-0.080 000 007 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.080 000 007(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
-0.080 000 007(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:

|-0.080 000 007| = 0.080 000 007


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: 0.080 000 007.

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) 0.080 000 007 × 2 = 0 + 0.160 000 014;
  • 2) 0.160 000 014 × 2 = 0 + 0.320 000 028;
  • 3) 0.320 000 028 × 2 = 0 + 0.640 000 056;
  • 4) 0.640 000 056 × 2 = 1 + 0.280 000 112;
  • 5) 0.280 000 112 × 2 = 0 + 0.560 000 224;
  • 6) 0.560 000 224 × 2 = 1 + 0.120 000 448;
  • 7) 0.120 000 448 × 2 = 0 + 0.240 000 896;
  • 8) 0.240 000 896 × 2 = 0 + 0.480 001 792;
  • 9) 0.480 001 792 × 2 = 0 + 0.960 003 584;
  • 10) 0.960 003 584 × 2 = 1 + 0.920 007 168;
  • 11) 0.920 007 168 × 2 = 1 + 0.840 014 336;
  • 12) 0.840 014 336 × 2 = 1 + 0.680 028 672;
  • 13) 0.680 028 672 × 2 = 1 + 0.360 057 344;
  • 14) 0.360 057 344 × 2 = 0 + 0.720 114 688;
  • 15) 0.720 114 688 × 2 = 1 + 0.440 229 376;
  • 16) 0.440 229 376 × 2 = 0 + 0.880 458 752;
  • 17) 0.880 458 752 × 2 = 1 + 0.760 917 504;
  • 18) 0.760 917 504 × 2 = 1 + 0.521 835 008;
  • 19) 0.521 835 008 × 2 = 1 + 0.043 670 016;
  • 20) 0.043 670 016 × 2 = 0 + 0.087 340 032;
  • 21) 0.087 340 032 × 2 = 0 + 0.174 680 064;
  • 22) 0.174 680 064 × 2 = 0 + 0.349 360 128;
  • 23) 0.349 360 128 × 2 = 0 + 0.698 720 256;
  • 24) 0.698 720 256 × 2 = 1 + 0.397 440 512;
  • 25) 0.397 440 512 × 2 = 0 + 0.794 881 024;
  • 26) 0.794 881 024 × 2 = 1 + 0.589 762 048;
  • 27) 0.589 762 048 × 2 = 1 + 0.179 524 096;
  • 28) 0.179 524 096 × 2 = 0 + 0.359 048 192;
  • 29) 0.359 048 192 × 2 = 0 + 0.718 096 384;
  • 30) 0.718 096 384 × 2 = 1 + 0.436 192 768;
  • 31) 0.436 192 768 × 2 = 0 + 0.872 385 536;
  • 32) 0.872 385 536 × 2 = 1 + 0.744 771 072;
  • 33) 0.744 771 072 × 2 = 1 + 0.489 542 144;
  • 34) 0.489 542 144 × 2 = 0 + 0.979 084 288;
  • 35) 0.979 084 288 × 2 = 1 + 0.958 168 576;
  • 36) 0.958 168 576 × 2 = 1 + 0.916 337 152;
  • 37) 0.916 337 152 × 2 = 1 + 0.832 674 304;
  • 38) 0.832 674 304 × 2 = 1 + 0.665 348 608;
  • 39) 0.665 348 608 × 2 = 1 + 0.330 697 216;
  • 40) 0.330 697 216 × 2 = 0 + 0.661 394 432;
  • 41) 0.661 394 432 × 2 = 1 + 0.322 788 864;
  • 42) 0.322 788 864 × 2 = 0 + 0.645 577 728;
  • 43) 0.645 577 728 × 2 = 1 + 0.291 155 456;
  • 44) 0.291 155 456 × 2 = 0 + 0.582 310 912;
  • 45) 0.582 310 912 × 2 = 1 + 0.164 621 824;
  • 46) 0.164 621 824 × 2 = 0 + 0.329 243 648;
  • 47) 0.329 243 648 × 2 = 0 + 0.658 487 296;
  • 48) 0.658 487 296 × 2 = 1 + 0.316 974 592;
  • 49) 0.316 974 592 × 2 = 0 + 0.633 949 184;
  • 50) 0.633 949 184 × 2 = 1 + 0.267 898 368;
  • 51) 0.267 898 368 × 2 = 0 + 0.535 796 736;
  • 52) 0.535 796 736 × 2 = 1 + 0.071 593 472;
  • 53) 0.071 593 472 × 2 = 0 + 0.143 186 944;
  • 54) 0.143 186 944 × 2 = 0 + 0.286 373 888;
  • 55) 0.286 373 888 × 2 = 0 + 0.572 747 776;
  • 56) 0.572 747 776 × 2 = 1 + 0.145 495 552;

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:


0.080 000 007(10) =

0.0001 0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001(2)

6. Positive number before normalization:

0.080 000 007(10) =

0.0001 0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.080 000 007(10) =

0.0001 0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001(2) =

0.0001 0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001(2) × 20 =

1.0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001(2) × 2-4


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):
1.0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 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) =

1019(10) =

011 1111 1011(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 0110 0101 1011 1110 1010 1001 0101 0001 =

0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001


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 1011

Mantissa (52 bits) =
0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001


Decimal number -0.080 000 007 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0100 0111 1010 1110 0001 0110 0101 1011 1110 1010 1001 0101 0001

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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