-3.528 861 624 096 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.528 861 624 096(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
-3.528 861 624 096(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:

|-3.528 861 624 096| = 3.528 861 624 096


2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.528 861 624 096.

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) 0.528 861 624 096 × 2 = 1 + 0.057 723 248 192;
  • 2) 0.057 723 248 192 × 2 = 0 + 0.115 446 496 384;
  • 3) 0.115 446 496 384 × 2 = 0 + 0.230 892 992 768;
  • 4) 0.230 892 992 768 × 2 = 0 + 0.461 785 985 536;
  • 5) 0.461 785 985 536 × 2 = 0 + 0.923 571 971 072;
  • 6) 0.923 571 971 072 × 2 = 1 + 0.847 143 942 144;
  • 7) 0.847 143 942 144 × 2 = 1 + 0.694 287 884 288;
  • 8) 0.694 287 884 288 × 2 = 1 + 0.388 575 768 576;
  • 9) 0.388 575 768 576 × 2 = 0 + 0.777 151 537 152;
  • 10) 0.777 151 537 152 × 2 = 1 + 0.554 303 074 304;
  • 11) 0.554 303 074 304 × 2 = 1 + 0.108 606 148 608;
  • 12) 0.108 606 148 608 × 2 = 0 + 0.217 212 297 216;
  • 13) 0.217 212 297 216 × 2 = 0 + 0.434 424 594 432;
  • 14) 0.434 424 594 432 × 2 = 0 + 0.868 849 188 864;
  • 15) 0.868 849 188 864 × 2 = 1 + 0.737 698 377 728;
  • 16) 0.737 698 377 728 × 2 = 1 + 0.475 396 755 456;
  • 17) 0.475 396 755 456 × 2 = 0 + 0.950 793 510 912;
  • 18) 0.950 793 510 912 × 2 = 1 + 0.901 587 021 824;
  • 19) 0.901 587 021 824 × 2 = 1 + 0.803 174 043 648;
  • 20) 0.803 174 043 648 × 2 = 1 + 0.606 348 087 296;
  • 21) 0.606 348 087 296 × 2 = 1 + 0.212 696 174 592;
  • 22) 0.212 696 174 592 × 2 = 0 + 0.425 392 349 184;
  • 23) 0.425 392 349 184 × 2 = 0 + 0.850 784 698 368;
  • 24) 0.850 784 698 368 × 2 = 1 + 0.701 569 396 736;
  • 25) 0.701 569 396 736 × 2 = 1 + 0.403 138 793 472;
  • 26) 0.403 138 793 472 × 2 = 0 + 0.806 277 586 944;
  • 27) 0.806 277 586 944 × 2 = 1 + 0.612 555 173 888;
  • 28) 0.612 555 173 888 × 2 = 1 + 0.225 110 347 776;
  • 29) 0.225 110 347 776 × 2 = 0 + 0.450 220 695 552;
  • 30) 0.450 220 695 552 × 2 = 0 + 0.900 441 391 104;
  • 31) 0.900 441 391 104 × 2 = 1 + 0.800 882 782 208;
  • 32) 0.800 882 782 208 × 2 = 1 + 0.601 765 564 416;
  • 33) 0.601 765 564 416 × 2 = 1 + 0.203 531 128 832;
  • 34) 0.203 531 128 832 × 2 = 0 + 0.407 062 257 664;
  • 35) 0.407 062 257 664 × 2 = 0 + 0.814 124 515 328;
  • 36) 0.814 124 515 328 × 2 = 1 + 0.628 249 030 656;
  • 37) 0.628 249 030 656 × 2 = 1 + 0.256 498 061 312;
  • 38) 0.256 498 061 312 × 2 = 0 + 0.512 996 122 624;
  • 39) 0.512 996 122 624 × 2 = 1 + 0.025 992 245 248;
  • 40) 0.025 992 245 248 × 2 = 0 + 0.051 984 490 496;
  • 41) 0.051 984 490 496 × 2 = 0 + 0.103 968 980 992;
  • 42) 0.103 968 980 992 × 2 = 0 + 0.207 937 961 984;
  • 43) 0.207 937 961 984 × 2 = 0 + 0.415 875 923 968;
  • 44) 0.415 875 923 968 × 2 = 0 + 0.831 751 847 936;
  • 45) 0.831 751 847 936 × 2 = 1 + 0.663 503 695 872;
  • 46) 0.663 503 695 872 × 2 = 1 + 0.327 007 391 744;
  • 47) 0.327 007 391 744 × 2 = 0 + 0.654 014 783 488;
  • 48) 0.654 014 783 488 × 2 = 1 + 0.308 029 566 976;
  • 49) 0.308 029 566 976 × 2 = 0 + 0.616 059 133 952;
  • 50) 0.616 059 133 952 × 2 = 1 + 0.232 118 267 904;
  • 51) 0.232 118 267 904 × 2 = 0 + 0.464 236 535 808;
  • 52) 0.464 236 535 808 × 2 = 0 + 0.928 473 071 616;
  • 53) 0.928 473 071 616 × 2 = 1 + 0.856 946 143 232;

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:


0.528 861 624 096(10) =

0.1000 0111 0110 0011 0111 1001 1011 0011 1001 1010 0000 1101 0100 1(2)

6. Positive number before normalization:

3.528 861 624 096(10) =

11.1000 0111 0110 0011 0111 1001 1011 0011 1001 1010 0000 1101 0100 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.528 861 624 096(10) =

11.1000 0111 0110 0011 0111 1001 1011 0011 1001 1010 0000 1101 0100 1(2) =

11.1000 0111 0110 0011 0111 1001 1011 0011 1001 1010 0000 1101 0100 1(2) × 20 =

1.1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010 01(2) × 21


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): 1

Mantissa (not normalized):
1.1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010 01


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010 01 =

1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010


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) =
100 0000 0000

Mantissa (52 bits) =
1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010


Decimal number -3.528 861 624 096 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1100 0011 1011 0001 1011 1100 1101 1001 1100 1101 0000 0110 1010

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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