-0.000 282 005 899 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 899(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
-0.000 282 005 899(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:

|-0.000 282 005 899| = 0.000 282 005 899


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: 0.000 282 005 899.

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) 0.000 282 005 899 × 2 = 0 + 0.000 564 011 798;
  • 2) 0.000 564 011 798 × 2 = 0 + 0.001 128 023 596;
  • 3) 0.001 128 023 596 × 2 = 0 + 0.002 256 047 192;
  • 4) 0.002 256 047 192 × 2 = 0 + 0.004 512 094 384;
  • 5) 0.004 512 094 384 × 2 = 0 + 0.009 024 188 768;
  • 6) 0.009 024 188 768 × 2 = 0 + 0.018 048 377 536;
  • 7) 0.018 048 377 536 × 2 = 0 + 0.036 096 755 072;
  • 8) 0.036 096 755 072 × 2 = 0 + 0.072 193 510 144;
  • 9) 0.072 193 510 144 × 2 = 0 + 0.144 387 020 288;
  • 10) 0.144 387 020 288 × 2 = 0 + 0.288 774 040 576;
  • 11) 0.288 774 040 576 × 2 = 0 + 0.577 548 081 152;
  • 12) 0.577 548 081 152 × 2 = 1 + 0.155 096 162 304;
  • 13) 0.155 096 162 304 × 2 = 0 + 0.310 192 324 608;
  • 14) 0.310 192 324 608 × 2 = 0 + 0.620 384 649 216;
  • 15) 0.620 384 649 216 × 2 = 1 + 0.240 769 298 432;
  • 16) 0.240 769 298 432 × 2 = 0 + 0.481 538 596 864;
  • 17) 0.481 538 596 864 × 2 = 0 + 0.963 077 193 728;
  • 18) 0.963 077 193 728 × 2 = 1 + 0.926 154 387 456;
  • 19) 0.926 154 387 456 × 2 = 1 + 0.852 308 774 912;
  • 20) 0.852 308 774 912 × 2 = 1 + 0.704 617 549 824;
  • 21) 0.704 617 549 824 × 2 = 1 + 0.409 235 099 648;
  • 22) 0.409 235 099 648 × 2 = 0 + 0.818 470 199 296;
  • 23) 0.818 470 199 296 × 2 = 1 + 0.636 940 398 592;
  • 24) 0.636 940 398 592 × 2 = 1 + 0.273 880 797 184;
  • 25) 0.273 880 797 184 × 2 = 0 + 0.547 761 594 368;
  • 26) 0.547 761 594 368 × 2 = 1 + 0.095 523 188 736;
  • 27) 0.095 523 188 736 × 2 = 0 + 0.191 046 377 472;
  • 28) 0.191 046 377 472 × 2 = 0 + 0.382 092 754 944;
  • 29) 0.382 092 754 944 × 2 = 0 + 0.764 185 509 888;
  • 30) 0.764 185 509 888 × 2 = 1 + 0.528 371 019 776;
  • 31) 0.528 371 019 776 × 2 = 1 + 0.056 742 039 552;
  • 32) 0.056 742 039 552 × 2 = 0 + 0.113 484 079 104;
  • 33) 0.113 484 079 104 × 2 = 0 + 0.226 968 158 208;
  • 34) 0.226 968 158 208 × 2 = 0 + 0.453 936 316 416;
  • 35) 0.453 936 316 416 × 2 = 0 + 0.907 872 632 832;
  • 36) 0.907 872 632 832 × 2 = 1 + 0.815 745 265 664;
  • 37) 0.815 745 265 664 × 2 = 1 + 0.631 490 531 328;
  • 38) 0.631 490 531 328 × 2 = 1 + 0.262 981 062 656;
  • 39) 0.262 981 062 656 × 2 = 0 + 0.525 962 125 312;
  • 40) 0.525 962 125 312 × 2 = 1 + 0.051 924 250 624;
  • 41) 0.051 924 250 624 × 2 = 0 + 0.103 848 501 248;
  • 42) 0.103 848 501 248 × 2 = 0 + 0.207 697 002 496;
  • 43) 0.207 697 002 496 × 2 = 0 + 0.415 394 004 992;
  • 44) 0.415 394 004 992 × 2 = 0 + 0.830 788 009 984;
  • 45) 0.830 788 009 984 × 2 = 1 + 0.661 576 019 968;
  • 46) 0.661 576 019 968 × 2 = 1 + 0.323 152 039 936;
  • 47) 0.323 152 039 936 × 2 = 0 + 0.646 304 079 872;
  • 48) 0.646 304 079 872 × 2 = 1 + 0.292 608 159 744;
  • 49) 0.292 608 159 744 × 2 = 0 + 0.585 216 319 488;
  • 50) 0.585 216 319 488 × 2 = 1 + 0.170 432 638 976;
  • 51) 0.170 432 638 976 × 2 = 0 + 0.340 865 277 952;
  • 52) 0.340 865 277 952 × 2 = 0 + 0.681 730 555 904;
  • 53) 0.681 730 555 904 × 2 = 1 + 0.363 461 111 808;
  • 54) 0.363 461 111 808 × 2 = 0 + 0.726 922 223 616;
  • 55) 0.726 922 223 616 × 2 = 1 + 0.453 844 447 232;
  • 56) 0.453 844 447 232 × 2 = 0 + 0.907 688 894 464;
  • 57) 0.907 688 894 464 × 2 = 1 + 0.815 377 788 928;
  • 58) 0.815 377 788 928 × 2 = 1 + 0.630 755 577 856;
  • 59) 0.630 755 577 856 × 2 = 1 + 0.261 511 155 712;
  • 60) 0.261 511 155 712 × 2 = 0 + 0.523 022 311 424;
  • 61) 0.523 022 311 424 × 2 = 1 + 0.046 044 622 848;
  • 62) 0.046 044 622 848 × 2 = 0 + 0.092 089 245 696;
  • 63) 0.092 089 245 696 × 2 = 0 + 0.184 178 491 392;
  • 64) 0.184 178 491 392 × 2 = 0 + 0.368 356 982 784;

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:


0.000 282 005 899(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000(2)

6. Positive number before normalization:

0.000 282 005 899(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000(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:


0.000 282 005 899(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000(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 0001 1101 0000 1101 0100 1010 1110 1000


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 0001 1101 0000 1101 0100 1010 1110 1000 =

0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000


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 0001 1101 0000 1101 0100 1010 1110 1000


Decimal number -0.000 282 005 899 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1101 0000 1101 0100 1010 1110 1000

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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