-0.000 282 006 357 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 357(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
-0.000 282 006 357(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:

|-0.000 282 006 357| = 0.000 282 006 357


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: 0.000 282 006 357.

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) 0.000 282 006 357 × 2 = 0 + 0.000 564 012 714;
  • 2) 0.000 564 012 714 × 2 = 0 + 0.001 128 025 428;
  • 3) 0.001 128 025 428 × 2 = 0 + 0.002 256 050 856;
  • 4) 0.002 256 050 856 × 2 = 0 + 0.004 512 101 712;
  • 5) 0.004 512 101 712 × 2 = 0 + 0.009 024 203 424;
  • 6) 0.009 024 203 424 × 2 = 0 + 0.018 048 406 848;
  • 7) 0.018 048 406 848 × 2 = 0 + 0.036 096 813 696;
  • 8) 0.036 096 813 696 × 2 = 0 + 0.072 193 627 392;
  • 9) 0.072 193 627 392 × 2 = 0 + 0.144 387 254 784;
  • 10) 0.144 387 254 784 × 2 = 0 + 0.288 774 509 568;
  • 11) 0.288 774 509 568 × 2 = 0 + 0.577 549 019 136;
  • 12) 0.577 549 019 136 × 2 = 1 + 0.155 098 038 272;
  • 13) 0.155 098 038 272 × 2 = 0 + 0.310 196 076 544;
  • 14) 0.310 196 076 544 × 2 = 0 + 0.620 392 153 088;
  • 15) 0.620 392 153 088 × 2 = 1 + 0.240 784 306 176;
  • 16) 0.240 784 306 176 × 2 = 0 + 0.481 568 612 352;
  • 17) 0.481 568 612 352 × 2 = 0 + 0.963 137 224 704;
  • 18) 0.963 137 224 704 × 2 = 1 + 0.926 274 449 408;
  • 19) 0.926 274 449 408 × 2 = 1 + 0.852 548 898 816;
  • 20) 0.852 548 898 816 × 2 = 1 + 0.705 097 797 632;
  • 21) 0.705 097 797 632 × 2 = 1 + 0.410 195 595 264;
  • 22) 0.410 195 595 264 × 2 = 0 + 0.820 391 190 528;
  • 23) 0.820 391 190 528 × 2 = 1 + 0.640 782 381 056;
  • 24) 0.640 782 381 056 × 2 = 1 + 0.281 564 762 112;
  • 25) 0.281 564 762 112 × 2 = 0 + 0.563 129 524 224;
  • 26) 0.563 129 524 224 × 2 = 1 + 0.126 259 048 448;
  • 27) 0.126 259 048 448 × 2 = 0 + 0.252 518 096 896;
  • 28) 0.252 518 096 896 × 2 = 0 + 0.505 036 193 792;
  • 29) 0.505 036 193 792 × 2 = 1 + 0.010 072 387 584;
  • 30) 0.010 072 387 584 × 2 = 0 + 0.020 144 775 168;
  • 31) 0.020 144 775 168 × 2 = 0 + 0.040 289 550 336;
  • 32) 0.040 289 550 336 × 2 = 0 + 0.080 579 100 672;
  • 33) 0.080 579 100 672 × 2 = 0 + 0.161 158 201 344;
  • 34) 0.161 158 201 344 × 2 = 0 + 0.322 316 402 688;
  • 35) 0.322 316 402 688 × 2 = 0 + 0.644 632 805 376;
  • 36) 0.644 632 805 376 × 2 = 1 + 0.289 265 610 752;
  • 37) 0.289 265 610 752 × 2 = 0 + 0.578 531 221 504;
  • 38) 0.578 531 221 504 × 2 = 1 + 0.157 062 443 008;
  • 39) 0.157 062 443 008 × 2 = 0 + 0.314 124 886 016;
  • 40) 0.314 124 886 016 × 2 = 0 + 0.628 249 772 032;
  • 41) 0.628 249 772 032 × 2 = 1 + 0.256 499 544 064;
  • 42) 0.256 499 544 064 × 2 = 0 + 0.512 999 088 128;
  • 43) 0.512 999 088 128 × 2 = 1 + 0.025 998 176 256;
  • 44) 0.025 998 176 256 × 2 = 0 + 0.051 996 352 512;
  • 45) 0.051 996 352 512 × 2 = 0 + 0.103 992 705 024;
  • 46) 0.103 992 705 024 × 2 = 0 + 0.207 985 410 048;
  • 47) 0.207 985 410 048 × 2 = 0 + 0.415 970 820 096;
  • 48) 0.415 970 820 096 × 2 = 0 + 0.831 941 640 192;
  • 49) 0.831 941 640 192 × 2 = 1 + 0.663 883 280 384;
  • 50) 0.663 883 280 384 × 2 = 1 + 0.327 766 560 768;
  • 51) 0.327 766 560 768 × 2 = 0 + 0.655 533 121 536;
  • 52) 0.655 533 121 536 × 2 = 1 + 0.311 066 243 072;
  • 53) 0.311 066 243 072 × 2 = 0 + 0.622 132 486 144;
  • 54) 0.622 132 486 144 × 2 = 1 + 0.244 264 972 288;
  • 55) 0.244 264 972 288 × 2 = 0 + 0.488 529 944 576;
  • 56) 0.488 529 944 576 × 2 = 0 + 0.977 059 889 152;
  • 57) 0.977 059 889 152 × 2 = 1 + 0.954 119 778 304;
  • 58) 0.954 119 778 304 × 2 = 1 + 0.908 239 556 608;
  • 59) 0.908 239 556 608 × 2 = 1 + 0.816 479 113 216;
  • 60) 0.816 479 113 216 × 2 = 1 + 0.632 958 226 432;
  • 61) 0.632 958 226 432 × 2 = 1 + 0.265 916 452 864;
  • 62) 0.265 916 452 864 × 2 = 0 + 0.531 832 905 728;
  • 63) 0.531 832 905 728 × 2 = 1 + 0.063 665 811 456;
  • 64) 0.063 665 811 456 × 2 = 0 + 0.127 331 622 912;

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:


0.000 282 006 357(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010(2)

6. Positive number before normalization:

0.000 282 006 357(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010(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:


0.000 282 006 357(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010(2) × 20 =

1.0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010(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 1000 0001 0100 1010 0000 1101 0100 1111 1010


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 1000 0001 0100 1010 0000 1101 0100 1111 1010 =

0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010


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 1000 0001 0100 1010 0000 1101 0100 1111 1010


Decimal number -0.000 282 006 357 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0001 0100 1010 0000 1101 0100 1111 1010

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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