-0.040 000 000 397 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.040 000 000 397(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
-0.040 000 000 397(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:

|-0.040 000 000 397| = 0.040 000 000 397


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: 0.040 000 000 397.

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) 0.040 000 000 397 × 2 = 0 + 0.080 000 000 794;
  • 2) 0.080 000 000 794 × 2 = 0 + 0.160 000 001 588;
  • 3) 0.160 000 001 588 × 2 = 0 + 0.320 000 003 176;
  • 4) 0.320 000 003 176 × 2 = 0 + 0.640 000 006 352;
  • 5) 0.640 000 006 352 × 2 = 1 + 0.280 000 012 704;
  • 6) 0.280 000 012 704 × 2 = 0 + 0.560 000 025 408;
  • 7) 0.560 000 025 408 × 2 = 1 + 0.120 000 050 816;
  • 8) 0.120 000 050 816 × 2 = 0 + 0.240 000 101 632;
  • 9) 0.240 000 101 632 × 2 = 0 + 0.480 000 203 264;
  • 10) 0.480 000 203 264 × 2 = 0 + 0.960 000 406 528;
  • 11) 0.960 000 406 528 × 2 = 1 + 0.920 000 813 056;
  • 12) 0.920 000 813 056 × 2 = 1 + 0.840 001 626 112;
  • 13) 0.840 001 626 112 × 2 = 1 + 0.680 003 252 224;
  • 14) 0.680 003 252 224 × 2 = 1 + 0.360 006 504 448;
  • 15) 0.360 006 504 448 × 2 = 0 + 0.720 013 008 896;
  • 16) 0.720 013 008 896 × 2 = 1 + 0.440 026 017 792;
  • 17) 0.440 026 017 792 × 2 = 0 + 0.880 052 035 584;
  • 18) 0.880 052 035 584 × 2 = 1 + 0.760 104 071 168;
  • 19) 0.760 104 071 168 × 2 = 1 + 0.520 208 142 336;
  • 20) 0.520 208 142 336 × 2 = 1 + 0.040 416 284 672;
  • 21) 0.040 416 284 672 × 2 = 0 + 0.080 832 569 344;
  • 22) 0.080 832 569 344 × 2 = 0 + 0.161 665 138 688;
  • 23) 0.161 665 138 688 × 2 = 0 + 0.323 330 277 376;
  • 24) 0.323 330 277 376 × 2 = 0 + 0.646 660 554 752;
  • 25) 0.646 660 554 752 × 2 = 1 + 0.293 321 109 504;
  • 26) 0.293 321 109 504 × 2 = 0 + 0.586 642 219 008;
  • 27) 0.586 642 219 008 × 2 = 1 + 0.173 284 438 016;
  • 28) 0.173 284 438 016 × 2 = 0 + 0.346 568 876 032;
  • 29) 0.346 568 876 032 × 2 = 0 + 0.693 137 752 064;
  • 30) 0.693 137 752 064 × 2 = 1 + 0.386 275 504 128;
  • 31) 0.386 275 504 128 × 2 = 0 + 0.772 551 008 256;
  • 32) 0.772 551 008 256 × 2 = 1 + 0.545 102 016 512;
  • 33) 0.545 102 016 512 × 2 = 1 + 0.090 204 033 024;
  • 34) 0.090 204 033 024 × 2 = 0 + 0.180 408 066 048;
  • 35) 0.180 408 066 048 × 2 = 0 + 0.360 816 132 096;
  • 36) 0.360 816 132 096 × 2 = 0 + 0.721 632 264 192;
  • 37) 0.721 632 264 192 × 2 = 1 + 0.443 264 528 384;
  • 38) 0.443 264 528 384 × 2 = 0 + 0.886 529 056 768;
  • 39) 0.886 529 056 768 × 2 = 1 + 0.773 058 113 536;
  • 40) 0.773 058 113 536 × 2 = 1 + 0.546 116 227 072;
  • 41) 0.546 116 227 072 × 2 = 1 + 0.092 232 454 144;
  • 42) 0.092 232 454 144 × 2 = 0 + 0.184 464 908 288;
  • 43) 0.184 464 908 288 × 2 = 0 + 0.368 929 816 576;
  • 44) 0.368 929 816 576 × 2 = 0 + 0.737 859 633 152;
  • 45) 0.737 859 633 152 × 2 = 1 + 0.475 719 266 304;
  • 46) 0.475 719 266 304 × 2 = 0 + 0.951 438 532 608;
  • 47) 0.951 438 532 608 × 2 = 1 + 0.902 877 065 216;
  • 48) 0.902 877 065 216 × 2 = 1 + 0.805 754 130 432;
  • 49) 0.805 754 130 432 × 2 = 1 + 0.611 508 260 864;
  • 50) 0.611 508 260 864 × 2 = 1 + 0.223 016 521 728;
  • 51) 0.223 016 521 728 × 2 = 0 + 0.446 033 043 456;
  • 52) 0.446 033 043 456 × 2 = 0 + 0.892 066 086 912;
  • 53) 0.892 066 086 912 × 2 = 1 + 0.784 132 173 824;
  • 54) 0.784 132 173 824 × 2 = 1 + 0.568 264 347 648;
  • 55) 0.568 264 347 648 × 2 = 1 + 0.136 528 695 296;
  • 56) 0.136 528 695 296 × 2 = 0 + 0.273 057 390 592;
  • 57) 0.273 057 390 592 × 2 = 0 + 0.546 114 781 184;

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:


0.040 000 000 397(10) =

0.0000 1010 0011 1101 0111 0000 1010 0101 1000 1011 1000 1011 1100 1110 0(2)

6. Positive number before normalization:

0.040 000 000 397(10) =

0.0000 1010 0011 1101 0111 0000 1010 0101 1000 1011 1000 1011 1100 1110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.040 000 000 397(10) =

0.0000 1010 0011 1101 0111 0000 1010 0101 1000 1011 1000 1011 1100 1110 0(2) =

0.0000 1010 0011 1101 0111 0000 1010 0101 1000 1011 1000 1011 1100 1110 0(2) × 20 =

1.0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100(2) × 2-5


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): -5

Mantissa (not normalized):
1.0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100 =

0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100


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 1010

Mantissa (52 bits) =
0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100


Decimal number -0.040 000 000 397 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0100 0111 1010 1110 0001 0100 1011 0001 0111 0001 0111 1001 1100

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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