-0.003 137 615 632 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.003 137 615 632(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
-0.003 137 615 632(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:

|-0.003 137 615 632| = 0.003 137 615 632


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: 0.003 137 615 632.

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) 0.003 137 615 632 × 2 = 0 + 0.006 275 231 264;
  • 2) 0.006 275 231 264 × 2 = 0 + 0.012 550 462 528;
  • 3) 0.012 550 462 528 × 2 = 0 + 0.025 100 925 056;
  • 4) 0.025 100 925 056 × 2 = 0 + 0.050 201 850 112;
  • 5) 0.050 201 850 112 × 2 = 0 + 0.100 403 700 224;
  • 6) 0.100 403 700 224 × 2 = 0 + 0.200 807 400 448;
  • 7) 0.200 807 400 448 × 2 = 0 + 0.401 614 800 896;
  • 8) 0.401 614 800 896 × 2 = 0 + 0.803 229 601 792;
  • 9) 0.803 229 601 792 × 2 = 1 + 0.606 459 203 584;
  • 10) 0.606 459 203 584 × 2 = 1 + 0.212 918 407 168;
  • 11) 0.212 918 407 168 × 2 = 0 + 0.425 836 814 336;
  • 12) 0.425 836 814 336 × 2 = 0 + 0.851 673 628 672;
  • 13) 0.851 673 628 672 × 2 = 1 + 0.703 347 257 344;
  • 14) 0.703 347 257 344 × 2 = 1 + 0.406 694 514 688;
  • 15) 0.406 694 514 688 × 2 = 0 + 0.813 389 029 376;
  • 16) 0.813 389 029 376 × 2 = 1 + 0.626 778 058 752;
  • 17) 0.626 778 058 752 × 2 = 1 + 0.253 556 117 504;
  • 18) 0.253 556 117 504 × 2 = 0 + 0.507 112 235 008;
  • 19) 0.507 112 235 008 × 2 = 1 + 0.014 224 470 016;
  • 20) 0.014 224 470 016 × 2 = 0 + 0.028 448 940 032;
  • 21) 0.028 448 940 032 × 2 = 0 + 0.056 897 880 064;
  • 22) 0.056 897 880 064 × 2 = 0 + 0.113 795 760 128;
  • 23) 0.113 795 760 128 × 2 = 0 + 0.227 591 520 256;
  • 24) 0.227 591 520 256 × 2 = 0 + 0.455 183 040 512;
  • 25) 0.455 183 040 512 × 2 = 0 + 0.910 366 081 024;
  • 26) 0.910 366 081 024 × 2 = 1 + 0.820 732 162 048;
  • 27) 0.820 732 162 048 × 2 = 1 + 0.641 464 324 096;
  • 28) 0.641 464 324 096 × 2 = 1 + 0.282 928 648 192;
  • 29) 0.282 928 648 192 × 2 = 0 + 0.565 857 296 384;
  • 30) 0.565 857 296 384 × 2 = 1 + 0.131 714 592 768;
  • 31) 0.131 714 592 768 × 2 = 0 + 0.263 429 185 536;
  • 32) 0.263 429 185 536 × 2 = 0 + 0.526 858 371 072;
  • 33) 0.526 858 371 072 × 2 = 1 + 0.053 716 742 144;
  • 34) 0.053 716 742 144 × 2 = 0 + 0.107 433 484 288;
  • 35) 0.107 433 484 288 × 2 = 0 + 0.214 866 968 576;
  • 36) 0.214 866 968 576 × 2 = 0 + 0.429 733 937 152;
  • 37) 0.429 733 937 152 × 2 = 0 + 0.859 467 874 304;
  • 38) 0.859 467 874 304 × 2 = 1 + 0.718 935 748 608;
  • 39) 0.718 935 748 608 × 2 = 1 + 0.437 871 497 216;
  • 40) 0.437 871 497 216 × 2 = 0 + 0.875 742 994 432;
  • 41) 0.875 742 994 432 × 2 = 1 + 0.751 485 988 864;
  • 42) 0.751 485 988 864 × 2 = 1 + 0.502 971 977 728;
  • 43) 0.502 971 977 728 × 2 = 1 + 0.005 943 955 456;
  • 44) 0.005 943 955 456 × 2 = 0 + 0.011 887 910 912;
  • 45) 0.011 887 910 912 × 2 = 0 + 0.023 775 821 824;
  • 46) 0.023 775 821 824 × 2 = 0 + 0.047 551 643 648;
  • 47) 0.047 551 643 648 × 2 = 0 + 0.095 103 287 296;
  • 48) 0.095 103 287 296 × 2 = 0 + 0.190 206 574 592;
  • 49) 0.190 206 574 592 × 2 = 0 + 0.380 413 149 184;
  • 50) 0.380 413 149 184 × 2 = 0 + 0.760 826 298 368;
  • 51) 0.760 826 298 368 × 2 = 1 + 0.521 652 596 736;
  • 52) 0.521 652 596 736 × 2 = 1 + 0.043 305 193 472;
  • 53) 0.043 305 193 472 × 2 = 0 + 0.086 610 386 944;
  • 54) 0.086 610 386 944 × 2 = 0 + 0.173 220 773 888;
  • 55) 0.173 220 773 888 × 2 = 0 + 0.346 441 547 776;
  • 56) 0.346 441 547 776 × 2 = 0 + 0.692 883 095 552;
  • 57) 0.692 883 095 552 × 2 = 1 + 0.385 766 191 104;
  • 58) 0.385 766 191 104 × 2 = 0 + 0.771 532 382 208;
  • 59) 0.771 532 382 208 × 2 = 1 + 0.543 064 764 416;
  • 60) 0.543 064 764 416 × 2 = 1 + 0.086 129 528 832;
  • 61) 0.086 129 528 832 × 2 = 0 + 0.172 259 057 664;

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:


0.003 137 615 632(10) =

0.0000 0000 1100 1101 1010 0000 0111 0100 1000 0110 1110 0000 0011 0000 1011 0(2)

6. Positive number before normalization:

0.003 137 615 632(10) =

0.0000 0000 1100 1101 1010 0000 0111 0100 1000 0110 1110 0000 0011 0000 1011 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.003 137 615 632(10) =

0.0000 0000 1100 1101 1010 0000 0111 0100 1000 0110 1110 0000 0011 0000 1011 0(2) =

0.0000 0000 1100 1101 1010 0000 0111 0100 1000 0110 1110 0000 0011 0000 1011 0(2) × 20 =

1.1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110(2) × 2-9


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): -9

Mantissa (not normalized):
1.1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(11-1) - 1 =

(-9 + 1 023)(10) =

1 014(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;
  • 1 014 ÷ 2 = 507 + 0;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 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) =

1014(10) =

011 1111 0110(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. 1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110 =

1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110


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 0110

Mantissa (52 bits) =
1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110


Decimal number -0.003 137 615 632 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0110 - 1001 1011 0100 0000 1110 1001 0000 1101 1100 0000 0110 0001 0110

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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