-0.000 282 005 860 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 860 5(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
-0.000 282 005 860 5(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:

|-0.000 282 005 860 5| = 0.000 282 005 860 5


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: 0.000 282 005 860 5.

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) 0.000 282 005 860 5 × 2 = 0 + 0.000 564 011 721;
  • 2) 0.000 564 011 721 × 2 = 0 + 0.001 128 023 442;
  • 3) 0.001 128 023 442 × 2 = 0 + 0.002 256 046 884;
  • 4) 0.002 256 046 884 × 2 = 0 + 0.004 512 093 768;
  • 5) 0.004 512 093 768 × 2 = 0 + 0.009 024 187 536;
  • 6) 0.009 024 187 536 × 2 = 0 + 0.018 048 375 072;
  • 7) 0.018 048 375 072 × 2 = 0 + 0.036 096 750 144;
  • 8) 0.036 096 750 144 × 2 = 0 + 0.072 193 500 288;
  • 9) 0.072 193 500 288 × 2 = 0 + 0.144 387 000 576;
  • 10) 0.144 387 000 576 × 2 = 0 + 0.288 774 001 152;
  • 11) 0.288 774 001 152 × 2 = 0 + 0.577 548 002 304;
  • 12) 0.577 548 002 304 × 2 = 1 + 0.155 096 004 608;
  • 13) 0.155 096 004 608 × 2 = 0 + 0.310 192 009 216;
  • 14) 0.310 192 009 216 × 2 = 0 + 0.620 384 018 432;
  • 15) 0.620 384 018 432 × 2 = 1 + 0.240 768 036 864;
  • 16) 0.240 768 036 864 × 2 = 0 + 0.481 536 073 728;
  • 17) 0.481 536 073 728 × 2 = 0 + 0.963 072 147 456;
  • 18) 0.963 072 147 456 × 2 = 1 + 0.926 144 294 912;
  • 19) 0.926 144 294 912 × 2 = 1 + 0.852 288 589 824;
  • 20) 0.852 288 589 824 × 2 = 1 + 0.704 577 179 648;
  • 21) 0.704 577 179 648 × 2 = 1 + 0.409 154 359 296;
  • 22) 0.409 154 359 296 × 2 = 0 + 0.818 308 718 592;
  • 23) 0.818 308 718 592 × 2 = 1 + 0.636 617 437 184;
  • 24) 0.636 617 437 184 × 2 = 1 + 0.273 234 874 368;
  • 25) 0.273 234 874 368 × 2 = 0 + 0.546 469 748 736;
  • 26) 0.546 469 748 736 × 2 = 1 + 0.092 939 497 472;
  • 27) 0.092 939 497 472 × 2 = 0 + 0.185 878 994 944;
  • 28) 0.185 878 994 944 × 2 = 0 + 0.371 757 989 888;
  • 29) 0.371 757 989 888 × 2 = 0 + 0.743 515 979 776;
  • 30) 0.743 515 979 776 × 2 = 1 + 0.487 031 959 552;
  • 31) 0.487 031 959 552 × 2 = 0 + 0.974 063 919 104;
  • 32) 0.974 063 919 104 × 2 = 1 + 0.948 127 838 208;
  • 33) 0.948 127 838 208 × 2 = 1 + 0.896 255 676 416;
  • 34) 0.896 255 676 416 × 2 = 1 + 0.792 511 352 832;
  • 35) 0.792 511 352 832 × 2 = 1 + 0.585 022 705 664;
  • 36) 0.585 022 705 664 × 2 = 1 + 0.170 045 411 328;
  • 37) 0.170 045 411 328 × 2 = 0 + 0.340 090 822 656;
  • 38) 0.340 090 822 656 × 2 = 0 + 0.680 181 645 312;
  • 39) 0.680 181 645 312 × 2 = 1 + 0.360 363 290 624;
  • 40) 0.360 363 290 624 × 2 = 0 + 0.720 726 581 248;
  • 41) 0.720 726 581 248 × 2 = 1 + 0.441 453 162 496;
  • 42) 0.441 453 162 496 × 2 = 0 + 0.882 906 324 992;
  • 43) 0.882 906 324 992 × 2 = 1 + 0.765 812 649 984;
  • 44) 0.765 812 649 984 × 2 = 1 + 0.531 625 299 968;
  • 45) 0.531 625 299 968 × 2 = 1 + 0.063 250 599 936;
  • 46) 0.063 250 599 936 × 2 = 0 + 0.126 501 199 872;
  • 47) 0.126 501 199 872 × 2 = 0 + 0.253 002 399 744;
  • 48) 0.253 002 399 744 × 2 = 0 + 0.506 004 799 488;
  • 49) 0.506 004 799 488 × 2 = 1 + 0.012 009 598 976;
  • 50) 0.012 009 598 976 × 2 = 0 + 0.024 019 197 952;
  • 51) 0.024 019 197 952 × 2 = 0 + 0.048 038 395 904;
  • 52) 0.048 038 395 904 × 2 = 0 + 0.096 076 791 808;
  • 53) 0.096 076 791 808 × 2 = 0 + 0.192 153 583 616;
  • 54) 0.192 153 583 616 × 2 = 0 + 0.384 307 167 232;
  • 55) 0.384 307 167 232 × 2 = 0 + 0.768 614 334 464;
  • 56) 0.768 614 334 464 × 2 = 1 + 0.537 228 668 928;
  • 57) 0.537 228 668 928 × 2 = 1 + 0.074 457 337 856;
  • 58) 0.074 457 337 856 × 2 = 0 + 0.148 914 675 712;
  • 59) 0.148 914 675 712 × 2 = 0 + 0.297 829 351 424;
  • 60) 0.297 829 351 424 × 2 = 0 + 0.595 658 702 848;
  • 61) 0.595 658 702 848 × 2 = 1 + 0.191 317 405 696;
  • 62) 0.191 317 405 696 × 2 = 0 + 0.382 634 811 392;
  • 63) 0.382 634 811 392 × 2 = 0 + 0.765 269 622 784;
  • 64) 0.765 269 622 784 × 2 = 1 + 0.530 539 245 568;

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:


0.000 282 005 860 5(10) =

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

6. Positive number before normalization:

0.000 282 005 860 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0010 1011 1000 1000 0001 1000 1001(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:


0.000 282 005 860 5(10) =

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

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

1.0010 0111 1011 0100 0101 1111 0010 1011 1000 1000 0001 1000 1001(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 1111 0010 1011 1000 1000 0001 1000 1001


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 1111 0010 1011 1000 1000 0001 1000 1001 =

0010 0111 1011 0100 0101 1111 0010 1011 1000 1000 0001 1000 1001


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 1111 0010 1011 1000 1000 0001 1000 1001


Decimal number -0.000 282 005 860 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 0010 1011 1000 1000 0001 1000 1001

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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