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 0010 0110 1111 0010 1001 0001 0100 0001


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 0010 0110 1111 0010 1001 0001 0100 0001 =

0010 0111 1011 0100 0110 0010 0110 1111 0010 1001 0001 0100 0001


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 0010 0110 1111 0010 1001 0001 0100 0001


Decimal number -0.000 282 005 908 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0110 1111 0010 1001 0001 0100 0001

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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