-0.000 282 008 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 72(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
-0.000 282 008 72(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:

|-0.000 282 008 72| = 0.000 282 008 72


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: 0.000 282 008 72.

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) 0.000 282 008 72 × 2 = 0 + 0.000 564 017 44;
  • 2) 0.000 564 017 44 × 2 = 0 + 0.001 128 034 88;
  • 3) 0.001 128 034 88 × 2 = 0 + 0.002 256 069 76;
  • 4) 0.002 256 069 76 × 2 = 0 + 0.004 512 139 52;
  • 5) 0.004 512 139 52 × 2 = 0 + 0.009 024 279 04;
  • 6) 0.009 024 279 04 × 2 = 0 + 0.018 048 558 08;
  • 7) 0.018 048 558 08 × 2 = 0 + 0.036 097 116 16;
  • 8) 0.036 097 116 16 × 2 = 0 + 0.072 194 232 32;
  • 9) 0.072 194 232 32 × 2 = 0 + 0.144 388 464 64;
  • 10) 0.144 388 464 64 × 2 = 0 + 0.288 776 929 28;
  • 11) 0.288 776 929 28 × 2 = 0 + 0.577 553 858 56;
  • 12) 0.577 553 858 56 × 2 = 1 + 0.155 107 717 12;
  • 13) 0.155 107 717 12 × 2 = 0 + 0.310 215 434 24;
  • 14) 0.310 215 434 24 × 2 = 0 + 0.620 430 868 48;
  • 15) 0.620 430 868 48 × 2 = 1 + 0.240 861 736 96;
  • 16) 0.240 861 736 96 × 2 = 0 + 0.481 723 473 92;
  • 17) 0.481 723 473 92 × 2 = 0 + 0.963 446 947 84;
  • 18) 0.963 446 947 84 × 2 = 1 + 0.926 893 895 68;
  • 19) 0.926 893 895 68 × 2 = 1 + 0.853 787 791 36;
  • 20) 0.853 787 791 36 × 2 = 1 + 0.707 575 582 72;
  • 21) 0.707 575 582 72 × 2 = 1 + 0.415 151 165 44;
  • 22) 0.415 151 165 44 × 2 = 0 + 0.830 302 330 88;
  • 23) 0.830 302 330 88 × 2 = 1 + 0.660 604 661 76;
  • 24) 0.660 604 661 76 × 2 = 1 + 0.321 209 323 52;
  • 25) 0.321 209 323 52 × 2 = 0 + 0.642 418 647 04;
  • 26) 0.642 418 647 04 × 2 = 1 + 0.284 837 294 08;
  • 27) 0.284 837 294 08 × 2 = 0 + 0.569 674 588 16;
  • 28) 0.569 674 588 16 × 2 = 1 + 0.139 349 176 32;
  • 29) 0.139 349 176 32 × 2 = 0 + 0.278 698 352 64;
  • 30) 0.278 698 352 64 × 2 = 0 + 0.557 396 705 28;
  • 31) 0.557 396 705 28 × 2 = 1 + 0.114 793 410 56;
  • 32) 0.114 793 410 56 × 2 = 0 + 0.229 586 821 12;
  • 33) 0.229 586 821 12 × 2 = 0 + 0.459 173 642 24;
  • 34) 0.459 173 642 24 × 2 = 0 + 0.918 347 284 48;
  • 35) 0.918 347 284 48 × 2 = 1 + 0.836 694 568 96;
  • 36) 0.836 694 568 96 × 2 = 1 + 0.673 389 137 92;
  • 37) 0.673 389 137 92 × 2 = 1 + 0.346 778 275 84;
  • 38) 0.346 778 275 84 × 2 = 0 + 0.693 556 551 68;
  • 39) 0.693 556 551 68 × 2 = 1 + 0.387 113 103 36;
  • 40) 0.387 113 103 36 × 2 = 0 + 0.774 226 206 72;
  • 41) 0.774 226 206 72 × 2 = 1 + 0.548 452 413 44;
  • 42) 0.548 452 413 44 × 2 = 1 + 0.096 904 826 88;
  • 43) 0.096 904 826 88 × 2 = 0 + 0.193 809 653 76;
  • 44) 0.193 809 653 76 × 2 = 0 + 0.387 619 307 52;
  • 45) 0.387 619 307 52 × 2 = 0 + 0.775 238 615 04;
  • 46) 0.775 238 615 04 × 2 = 1 + 0.550 477 230 08;
  • 47) 0.550 477 230 08 × 2 = 1 + 0.100 954 460 16;
  • 48) 0.100 954 460 16 × 2 = 0 + 0.201 908 920 32;
  • 49) 0.201 908 920 32 × 2 = 0 + 0.403 817 840 64;
  • 50) 0.403 817 840 64 × 2 = 0 + 0.807 635 681 28;
  • 51) 0.807 635 681 28 × 2 = 1 + 0.615 271 362 56;
  • 52) 0.615 271 362 56 × 2 = 1 + 0.230 542 725 12;
  • 53) 0.230 542 725 12 × 2 = 0 + 0.461 085 450 24;
  • 54) 0.461 085 450 24 × 2 = 0 + 0.922 170 900 48;
  • 55) 0.922 170 900 48 × 2 = 1 + 0.844 341 800 96;
  • 56) 0.844 341 800 96 × 2 = 1 + 0.688 683 601 92;
  • 57) 0.688 683 601 92 × 2 = 1 + 0.377 367 203 84;
  • 58) 0.377 367 203 84 × 2 = 0 + 0.754 734 407 68;
  • 59) 0.754 734 407 68 × 2 = 1 + 0.509 468 815 36;
  • 60) 0.509 468 815 36 × 2 = 1 + 0.018 937 630 72;
  • 61) 0.018 937 630 72 × 2 = 0 + 0.037 875 261 44;
  • 62) 0.037 875 261 44 × 2 = 0 + 0.075 750 522 88;
  • 63) 0.075 750 522 88 × 2 = 0 + 0.151 501 045 76;
  • 64) 0.151 501 045 76 × 2 = 0 + 0.303 002 091 52;

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:


0.000 282 008 72(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000(2)

6. Positive number before normalization:

0.000 282 008 72(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000(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:


0.000 282 008 72(10) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000(2) =

0.0000 0000 0001 0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000(2) × 20 =

1.0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000(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 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000


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 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000 =

0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000


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 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000


Decimal number -0.000 282 008 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0010 0011 1010 1100 0110 0011 0011 1011 0000

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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