-0.000 282 005 788 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 788(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
-0.000 282 005 788(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:

|-0.000 282 005 788| = 0.000 282 005 788


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: 0.000 282 005 788.

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) 0.000 282 005 788 × 2 = 0 + 0.000 564 011 576;
  • 2) 0.000 564 011 576 × 2 = 0 + 0.001 128 023 152;
  • 3) 0.001 128 023 152 × 2 = 0 + 0.002 256 046 304;
  • 4) 0.002 256 046 304 × 2 = 0 + 0.004 512 092 608;
  • 5) 0.004 512 092 608 × 2 = 0 + 0.009 024 185 216;
  • 6) 0.009 024 185 216 × 2 = 0 + 0.018 048 370 432;
  • 7) 0.018 048 370 432 × 2 = 0 + 0.036 096 740 864;
  • 8) 0.036 096 740 864 × 2 = 0 + 0.072 193 481 728;
  • 9) 0.072 193 481 728 × 2 = 0 + 0.144 386 963 456;
  • 10) 0.144 386 963 456 × 2 = 0 + 0.288 773 926 912;
  • 11) 0.288 773 926 912 × 2 = 0 + 0.577 547 853 824;
  • 12) 0.577 547 853 824 × 2 = 1 + 0.155 095 707 648;
  • 13) 0.155 095 707 648 × 2 = 0 + 0.310 191 415 296;
  • 14) 0.310 191 415 296 × 2 = 0 + 0.620 382 830 592;
  • 15) 0.620 382 830 592 × 2 = 1 + 0.240 765 661 184;
  • 16) 0.240 765 661 184 × 2 = 0 + 0.481 531 322 368;
  • 17) 0.481 531 322 368 × 2 = 0 + 0.963 062 644 736;
  • 18) 0.963 062 644 736 × 2 = 1 + 0.926 125 289 472;
  • 19) 0.926 125 289 472 × 2 = 1 + 0.852 250 578 944;
  • 20) 0.852 250 578 944 × 2 = 1 + 0.704 501 157 888;
  • 21) 0.704 501 157 888 × 2 = 1 + 0.409 002 315 776;
  • 22) 0.409 002 315 776 × 2 = 0 + 0.818 004 631 552;
  • 23) 0.818 004 631 552 × 2 = 1 + 0.636 009 263 104;
  • 24) 0.636 009 263 104 × 2 = 1 + 0.272 018 526 208;
  • 25) 0.272 018 526 208 × 2 = 0 + 0.544 037 052 416;
  • 26) 0.544 037 052 416 × 2 = 1 + 0.088 074 104 832;
  • 27) 0.088 074 104 832 × 2 = 0 + 0.176 148 209 664;
  • 28) 0.176 148 209 664 × 2 = 0 + 0.352 296 419 328;
  • 29) 0.352 296 419 328 × 2 = 0 + 0.704 592 838 656;
  • 30) 0.704 592 838 656 × 2 = 1 + 0.409 185 677 312;
  • 31) 0.409 185 677 312 × 2 = 0 + 0.818 371 354 624;
  • 32) 0.818 371 354 624 × 2 = 1 + 0.636 742 709 248;
  • 33) 0.636 742 709 248 × 2 = 1 + 0.273 485 418 496;
  • 34) 0.273 485 418 496 × 2 = 0 + 0.546 970 836 992;
  • 35) 0.546 970 836 992 × 2 = 1 + 0.093 941 673 984;
  • 36) 0.093 941 673 984 × 2 = 0 + 0.187 883 347 968;
  • 37) 0.187 883 347 968 × 2 = 0 + 0.375 766 695 936;
  • 38) 0.375 766 695 936 × 2 = 0 + 0.751 533 391 872;
  • 39) 0.751 533 391 872 × 2 = 1 + 0.503 066 783 744;
  • 40) 0.503 066 783 744 × 2 = 1 + 0.006 133 567 488;
  • 41) 0.006 133 567 488 × 2 = 0 + 0.012 267 134 976;
  • 42) 0.012 267 134 976 × 2 = 0 + 0.024 534 269 952;
  • 43) 0.024 534 269 952 × 2 = 0 + 0.049 068 539 904;
  • 44) 0.049 068 539 904 × 2 = 0 + 0.098 137 079 808;
  • 45) 0.098 137 079 808 × 2 = 0 + 0.196 274 159 616;
  • 46) 0.196 274 159 616 × 2 = 0 + 0.392 548 319 232;
  • 47) 0.392 548 319 232 × 2 = 0 + 0.785 096 638 464;
  • 48) 0.785 096 638 464 × 2 = 1 + 0.570 193 276 928;
  • 49) 0.570 193 276 928 × 2 = 1 + 0.140 386 553 856;
  • 50) 0.140 386 553 856 × 2 = 0 + 0.280 773 107 712;
  • 51) 0.280 773 107 712 × 2 = 0 + 0.561 546 215 424;
  • 52) 0.561 546 215 424 × 2 = 1 + 0.123 092 430 848;
  • 53) 0.123 092 430 848 × 2 = 0 + 0.246 184 861 696;
  • 54) 0.246 184 861 696 × 2 = 0 + 0.492 369 723 392;
  • 55) 0.492 369 723 392 × 2 = 0 + 0.984 739 446 784;
  • 56) 0.984 739 446 784 × 2 = 1 + 0.969 478 893 568;
  • 57) 0.969 478 893 568 × 2 = 1 + 0.938 957 787 136;
  • 58) 0.938 957 787 136 × 2 = 1 + 0.877 915 574 272;
  • 59) 0.877 915 574 272 × 2 = 1 + 0.755 831 148 544;
  • 60) 0.755 831 148 544 × 2 = 1 + 0.511 662 297 088;
  • 61) 0.511 662 297 088 × 2 = 1 + 0.023 324 594 176;
  • 62) 0.023 324 594 176 × 2 = 0 + 0.046 649 188 352;
  • 63) 0.046 649 188 352 × 2 = 0 + 0.093 298 376 704;
  • 64) 0.093 298 376 704 × 2 = 0 + 0.186 596 753 408;

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:


0.000 282 005 788(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 1000(2)

6. Positive number before normalization:

0.000 282 005 788(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 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:


0.000 282 005 788(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 1000(2) × 20 =

1.0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 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 0101 1010 0011 0000 0001 1001 0001 1111 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 0101 1010 0011 0000 0001 1001 0001 1111 1000 =

0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 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 0101 1010 0011 0000 0001 1001 0001 1111 1000


Decimal number -0.000 282 005 788 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1010 0011 0000 0001 1001 0001 1111 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