-0.000 282 005 914 428 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 428(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
-0.000 282 005 914 428(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:

|-0.000 282 005 914 428| = 0.000 282 005 914 428


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: 0.000 282 005 914 428.

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) 0.000 282 005 914 428 × 2 = 0 + 0.000 564 011 828 856;
  • 2) 0.000 564 011 828 856 × 2 = 0 + 0.001 128 023 657 712;
  • 3) 0.001 128 023 657 712 × 2 = 0 + 0.002 256 047 315 424;
  • 4) 0.002 256 047 315 424 × 2 = 0 + 0.004 512 094 630 848;
  • 5) 0.004 512 094 630 848 × 2 = 0 + 0.009 024 189 261 696;
  • 6) 0.009 024 189 261 696 × 2 = 0 + 0.018 048 378 523 392;
  • 7) 0.018 048 378 523 392 × 2 = 0 + 0.036 096 757 046 784;
  • 8) 0.036 096 757 046 784 × 2 = 0 + 0.072 193 514 093 568;
  • 9) 0.072 193 514 093 568 × 2 = 0 + 0.144 387 028 187 136;
  • 10) 0.144 387 028 187 136 × 2 = 0 + 0.288 774 056 374 272;
  • 11) 0.288 774 056 374 272 × 2 = 0 + 0.577 548 112 748 544;
  • 12) 0.577 548 112 748 544 × 2 = 1 + 0.155 096 225 497 088;
  • 13) 0.155 096 225 497 088 × 2 = 0 + 0.310 192 450 994 176;
  • 14) 0.310 192 450 994 176 × 2 = 0 + 0.620 384 901 988 352;
  • 15) 0.620 384 901 988 352 × 2 = 1 + 0.240 769 803 976 704;
  • 16) 0.240 769 803 976 704 × 2 = 0 + 0.481 539 607 953 408;
  • 17) 0.481 539 607 953 408 × 2 = 0 + 0.963 079 215 906 816;
  • 18) 0.963 079 215 906 816 × 2 = 1 + 0.926 158 431 813 632;
  • 19) 0.926 158 431 813 632 × 2 = 1 + 0.852 316 863 627 264;
  • 20) 0.852 316 863 627 264 × 2 = 1 + 0.704 633 727 254 528;
  • 21) 0.704 633 727 254 528 × 2 = 1 + 0.409 267 454 509 056;
  • 22) 0.409 267 454 509 056 × 2 = 0 + 0.818 534 909 018 112;
  • 23) 0.818 534 909 018 112 × 2 = 1 + 0.637 069 818 036 224;
  • 24) 0.637 069 818 036 224 × 2 = 1 + 0.274 139 636 072 448;
  • 25) 0.274 139 636 072 448 × 2 = 0 + 0.548 279 272 144 896;
  • 26) 0.548 279 272 144 896 × 2 = 1 + 0.096 558 544 289 792;
  • 27) 0.096 558 544 289 792 × 2 = 0 + 0.193 117 088 579 584;
  • 28) 0.193 117 088 579 584 × 2 = 0 + 0.386 234 177 159 168;
  • 29) 0.386 234 177 159 168 × 2 = 0 + 0.772 468 354 318 336;
  • 30) 0.772 468 354 318 336 × 2 = 1 + 0.544 936 708 636 672;
  • 31) 0.544 936 708 636 672 × 2 = 1 + 0.089 873 417 273 344;
  • 32) 0.089 873 417 273 344 × 2 = 0 + 0.179 746 834 546 688;
  • 33) 0.179 746 834 546 688 × 2 = 0 + 0.359 493 669 093 376;
  • 34) 0.359 493 669 093 376 × 2 = 0 + 0.718 987 338 186 752;
  • 35) 0.718 987 338 186 752 × 2 = 1 + 0.437 974 676 373 504;
  • 36) 0.437 974 676 373 504 × 2 = 0 + 0.875 949 352 747 008;
  • 37) 0.875 949 352 747 008 × 2 = 1 + 0.751 898 705 494 016;
  • 38) 0.751 898 705 494 016 × 2 = 1 + 0.503 797 410 988 032;
  • 39) 0.503 797 410 988 032 × 2 = 1 + 0.007 594 821 976 064;
  • 40) 0.007 594 821 976 064 × 2 = 0 + 0.015 189 643 952 128;
  • 41) 0.015 189 643 952 128 × 2 = 0 + 0.030 379 287 904 256;
  • 42) 0.030 379 287 904 256 × 2 = 0 + 0.060 758 575 808 512;
  • 43) 0.060 758 575 808 512 × 2 = 0 + 0.121 517 151 617 024;
  • 44) 0.121 517 151 617 024 × 2 = 0 + 0.243 034 303 234 048;
  • 45) 0.243 034 303 234 048 × 2 = 0 + 0.486 068 606 468 096;
  • 46) 0.486 068 606 468 096 × 2 = 0 + 0.972 137 212 936 192;
  • 47) 0.972 137 212 936 192 × 2 = 1 + 0.944 274 425 872 384;
  • 48) 0.944 274 425 872 384 × 2 = 1 + 0.888 548 851 744 768;
  • 49) 0.888 548 851 744 768 × 2 = 1 + 0.777 097 703 489 536;
  • 50) 0.777 097 703 489 536 × 2 = 1 + 0.554 195 406 979 072;
  • 51) 0.554 195 406 979 072 × 2 = 1 + 0.108 390 813 958 144;
  • 52) 0.108 390 813 958 144 × 2 = 0 + 0.216 781 627 916 288;
  • 53) 0.216 781 627 916 288 × 2 = 0 + 0.433 563 255 832 576;
  • 54) 0.433 563 255 832 576 × 2 = 0 + 0.867 126 511 665 152;
  • 55) 0.867 126 511 665 152 × 2 = 1 + 0.734 253 023 330 304;
  • 56) 0.734 253 023 330 304 × 2 = 1 + 0.468 506 046 660 608;
  • 57) 0.468 506 046 660 608 × 2 = 0 + 0.937 012 093 321 216;
  • 58) 0.937 012 093 321 216 × 2 = 1 + 0.874 024 186 642 432;
  • 59) 0.874 024 186 642 432 × 2 = 1 + 0.748 048 373 284 864;
  • 60) 0.748 048 373 284 864 × 2 = 1 + 0.496 096 746 569 728;
  • 61) 0.496 096 746 569 728 × 2 = 0 + 0.992 193 493 139 456;
  • 62) 0.992 193 493 139 456 × 2 = 1 + 0.984 386 986 278 912;
  • 63) 0.984 386 986 278 912 × 2 = 1 + 0.968 773 972 557 824;
  • 64) 0.968 773 972 557 824 × 2 = 1 + 0.937 547 945 115 648;

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:


0.000 282 005 914 428(10) =

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

6. Positive number before normalization:

0.000 282 005 914 428(10) =

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


0.000 282 005 914 428(10) =

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

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

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


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 1110 0000 0011 1110 0011 0111 0111 =

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


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 1110 0000 0011 1110 0011 0111 0111


Decimal number -0.000 282 005 914 428 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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