-0.000 282 005 917 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 917 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
-0.000 282 005 917 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:

|-0.000 282 005 917 3| = 0.000 282 005 917 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: 0.000 282 005 917 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) 0.000 282 005 917 3 × 2 = 0 + 0.000 564 011 834 6;
  • 2) 0.000 564 011 834 6 × 2 = 0 + 0.001 128 023 669 2;
  • 3) 0.001 128 023 669 2 × 2 = 0 + 0.002 256 047 338 4;
  • 4) 0.002 256 047 338 4 × 2 = 0 + 0.004 512 094 676 8;
  • 5) 0.004 512 094 676 8 × 2 = 0 + 0.009 024 189 353 6;
  • 6) 0.009 024 189 353 6 × 2 = 0 + 0.018 048 378 707 2;
  • 7) 0.018 048 378 707 2 × 2 = 0 + 0.036 096 757 414 4;
  • 8) 0.036 096 757 414 4 × 2 = 0 + 0.072 193 514 828 8;
  • 9) 0.072 193 514 828 8 × 2 = 0 + 0.144 387 029 657 6;
  • 10) 0.144 387 029 657 6 × 2 = 0 + 0.288 774 059 315 2;
  • 11) 0.288 774 059 315 2 × 2 = 0 + 0.577 548 118 630 4;
  • 12) 0.577 548 118 630 4 × 2 = 1 + 0.155 096 237 260 8;
  • 13) 0.155 096 237 260 8 × 2 = 0 + 0.310 192 474 521 6;
  • 14) 0.310 192 474 521 6 × 2 = 0 + 0.620 384 949 043 2;
  • 15) 0.620 384 949 043 2 × 2 = 1 + 0.240 769 898 086 4;
  • 16) 0.240 769 898 086 4 × 2 = 0 + 0.481 539 796 172 8;
  • 17) 0.481 539 796 172 8 × 2 = 0 + 0.963 079 592 345 6;
  • 18) 0.963 079 592 345 6 × 2 = 1 + 0.926 159 184 691 2;
  • 19) 0.926 159 184 691 2 × 2 = 1 + 0.852 318 369 382 4;
  • 20) 0.852 318 369 382 4 × 2 = 1 + 0.704 636 738 764 8;
  • 21) 0.704 636 738 764 8 × 2 = 1 + 0.409 273 477 529 6;
  • 22) 0.409 273 477 529 6 × 2 = 0 + 0.818 546 955 059 2;
  • 23) 0.818 546 955 059 2 × 2 = 1 + 0.637 093 910 118 4;
  • 24) 0.637 093 910 118 4 × 2 = 1 + 0.274 187 820 236 8;
  • 25) 0.274 187 820 236 8 × 2 = 0 + 0.548 375 640 473 6;
  • 26) 0.548 375 640 473 6 × 2 = 1 + 0.096 751 280 947 2;
  • 27) 0.096 751 280 947 2 × 2 = 0 + 0.193 502 561 894 4;
  • 28) 0.193 502 561 894 4 × 2 = 0 + 0.387 005 123 788 8;
  • 29) 0.387 005 123 788 8 × 2 = 0 + 0.774 010 247 577 6;
  • 30) 0.774 010 247 577 6 × 2 = 1 + 0.548 020 495 155 2;
  • 31) 0.548 020 495 155 2 × 2 = 1 + 0.096 040 990 310 4;
  • 32) 0.096 040 990 310 4 × 2 = 0 + 0.192 081 980 620 8;
  • 33) 0.192 081 980 620 8 × 2 = 0 + 0.384 163 961 241 6;
  • 34) 0.384 163 961 241 6 × 2 = 0 + 0.768 327 922 483 2;
  • 35) 0.768 327 922 483 2 × 2 = 1 + 0.536 655 844 966 4;
  • 36) 0.536 655 844 966 4 × 2 = 1 + 0.073 311 689 932 8;
  • 37) 0.073 311 689 932 8 × 2 = 0 + 0.146 623 379 865 6;
  • 38) 0.146 623 379 865 6 × 2 = 0 + 0.293 246 759 731 2;
  • 39) 0.293 246 759 731 2 × 2 = 0 + 0.586 493 519 462 4;
  • 40) 0.586 493 519 462 4 × 2 = 1 + 0.172 987 038 924 8;
  • 41) 0.172 987 038 924 8 × 2 = 0 + 0.345 974 077 849 6;
  • 42) 0.345 974 077 849 6 × 2 = 0 + 0.691 948 155 699 2;
  • 43) 0.691 948 155 699 2 × 2 = 1 + 0.383 896 311 398 4;
  • 44) 0.383 896 311 398 4 × 2 = 0 + 0.767 792 622 796 8;
  • 45) 0.767 792 622 796 8 × 2 = 1 + 0.535 585 245 593 6;
  • 46) 0.535 585 245 593 6 × 2 = 1 + 0.071 170 491 187 2;
  • 47) 0.071 170 491 187 2 × 2 = 0 + 0.142 340 982 374 4;
  • 48) 0.142 340 982 374 4 × 2 = 0 + 0.284 681 964 748 8;
  • 49) 0.284 681 964 748 8 × 2 = 0 + 0.569 363 929 497 6;
  • 50) 0.569 363 929 497 6 × 2 = 1 + 0.138 727 858 995 2;
  • 51) 0.138 727 858 995 2 × 2 = 0 + 0.277 455 717 990 4;
  • 52) 0.277 455 717 990 4 × 2 = 0 + 0.554 911 435 980 8;
  • 53) 0.554 911 435 980 8 × 2 = 1 + 0.109 822 871 961 6;
  • 54) 0.109 822 871 961 6 × 2 = 0 + 0.219 645 743 923 2;
  • 55) 0.219 645 743 923 2 × 2 = 0 + 0.439 291 487 846 4;
  • 56) 0.439 291 487 846 4 × 2 = 0 + 0.878 582 975 692 8;
  • 57) 0.878 582 975 692 8 × 2 = 1 + 0.757 165 951 385 6;
  • 58) 0.757 165 951 385 6 × 2 = 1 + 0.514 331 902 771 2;
  • 59) 0.514 331 902 771 2 × 2 = 1 + 0.028 663 805 542 4;
  • 60) 0.028 663 805 542 4 × 2 = 0 + 0.057 327 611 084 8;
  • 61) 0.057 327 611 084 8 × 2 = 0 + 0.114 655 222 169 6;
  • 62) 0.114 655 222 169 6 × 2 = 0 + 0.229 310 444 339 2;
  • 63) 0.229 310 444 339 2 × 2 = 0 + 0.458 620 888 678 4;
  • 64) 0.458 620 888 678 4 × 2 = 0 + 0.917 241 777 356 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:


0.000 282 005 917 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 0000(2)

6. Positive number before normalization:

0.000 282 005 917 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 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:


0.000 282 005 917 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 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 0100 0110 0011 0001 0010 1100 0100 1000 1110 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 0100 0110 0011 0001 0010 1100 0100 1000 1110 0000 =

0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 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 0100 0110 0011 0001 0010 1100 0100 1000 1110 0000


Decimal number -0.000 282 005 917 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0001 0010 1100 0100 1000 1110 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