-0.000 282 005 933 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 933(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
-0.000 282 005 933(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:

|-0.000 282 005 933| = 0.000 282 005 933


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: 0.000 282 005 933.

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) 0.000 282 005 933 × 2 = 0 + 0.000 564 011 866;
  • 2) 0.000 564 011 866 × 2 = 0 + 0.001 128 023 732;
  • 3) 0.001 128 023 732 × 2 = 0 + 0.002 256 047 464;
  • 4) 0.002 256 047 464 × 2 = 0 + 0.004 512 094 928;
  • 5) 0.004 512 094 928 × 2 = 0 + 0.009 024 189 856;
  • 6) 0.009 024 189 856 × 2 = 0 + 0.018 048 379 712;
  • 7) 0.018 048 379 712 × 2 = 0 + 0.036 096 759 424;
  • 8) 0.036 096 759 424 × 2 = 0 + 0.072 193 518 848;
  • 9) 0.072 193 518 848 × 2 = 0 + 0.144 387 037 696;
  • 10) 0.144 387 037 696 × 2 = 0 + 0.288 774 075 392;
  • 11) 0.288 774 075 392 × 2 = 0 + 0.577 548 150 784;
  • 12) 0.577 548 150 784 × 2 = 1 + 0.155 096 301 568;
  • 13) 0.155 096 301 568 × 2 = 0 + 0.310 192 603 136;
  • 14) 0.310 192 603 136 × 2 = 0 + 0.620 385 206 272;
  • 15) 0.620 385 206 272 × 2 = 1 + 0.240 770 412 544;
  • 16) 0.240 770 412 544 × 2 = 0 + 0.481 540 825 088;
  • 17) 0.481 540 825 088 × 2 = 0 + 0.963 081 650 176;
  • 18) 0.963 081 650 176 × 2 = 1 + 0.926 163 300 352;
  • 19) 0.926 163 300 352 × 2 = 1 + 0.852 326 600 704;
  • 20) 0.852 326 600 704 × 2 = 1 + 0.704 653 201 408;
  • 21) 0.704 653 201 408 × 2 = 1 + 0.409 306 402 816;
  • 22) 0.409 306 402 816 × 2 = 0 + 0.818 612 805 632;
  • 23) 0.818 612 805 632 × 2 = 1 + 0.637 225 611 264;
  • 24) 0.637 225 611 264 × 2 = 1 + 0.274 451 222 528;
  • 25) 0.274 451 222 528 × 2 = 0 + 0.548 902 445 056;
  • 26) 0.548 902 445 056 × 2 = 1 + 0.097 804 890 112;
  • 27) 0.097 804 890 112 × 2 = 0 + 0.195 609 780 224;
  • 28) 0.195 609 780 224 × 2 = 0 + 0.391 219 560 448;
  • 29) 0.391 219 560 448 × 2 = 0 + 0.782 439 120 896;
  • 30) 0.782 439 120 896 × 2 = 1 + 0.564 878 241 792;
  • 31) 0.564 878 241 792 × 2 = 1 + 0.129 756 483 584;
  • 32) 0.129 756 483 584 × 2 = 0 + 0.259 512 967 168;
  • 33) 0.259 512 967 168 × 2 = 0 + 0.519 025 934 336;
  • 34) 0.519 025 934 336 × 2 = 1 + 0.038 051 868 672;
  • 35) 0.038 051 868 672 × 2 = 0 + 0.076 103 737 344;
  • 36) 0.076 103 737 344 × 2 = 0 + 0.152 207 474 688;
  • 37) 0.152 207 474 688 × 2 = 0 + 0.304 414 949 376;
  • 38) 0.304 414 949 376 × 2 = 0 + 0.608 829 898 752;
  • 39) 0.608 829 898 752 × 2 = 1 + 0.217 659 797 504;
  • 40) 0.217 659 797 504 × 2 = 0 + 0.435 319 595 008;
  • 41) 0.435 319 595 008 × 2 = 0 + 0.870 639 190 016;
  • 42) 0.870 639 190 016 × 2 = 1 + 0.741 278 380 032;
  • 43) 0.741 278 380 032 × 2 = 1 + 0.482 556 760 064;
  • 44) 0.482 556 760 064 × 2 = 0 + 0.965 113 520 128;
  • 45) 0.965 113 520 128 × 2 = 1 + 0.930 227 040 256;
  • 46) 0.930 227 040 256 × 2 = 1 + 0.860 454 080 512;
  • 47) 0.860 454 080 512 × 2 = 1 + 0.720 908 161 024;
  • 48) 0.720 908 161 024 × 2 = 1 + 0.441 816 322 048;
  • 49) 0.441 816 322 048 × 2 = 0 + 0.883 632 644 096;
  • 50) 0.883 632 644 096 × 2 = 1 + 0.767 265 288 192;
  • 51) 0.767 265 288 192 × 2 = 1 + 0.534 530 576 384;
  • 52) 0.534 530 576 384 × 2 = 1 + 0.069 061 152 768;
  • 53) 0.069 061 152 768 × 2 = 0 + 0.138 122 305 536;
  • 54) 0.138 122 305 536 × 2 = 0 + 0.276 244 611 072;
  • 55) 0.276 244 611 072 × 2 = 0 + 0.552 489 222 144;
  • 56) 0.552 489 222 144 × 2 = 1 + 0.104 978 444 288;
  • 57) 0.104 978 444 288 × 2 = 0 + 0.209 956 888 576;
  • 58) 0.209 956 888 576 × 2 = 0 + 0.419 913 777 152;
  • 59) 0.419 913 777 152 × 2 = 0 + 0.839 827 554 304;
  • 60) 0.839 827 554 304 × 2 = 1 + 0.679 655 108 608;
  • 61) 0.679 655 108 608 × 2 = 1 + 0.359 310 217 216;
  • 62) 0.359 310 217 216 × 2 = 0 + 0.718 620 434 432;
  • 63) 0.718 620 434 432 × 2 = 1 + 0.437 240 868 864;
  • 64) 0.437 240 868 864 × 2 = 0 + 0.874 481 737 728;

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:


0.000 282 005 933(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0010 0110 1111 0111 0001 0001 1010(2)

6. Positive number before normalization:

0.000 282 005 933(10) =

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


0.000 282 005 933(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0010 0110 1111 0111 0001 0001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0010 0110 1111 0111 0001 0001 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0100 0010 0110 1111 0111 0001 0001 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 0110 0100 0010 0110 1111 0111 0001 0001 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 0110 0100 0010 0110 1111 0111 0001 0001 1010 =

0010 0111 1011 0100 0110 0100 0010 0110 1111 0111 0001 0001 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 0110 0100 0010 0110 1111 0111 0001 0001 1010


Decimal number -0.000 282 005 933 converted to 64 bit double precision IEEE 754 binary floating point representation:

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