-0.000 282 005 914 372 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 372(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
-0.000 282 005 914 372(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:

|-0.000 282 005 914 372| = 0.000 282 005 914 372


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: 0.000 282 005 914 372.

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) 0.000 282 005 914 372 × 2 = 0 + 0.000 564 011 828 744;
  • 2) 0.000 564 011 828 744 × 2 = 0 + 0.001 128 023 657 488;
  • 3) 0.001 128 023 657 488 × 2 = 0 + 0.002 256 047 314 976;
  • 4) 0.002 256 047 314 976 × 2 = 0 + 0.004 512 094 629 952;
  • 5) 0.004 512 094 629 952 × 2 = 0 + 0.009 024 189 259 904;
  • 6) 0.009 024 189 259 904 × 2 = 0 + 0.018 048 378 519 808;
  • 7) 0.018 048 378 519 808 × 2 = 0 + 0.036 096 757 039 616;
  • 8) 0.036 096 757 039 616 × 2 = 0 + 0.072 193 514 079 232;
  • 9) 0.072 193 514 079 232 × 2 = 0 + 0.144 387 028 158 464;
  • 10) 0.144 387 028 158 464 × 2 = 0 + 0.288 774 056 316 928;
  • 11) 0.288 774 056 316 928 × 2 = 0 + 0.577 548 112 633 856;
  • 12) 0.577 548 112 633 856 × 2 = 1 + 0.155 096 225 267 712;
  • 13) 0.155 096 225 267 712 × 2 = 0 + 0.310 192 450 535 424;
  • 14) 0.310 192 450 535 424 × 2 = 0 + 0.620 384 901 070 848;
  • 15) 0.620 384 901 070 848 × 2 = 1 + 0.240 769 802 141 696;
  • 16) 0.240 769 802 141 696 × 2 = 0 + 0.481 539 604 283 392;
  • 17) 0.481 539 604 283 392 × 2 = 0 + 0.963 079 208 566 784;
  • 18) 0.963 079 208 566 784 × 2 = 1 + 0.926 158 417 133 568;
  • 19) 0.926 158 417 133 568 × 2 = 1 + 0.852 316 834 267 136;
  • 20) 0.852 316 834 267 136 × 2 = 1 + 0.704 633 668 534 272;
  • 21) 0.704 633 668 534 272 × 2 = 1 + 0.409 267 337 068 544;
  • 22) 0.409 267 337 068 544 × 2 = 0 + 0.818 534 674 137 088;
  • 23) 0.818 534 674 137 088 × 2 = 1 + 0.637 069 348 274 176;
  • 24) 0.637 069 348 274 176 × 2 = 1 + 0.274 138 696 548 352;
  • 25) 0.274 138 696 548 352 × 2 = 0 + 0.548 277 393 096 704;
  • 26) 0.548 277 393 096 704 × 2 = 1 + 0.096 554 786 193 408;
  • 27) 0.096 554 786 193 408 × 2 = 0 + 0.193 109 572 386 816;
  • 28) 0.193 109 572 386 816 × 2 = 0 + 0.386 219 144 773 632;
  • 29) 0.386 219 144 773 632 × 2 = 0 + 0.772 438 289 547 264;
  • 30) 0.772 438 289 547 264 × 2 = 1 + 0.544 876 579 094 528;
  • 31) 0.544 876 579 094 528 × 2 = 1 + 0.089 753 158 189 056;
  • 32) 0.089 753 158 189 056 × 2 = 0 + 0.179 506 316 378 112;
  • 33) 0.179 506 316 378 112 × 2 = 0 + 0.359 012 632 756 224;
  • 34) 0.359 012 632 756 224 × 2 = 0 + 0.718 025 265 512 448;
  • 35) 0.718 025 265 512 448 × 2 = 1 + 0.436 050 531 024 896;
  • 36) 0.436 050 531 024 896 × 2 = 0 + 0.872 101 062 049 792;
  • 37) 0.872 101 062 049 792 × 2 = 1 + 0.744 202 124 099 584;
  • 38) 0.744 202 124 099 584 × 2 = 1 + 0.488 404 248 199 168;
  • 39) 0.488 404 248 199 168 × 2 = 0 + 0.976 808 496 398 336;
  • 40) 0.976 808 496 398 336 × 2 = 1 + 0.953 616 992 796 672;
  • 41) 0.953 616 992 796 672 × 2 = 1 + 0.907 233 985 593 344;
  • 42) 0.907 233 985 593 344 × 2 = 1 + 0.814 467 971 186 688;
  • 43) 0.814 467 971 186 688 × 2 = 1 + 0.628 935 942 373 376;
  • 44) 0.628 935 942 373 376 × 2 = 1 + 0.257 871 884 746 752;
  • 45) 0.257 871 884 746 752 × 2 = 0 + 0.515 743 769 493 504;
  • 46) 0.515 743 769 493 504 × 2 = 1 + 0.031 487 538 987 008;
  • 47) 0.031 487 538 987 008 × 2 = 0 + 0.062 975 077 974 016;
  • 48) 0.062 975 077 974 016 × 2 = 0 + 0.125 950 155 948 032;
  • 49) 0.125 950 155 948 032 × 2 = 0 + 0.251 900 311 896 064;
  • 50) 0.251 900 311 896 064 × 2 = 0 + 0.503 800 623 792 128;
  • 51) 0.503 800 623 792 128 × 2 = 1 + 0.007 601 247 584 256;
  • 52) 0.007 601 247 584 256 × 2 = 0 + 0.015 202 495 168 512;
  • 53) 0.015 202 495 168 512 × 2 = 0 + 0.030 404 990 337 024;
  • 54) 0.030 404 990 337 024 × 2 = 0 + 0.060 809 980 674 048;
  • 55) 0.060 809 980 674 048 × 2 = 0 + 0.121 619 961 348 096;
  • 56) 0.121 619 961 348 096 × 2 = 0 + 0.243 239 922 696 192;
  • 57) 0.243 239 922 696 192 × 2 = 0 + 0.486 479 845 392 384;
  • 58) 0.486 479 845 392 384 × 2 = 0 + 0.972 959 690 784 768;
  • 59) 0.972 959 690 784 768 × 2 = 1 + 0.945 919 381 569 536;
  • 60) 0.945 919 381 569 536 × 2 = 1 + 0.891 838 763 139 072;
  • 61) 0.891 838 763 139 072 × 2 = 1 + 0.783 677 526 278 144;
  • 62) 0.783 677 526 278 144 × 2 = 1 + 0.567 355 052 556 288;
  • 63) 0.567 355 052 556 288 × 2 = 1 + 0.134 710 105 112 576;
  • 64) 0.134 710 105 112 576 × 2 = 0 + 0.269 420 210 225 152;

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:


0.000 282 005 914 372(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110(2)

6. Positive number before normalization:

0.000 282 005 914 372(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110(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:


0.000 282 005 914 372(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110(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 1101 1111 0100 0010 0000 0011 1110


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 1101 1111 0100 0010 0000 0011 1110 =

0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110


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 1101 1111 0100 0010 0000 0011 1110


Decimal number -0.000 282 005 914 372 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0100 0010 0000 0011 1110

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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