-0.000 038 147 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 038 147(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
-0.000 038 147(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:

|-0.000 038 147| = 0.000 038 147


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: 0.000 038 147.

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) 0.000 038 147 × 2 = 0 + 0.000 076 294;
  • 2) 0.000 076 294 × 2 = 0 + 0.000 152 588;
  • 3) 0.000 152 588 × 2 = 0 + 0.000 305 176;
  • 4) 0.000 305 176 × 2 = 0 + 0.000 610 352;
  • 5) 0.000 610 352 × 2 = 0 + 0.001 220 704;
  • 6) 0.001 220 704 × 2 = 0 + 0.002 441 408;
  • 7) 0.002 441 408 × 2 = 0 + 0.004 882 816;
  • 8) 0.004 882 816 × 2 = 0 + 0.009 765 632;
  • 9) 0.009 765 632 × 2 = 0 + 0.019 531 264;
  • 10) 0.019 531 264 × 2 = 0 + 0.039 062 528;
  • 11) 0.039 062 528 × 2 = 0 + 0.078 125 056;
  • 12) 0.078 125 056 × 2 = 0 + 0.156 250 112;
  • 13) 0.156 250 112 × 2 = 0 + 0.312 500 224;
  • 14) 0.312 500 224 × 2 = 0 + 0.625 000 448;
  • 15) 0.625 000 448 × 2 = 1 + 0.250 000 896;
  • 16) 0.250 000 896 × 2 = 0 + 0.500 001 792;
  • 17) 0.500 001 792 × 2 = 1 + 0.000 003 584;
  • 18) 0.000 003 584 × 2 = 0 + 0.000 007 168;
  • 19) 0.000 007 168 × 2 = 0 + 0.000 014 336;
  • 20) 0.000 014 336 × 2 = 0 + 0.000 028 672;
  • 21) 0.000 028 672 × 2 = 0 + 0.000 057 344;
  • 22) 0.000 057 344 × 2 = 0 + 0.000 114 688;
  • 23) 0.000 114 688 × 2 = 0 + 0.000 229 376;
  • 24) 0.000 229 376 × 2 = 0 + 0.000 458 752;
  • 25) 0.000 458 752 × 2 = 0 + 0.000 917 504;
  • 26) 0.000 917 504 × 2 = 0 + 0.001 835 008;
  • 27) 0.001 835 008 × 2 = 0 + 0.003 670 016;
  • 28) 0.003 670 016 × 2 = 0 + 0.007 340 032;
  • 29) 0.007 340 032 × 2 = 0 + 0.014 680 064;
  • 30) 0.014 680 064 × 2 = 0 + 0.029 360 128;
  • 31) 0.029 360 128 × 2 = 0 + 0.058 720 256;
  • 32) 0.058 720 256 × 2 = 0 + 0.117 440 512;
  • 33) 0.117 440 512 × 2 = 0 + 0.234 881 024;
  • 34) 0.234 881 024 × 2 = 0 + 0.469 762 048;
  • 35) 0.469 762 048 × 2 = 0 + 0.939 524 096;
  • 36) 0.939 524 096 × 2 = 1 + 0.879 048 192;
  • 37) 0.879 048 192 × 2 = 1 + 0.758 096 384;
  • 38) 0.758 096 384 × 2 = 1 + 0.516 192 768;
  • 39) 0.516 192 768 × 2 = 1 + 0.032 385 536;
  • 40) 0.032 385 536 × 2 = 0 + 0.064 771 072;
  • 41) 0.064 771 072 × 2 = 0 + 0.129 542 144;
  • 42) 0.129 542 144 × 2 = 0 + 0.259 084 288;
  • 43) 0.259 084 288 × 2 = 0 + 0.518 168 576;
  • 44) 0.518 168 576 × 2 = 1 + 0.036 337 152;
  • 45) 0.036 337 152 × 2 = 0 + 0.072 674 304;
  • 46) 0.072 674 304 × 2 = 0 + 0.145 348 608;
  • 47) 0.145 348 608 × 2 = 0 + 0.290 697 216;
  • 48) 0.290 697 216 × 2 = 0 + 0.581 394 432;
  • 49) 0.581 394 432 × 2 = 1 + 0.162 788 864;
  • 50) 0.162 788 864 × 2 = 0 + 0.325 577 728;
  • 51) 0.325 577 728 × 2 = 0 + 0.651 155 456;
  • 52) 0.651 155 456 × 2 = 1 + 0.302 310 912;
  • 53) 0.302 310 912 × 2 = 0 + 0.604 621 824;
  • 54) 0.604 621 824 × 2 = 1 + 0.209 243 648;
  • 55) 0.209 243 648 × 2 = 0 + 0.418 487 296;
  • 56) 0.418 487 296 × 2 = 0 + 0.836 974 592;
  • 57) 0.836 974 592 × 2 = 1 + 0.673 949 184;
  • 58) 0.673 949 184 × 2 = 1 + 0.347 898 368;
  • 59) 0.347 898 368 × 2 = 0 + 0.695 796 736;
  • 60) 0.695 796 736 × 2 = 1 + 0.391 593 472;
  • 61) 0.391 593 472 × 2 = 0 + 0.783 186 944;
  • 62) 0.783 186 944 × 2 = 1 + 0.566 373 888;
  • 63) 0.566 373 888 × 2 = 1 + 0.132 747 776;
  • 64) 0.132 747 776 × 2 = 0 + 0.265 495 552;
  • 65) 0.265 495 552 × 2 = 0 + 0.530 991 104;
  • 66) 0.530 991 104 × 2 = 1 + 0.061 982 208;
  • 67) 0.061 982 208 × 2 = 0 + 0.123 964 416;

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:


0.000 038 147(10) =

0.0000 0000 0000 0010 1000 0000 0000 0000 0001 1110 0001 0000 1001 0100 1101 0110 010(2)

6. Positive number before normalization:

0.000 038 147(10) =

0.0000 0000 0000 0010 1000 0000 0000 0000 0001 1110 0001 0000 1001 0100 1101 0110 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 038 147(10) =

0.0000 0000 0000 0010 1000 0000 0000 0000 0001 1110 0001 0000 1001 0100 1101 0110 010(2) =

0.0000 0000 0000 0010 1000 0000 0000 0000 0001 1110 0001 0000 1001 0100 1101 0110 010(2) × 20 =

1.0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010(2) × 2-15


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): -15

Mantissa (not normalized):
1.0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010 =

0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010


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 0000

Mantissa (52 bits) =
0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010


Decimal number -0.000 038 147 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0100 0000 0000 0000 0000 1111 0000 1000 0100 1010 0110 1011 0010

» Convert decimal -0.000 038 156 to 64 bit double precision IEEE 754 binary floating point representation standard

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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