-0.000 282 005 868 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 868 3(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
-0.000 282 005 868 3(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:

|-0.000 282 005 868 3| = 0.000 282 005 868 3


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: 0.000 282 005 868 3.

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) 0.000 282 005 868 3 × 2 = 0 + 0.000 564 011 736 6;
  • 2) 0.000 564 011 736 6 × 2 = 0 + 0.001 128 023 473 2;
  • 3) 0.001 128 023 473 2 × 2 = 0 + 0.002 256 046 946 4;
  • 4) 0.002 256 046 946 4 × 2 = 0 + 0.004 512 093 892 8;
  • 5) 0.004 512 093 892 8 × 2 = 0 + 0.009 024 187 785 6;
  • 6) 0.009 024 187 785 6 × 2 = 0 + 0.018 048 375 571 2;
  • 7) 0.018 048 375 571 2 × 2 = 0 + 0.036 096 751 142 4;
  • 8) 0.036 096 751 142 4 × 2 = 0 + 0.072 193 502 284 8;
  • 9) 0.072 193 502 284 8 × 2 = 0 + 0.144 387 004 569 6;
  • 10) 0.144 387 004 569 6 × 2 = 0 + 0.288 774 009 139 2;
  • 11) 0.288 774 009 139 2 × 2 = 0 + 0.577 548 018 278 4;
  • 12) 0.577 548 018 278 4 × 2 = 1 + 0.155 096 036 556 8;
  • 13) 0.155 096 036 556 8 × 2 = 0 + 0.310 192 073 113 6;
  • 14) 0.310 192 073 113 6 × 2 = 0 + 0.620 384 146 227 2;
  • 15) 0.620 384 146 227 2 × 2 = 1 + 0.240 768 292 454 4;
  • 16) 0.240 768 292 454 4 × 2 = 0 + 0.481 536 584 908 8;
  • 17) 0.481 536 584 908 8 × 2 = 0 + 0.963 073 169 817 6;
  • 18) 0.963 073 169 817 6 × 2 = 1 + 0.926 146 339 635 2;
  • 19) 0.926 146 339 635 2 × 2 = 1 + 0.852 292 679 270 4;
  • 20) 0.852 292 679 270 4 × 2 = 1 + 0.704 585 358 540 8;
  • 21) 0.704 585 358 540 8 × 2 = 1 + 0.409 170 717 081 6;
  • 22) 0.409 170 717 081 6 × 2 = 0 + 0.818 341 434 163 2;
  • 23) 0.818 341 434 163 2 × 2 = 1 + 0.636 682 868 326 4;
  • 24) 0.636 682 868 326 4 × 2 = 1 + 0.273 365 736 652 8;
  • 25) 0.273 365 736 652 8 × 2 = 0 + 0.546 731 473 305 6;
  • 26) 0.546 731 473 305 6 × 2 = 1 + 0.093 462 946 611 2;
  • 27) 0.093 462 946 611 2 × 2 = 0 + 0.186 925 893 222 4;
  • 28) 0.186 925 893 222 4 × 2 = 0 + 0.373 851 786 444 8;
  • 29) 0.373 851 786 444 8 × 2 = 0 + 0.747 703 572 889 6;
  • 30) 0.747 703 572 889 6 × 2 = 1 + 0.495 407 145 779 2;
  • 31) 0.495 407 145 779 2 × 2 = 0 + 0.990 814 291 558 4;
  • 32) 0.990 814 291 558 4 × 2 = 1 + 0.981 628 583 116 8;
  • 33) 0.981 628 583 116 8 × 2 = 1 + 0.963 257 166 233 6;
  • 34) 0.963 257 166 233 6 × 2 = 1 + 0.926 514 332 467 2;
  • 35) 0.926 514 332 467 2 × 2 = 1 + 0.853 028 664 934 4;
  • 36) 0.853 028 664 934 4 × 2 = 1 + 0.706 057 329 868 8;
  • 37) 0.706 057 329 868 8 × 2 = 1 + 0.412 114 659 737 6;
  • 38) 0.412 114 659 737 6 × 2 = 0 + 0.824 229 319 475 2;
  • 39) 0.824 229 319 475 2 × 2 = 1 + 0.648 458 638 950 4;
  • 40) 0.648 458 638 950 4 × 2 = 1 + 0.296 917 277 900 8;
  • 41) 0.296 917 277 900 8 × 2 = 0 + 0.593 834 555 801 6;
  • 42) 0.593 834 555 801 6 × 2 = 1 + 0.187 669 111 603 2;
  • 43) 0.187 669 111 603 2 × 2 = 0 + 0.375 338 223 206 4;
  • 44) 0.375 338 223 206 4 × 2 = 0 + 0.750 676 446 412 8;
  • 45) 0.750 676 446 412 8 × 2 = 1 + 0.501 352 892 825 6;
  • 46) 0.501 352 892 825 6 × 2 = 1 + 0.002 705 785 651 2;
  • 47) 0.002 705 785 651 2 × 2 = 0 + 0.005 411 571 302 4;
  • 48) 0.005 411 571 302 4 × 2 = 0 + 0.010 823 142 604 8;
  • 49) 0.010 823 142 604 8 × 2 = 0 + 0.021 646 285 209 6;
  • 50) 0.021 646 285 209 6 × 2 = 0 + 0.043 292 570 419 2;
  • 51) 0.043 292 570 419 2 × 2 = 0 + 0.086 585 140 838 4;
  • 52) 0.086 585 140 838 4 × 2 = 0 + 0.173 170 281 676 8;
  • 53) 0.173 170 281 676 8 × 2 = 0 + 0.346 340 563 353 6;
  • 54) 0.346 340 563 353 6 × 2 = 0 + 0.692 681 126 707 2;
  • 55) 0.692 681 126 707 2 × 2 = 1 + 0.385 362 253 414 4;
  • 56) 0.385 362 253 414 4 × 2 = 0 + 0.770 724 506 828 8;
  • 57) 0.770 724 506 828 8 × 2 = 1 + 0.541 449 013 657 6;
  • 58) 0.541 449 013 657 6 × 2 = 1 + 0.082 898 027 315 2;
  • 59) 0.082 898 027 315 2 × 2 = 0 + 0.165 796 054 630 4;
  • 60) 0.165 796 054 630 4 × 2 = 0 + 0.331 592 109 260 8;
  • 61) 0.331 592 109 260 8 × 2 = 0 + 0.663 184 218 521 6;
  • 62) 0.663 184 218 521 6 × 2 = 1 + 0.326 368 437 043 2;
  • 63) 0.326 368 437 043 2 × 2 = 0 + 0.652 736 874 086 4;
  • 64) 0.652 736 874 086 4 × 2 = 1 + 0.305 473 748 172 8;

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:


0.000 282 005 868 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101(2)

6. Positive number before normalization:

0.000 282 005 868 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101(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:


0.000 282 005 868 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101(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 0101 1111 1011 0100 1100 0000 0010 1100 0101


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 0101 1111 1011 0100 1100 0000 0010 1100 0101 =

0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101


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 0101 1111 1011 0100 1100 0000 0010 1100 0101


Decimal number -0.000 282 005 868 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1011 0100 1100 0000 0010 1100 0101

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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