-0.000 282 005 902 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 902 7(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
-0.000 282 005 902 7(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:

|-0.000 282 005 902 7| = 0.000 282 005 902 7


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: 0.000 282 005 902 7.

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) 0.000 282 005 902 7 × 2 = 0 + 0.000 564 011 805 4;
  • 2) 0.000 564 011 805 4 × 2 = 0 + 0.001 128 023 610 8;
  • 3) 0.001 128 023 610 8 × 2 = 0 + 0.002 256 047 221 6;
  • 4) 0.002 256 047 221 6 × 2 = 0 + 0.004 512 094 443 2;
  • 5) 0.004 512 094 443 2 × 2 = 0 + 0.009 024 188 886 4;
  • 6) 0.009 024 188 886 4 × 2 = 0 + 0.018 048 377 772 8;
  • 7) 0.018 048 377 772 8 × 2 = 0 + 0.036 096 755 545 6;
  • 8) 0.036 096 755 545 6 × 2 = 0 + 0.072 193 511 091 2;
  • 9) 0.072 193 511 091 2 × 2 = 0 + 0.144 387 022 182 4;
  • 10) 0.144 387 022 182 4 × 2 = 0 + 0.288 774 044 364 8;
  • 11) 0.288 774 044 364 8 × 2 = 0 + 0.577 548 088 729 6;
  • 12) 0.577 548 088 729 6 × 2 = 1 + 0.155 096 177 459 2;
  • 13) 0.155 096 177 459 2 × 2 = 0 + 0.310 192 354 918 4;
  • 14) 0.310 192 354 918 4 × 2 = 0 + 0.620 384 709 836 8;
  • 15) 0.620 384 709 836 8 × 2 = 1 + 0.240 769 419 673 6;
  • 16) 0.240 769 419 673 6 × 2 = 0 + 0.481 538 839 347 2;
  • 17) 0.481 538 839 347 2 × 2 = 0 + 0.963 077 678 694 4;
  • 18) 0.963 077 678 694 4 × 2 = 1 + 0.926 155 357 388 8;
  • 19) 0.926 155 357 388 8 × 2 = 1 + 0.852 310 714 777 6;
  • 20) 0.852 310 714 777 6 × 2 = 1 + 0.704 621 429 555 2;
  • 21) 0.704 621 429 555 2 × 2 = 1 + 0.409 242 859 110 4;
  • 22) 0.409 242 859 110 4 × 2 = 0 + 0.818 485 718 220 8;
  • 23) 0.818 485 718 220 8 × 2 = 1 + 0.636 971 436 441 6;
  • 24) 0.636 971 436 441 6 × 2 = 1 + 0.273 942 872 883 2;
  • 25) 0.273 942 872 883 2 × 2 = 0 + 0.547 885 745 766 4;
  • 26) 0.547 885 745 766 4 × 2 = 1 + 0.095 771 491 532 8;
  • 27) 0.095 771 491 532 8 × 2 = 0 + 0.191 542 983 065 6;
  • 28) 0.191 542 983 065 6 × 2 = 0 + 0.383 085 966 131 2;
  • 29) 0.383 085 966 131 2 × 2 = 0 + 0.766 171 932 262 4;
  • 30) 0.766 171 932 262 4 × 2 = 1 + 0.532 343 864 524 8;
  • 31) 0.532 343 864 524 8 × 2 = 1 + 0.064 687 729 049 6;
  • 32) 0.064 687 729 049 6 × 2 = 0 + 0.129 375 458 099 2;
  • 33) 0.129 375 458 099 2 × 2 = 0 + 0.258 750 916 198 4;
  • 34) 0.258 750 916 198 4 × 2 = 0 + 0.517 501 832 396 8;
  • 35) 0.517 501 832 396 8 × 2 = 1 + 0.035 003 664 793 6;
  • 36) 0.035 003 664 793 6 × 2 = 0 + 0.070 007 329 587 2;
  • 37) 0.070 007 329 587 2 × 2 = 0 + 0.140 014 659 174 4;
  • 38) 0.140 014 659 174 4 × 2 = 0 + 0.280 029 318 348 8;
  • 39) 0.280 029 318 348 8 × 2 = 0 + 0.560 058 636 697 6;
  • 40) 0.560 058 636 697 6 × 2 = 1 + 0.120 117 273 395 2;
  • 41) 0.120 117 273 395 2 × 2 = 0 + 0.240 234 546 790 4;
  • 42) 0.240 234 546 790 4 × 2 = 0 + 0.480 469 093 580 8;
  • 43) 0.480 469 093 580 8 × 2 = 0 + 0.960 938 187 161 6;
  • 44) 0.960 938 187 161 6 × 2 = 1 + 0.921 876 374 323 2;
  • 45) 0.921 876 374 323 2 × 2 = 1 + 0.843 752 748 646 4;
  • 46) 0.843 752 748 646 4 × 2 = 1 + 0.687 505 497 292 8;
  • 47) 0.687 505 497 292 8 × 2 = 1 + 0.375 010 994 585 6;
  • 48) 0.375 010 994 585 6 × 2 = 0 + 0.750 021 989 171 2;
  • 49) 0.750 021 989 171 2 × 2 = 1 + 0.500 043 978 342 4;
  • 50) 0.500 043 978 342 4 × 2 = 1 + 0.000 087 956 684 8;
  • 51) 0.000 087 956 684 8 × 2 = 0 + 0.000 175 913 369 6;
  • 52) 0.000 175 913 369 6 × 2 = 0 + 0.000 351 826 739 2;
  • 53) 0.000 351 826 739 2 × 2 = 0 + 0.000 703 653 478 4;
  • 54) 0.000 703 653 478 4 × 2 = 0 + 0.001 407 306 956 8;
  • 55) 0.001 407 306 956 8 × 2 = 0 + 0.002 814 613 913 6;
  • 56) 0.002 814 613 913 6 × 2 = 0 + 0.005 629 227 827 2;
  • 57) 0.005 629 227 827 2 × 2 = 0 + 0.011 258 455 654 4;
  • 58) 0.011 258 455 654 4 × 2 = 0 + 0.022 516 911 308 8;
  • 59) 0.022 516 911 308 8 × 2 = 0 + 0.045 033 822 617 6;
  • 60) 0.045 033 822 617 6 × 2 = 0 + 0.090 067 645 235 2;
  • 61) 0.090 067 645 235 2 × 2 = 0 + 0.180 135 290 470 4;
  • 62) 0.180 135 290 470 4 × 2 = 0 + 0.360 270 580 940 8;
  • 63) 0.360 270 580 940 8 × 2 = 0 + 0.720 541 161 881 6;
  • 64) 0.720 541 161 881 6 × 2 = 1 + 0.441 082 323 763 2;

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:


0.000 282 005 902 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001(2)

6. Positive number before normalization:

0.000 282 005 902 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001(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:


0.000 282 005 902 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001(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 0001 0001 1110 1100 0000 0000 0001


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 0001 0001 1110 1100 0000 0000 0001 =

0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001


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 0001 0001 1110 1100 0000 0000 0001


Decimal number -0.000 282 005 902 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0001 0001 1110 1100 0000 0000 0001

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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