-0.000 282 005 906 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 906 4(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
-0.000 282 005 906 4(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:

|-0.000 282 005 906 4| = 0.000 282 005 906 4


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: 0.000 282 005 906 4.

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) 0.000 282 005 906 4 × 2 = 0 + 0.000 564 011 812 8;
  • 2) 0.000 564 011 812 8 × 2 = 0 + 0.001 128 023 625 6;
  • 3) 0.001 128 023 625 6 × 2 = 0 + 0.002 256 047 251 2;
  • 4) 0.002 256 047 251 2 × 2 = 0 + 0.004 512 094 502 4;
  • 5) 0.004 512 094 502 4 × 2 = 0 + 0.009 024 189 004 8;
  • 6) 0.009 024 189 004 8 × 2 = 0 + 0.018 048 378 009 6;
  • 7) 0.018 048 378 009 6 × 2 = 0 + 0.036 096 756 019 2;
  • 8) 0.036 096 756 019 2 × 2 = 0 + 0.072 193 512 038 4;
  • 9) 0.072 193 512 038 4 × 2 = 0 + 0.144 387 024 076 8;
  • 10) 0.144 387 024 076 8 × 2 = 0 + 0.288 774 048 153 6;
  • 11) 0.288 774 048 153 6 × 2 = 0 + 0.577 548 096 307 2;
  • 12) 0.577 548 096 307 2 × 2 = 1 + 0.155 096 192 614 4;
  • 13) 0.155 096 192 614 4 × 2 = 0 + 0.310 192 385 228 8;
  • 14) 0.310 192 385 228 8 × 2 = 0 + 0.620 384 770 457 6;
  • 15) 0.620 384 770 457 6 × 2 = 1 + 0.240 769 540 915 2;
  • 16) 0.240 769 540 915 2 × 2 = 0 + 0.481 539 081 830 4;
  • 17) 0.481 539 081 830 4 × 2 = 0 + 0.963 078 163 660 8;
  • 18) 0.963 078 163 660 8 × 2 = 1 + 0.926 156 327 321 6;
  • 19) 0.926 156 327 321 6 × 2 = 1 + 0.852 312 654 643 2;
  • 20) 0.852 312 654 643 2 × 2 = 1 + 0.704 625 309 286 4;
  • 21) 0.704 625 309 286 4 × 2 = 1 + 0.409 250 618 572 8;
  • 22) 0.409 250 618 572 8 × 2 = 0 + 0.818 501 237 145 6;
  • 23) 0.818 501 237 145 6 × 2 = 1 + 0.637 002 474 291 2;
  • 24) 0.637 002 474 291 2 × 2 = 1 + 0.274 004 948 582 4;
  • 25) 0.274 004 948 582 4 × 2 = 0 + 0.548 009 897 164 8;
  • 26) 0.548 009 897 164 8 × 2 = 1 + 0.096 019 794 329 6;
  • 27) 0.096 019 794 329 6 × 2 = 0 + 0.192 039 588 659 2;
  • 28) 0.192 039 588 659 2 × 2 = 0 + 0.384 079 177 318 4;
  • 29) 0.384 079 177 318 4 × 2 = 0 + 0.768 158 354 636 8;
  • 30) 0.768 158 354 636 8 × 2 = 1 + 0.536 316 709 273 6;
  • 31) 0.536 316 709 273 6 × 2 = 1 + 0.072 633 418 547 2;
  • 32) 0.072 633 418 547 2 × 2 = 0 + 0.145 266 837 094 4;
  • 33) 0.145 266 837 094 4 × 2 = 0 + 0.290 533 674 188 8;
  • 34) 0.290 533 674 188 8 × 2 = 0 + 0.581 067 348 377 6;
  • 35) 0.581 067 348 377 6 × 2 = 1 + 0.162 134 696 755 2;
  • 36) 0.162 134 696 755 2 × 2 = 0 + 0.324 269 393 510 4;
  • 37) 0.324 269 393 510 4 × 2 = 0 + 0.648 538 787 020 8;
  • 38) 0.648 538 787 020 8 × 2 = 1 + 0.297 077 574 041 6;
  • 39) 0.297 077 574 041 6 × 2 = 0 + 0.594 155 148 083 2;
  • 40) 0.594 155 148 083 2 × 2 = 1 + 0.188 310 296 166 4;
  • 41) 0.188 310 296 166 4 × 2 = 0 + 0.376 620 592 332 8;
  • 42) 0.376 620 592 332 8 × 2 = 0 + 0.753 241 184 665 6;
  • 43) 0.753 241 184 665 6 × 2 = 1 + 0.506 482 369 331 2;
  • 44) 0.506 482 369 331 2 × 2 = 1 + 0.012 964 738 662 4;
  • 45) 0.012 964 738 662 4 × 2 = 0 + 0.025 929 477 324 8;
  • 46) 0.025 929 477 324 8 × 2 = 0 + 0.051 858 954 649 6;
  • 47) 0.051 858 954 649 6 × 2 = 0 + 0.103 717 909 299 2;
  • 48) 0.103 717 909 299 2 × 2 = 0 + 0.207 435 818 598 4;
  • 49) 0.207 435 818 598 4 × 2 = 0 + 0.414 871 637 196 8;
  • 50) 0.414 871 637 196 8 × 2 = 0 + 0.829 743 274 393 6;
  • 51) 0.829 743 274 393 6 × 2 = 1 + 0.659 486 548 787 2;
  • 52) 0.659 486 548 787 2 × 2 = 1 + 0.318 973 097 574 4;
  • 53) 0.318 973 097 574 4 × 2 = 0 + 0.637 946 195 148 8;
  • 54) 0.637 946 195 148 8 × 2 = 1 + 0.275 892 390 297 6;
  • 55) 0.275 892 390 297 6 × 2 = 0 + 0.551 784 780 595 2;
  • 56) 0.551 784 780 595 2 × 2 = 1 + 0.103 569 561 190 4;
  • 57) 0.103 569 561 190 4 × 2 = 0 + 0.207 139 122 380 8;
  • 58) 0.207 139 122 380 8 × 2 = 0 + 0.414 278 244 761 6;
  • 59) 0.414 278 244 761 6 × 2 = 0 + 0.828 556 489 523 2;
  • 60) 0.828 556 489 523 2 × 2 = 1 + 0.657 112 979 046 4;
  • 61) 0.657 112 979 046 4 × 2 = 1 + 0.314 225 958 092 8;
  • 62) 0.314 225 958 092 8 × 2 = 0 + 0.628 451 916 185 6;
  • 63) 0.628 451 916 185 6 × 2 = 1 + 0.256 903 832 371 2;
  • 64) 0.256 903 832 371 2 × 2 = 0 + 0.513 807 664 742 4;

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:


0.000 282 005 906 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 0001 1010(2)

6. Positive number before normalization:

0.000 282 005 906 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 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:


0.000 282 005 906 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 0001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 0001 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 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 0010 0101 0011 0000 0011 0101 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 0010 0101 0011 0000 0011 0101 0001 1010 =

0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 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 0010 0101 0011 0000 0011 0101 0001 1010


Decimal number -0.000 282 005 906 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0101 0011 0000 0011 0101 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