-0.180 000 000 036 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 036 5(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
-0.180 000 000 036 5(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:

|-0.180 000 000 036 5| = 0.180 000 000 036 5


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: 0.180 000 000 036 5.

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) 0.180 000 000 036 5 × 2 = 0 + 0.360 000 000 073;
  • 2) 0.360 000 000 073 × 2 = 0 + 0.720 000 000 146;
  • 3) 0.720 000 000 146 × 2 = 1 + 0.440 000 000 292;
  • 4) 0.440 000 000 292 × 2 = 0 + 0.880 000 000 584;
  • 5) 0.880 000 000 584 × 2 = 1 + 0.760 000 001 168;
  • 6) 0.760 000 001 168 × 2 = 1 + 0.520 000 002 336;
  • 7) 0.520 000 002 336 × 2 = 1 + 0.040 000 004 672;
  • 8) 0.040 000 004 672 × 2 = 0 + 0.080 000 009 344;
  • 9) 0.080 000 009 344 × 2 = 0 + 0.160 000 018 688;
  • 10) 0.160 000 018 688 × 2 = 0 + 0.320 000 037 376;
  • 11) 0.320 000 037 376 × 2 = 0 + 0.640 000 074 752;
  • 12) 0.640 000 074 752 × 2 = 1 + 0.280 000 149 504;
  • 13) 0.280 000 149 504 × 2 = 0 + 0.560 000 299 008;
  • 14) 0.560 000 299 008 × 2 = 1 + 0.120 000 598 016;
  • 15) 0.120 000 598 016 × 2 = 0 + 0.240 001 196 032;
  • 16) 0.240 001 196 032 × 2 = 0 + 0.480 002 392 064;
  • 17) 0.480 002 392 064 × 2 = 0 + 0.960 004 784 128;
  • 18) 0.960 004 784 128 × 2 = 1 + 0.920 009 568 256;
  • 19) 0.920 009 568 256 × 2 = 1 + 0.840 019 136 512;
  • 20) 0.840 019 136 512 × 2 = 1 + 0.680 038 273 024;
  • 21) 0.680 038 273 024 × 2 = 1 + 0.360 076 546 048;
  • 22) 0.360 076 546 048 × 2 = 0 + 0.720 153 092 096;
  • 23) 0.720 153 092 096 × 2 = 1 + 0.440 306 184 192;
  • 24) 0.440 306 184 192 × 2 = 0 + 0.880 612 368 384;
  • 25) 0.880 612 368 384 × 2 = 1 + 0.761 224 736 768;
  • 26) 0.761 224 736 768 × 2 = 1 + 0.522 449 473 536;
  • 27) 0.522 449 473 536 × 2 = 1 + 0.044 898 947 072;
  • 28) 0.044 898 947 072 × 2 = 0 + 0.089 797 894 144;
  • 29) 0.089 797 894 144 × 2 = 0 + 0.179 595 788 288;
  • 30) 0.179 595 788 288 × 2 = 0 + 0.359 191 576 576;
  • 31) 0.359 191 576 576 × 2 = 0 + 0.718 383 153 152;
  • 32) 0.718 383 153 152 × 2 = 1 + 0.436 766 306 304;
  • 33) 0.436 766 306 304 × 2 = 0 + 0.873 532 612 608;
  • 34) 0.873 532 612 608 × 2 = 1 + 0.747 065 225 216;
  • 35) 0.747 065 225 216 × 2 = 1 + 0.494 130 450 432;
  • 36) 0.494 130 450 432 × 2 = 0 + 0.988 260 900 864;
  • 37) 0.988 260 900 864 × 2 = 1 + 0.976 521 801 728;
  • 38) 0.976 521 801 728 × 2 = 1 + 0.953 043 603 456;
  • 39) 0.953 043 603 456 × 2 = 1 + 0.906 087 206 912;
  • 40) 0.906 087 206 912 × 2 = 1 + 0.812 174 413 824;
  • 41) 0.812 174 413 824 × 2 = 1 + 0.624 348 827 648;
  • 42) 0.624 348 827 648 × 2 = 1 + 0.248 697 655 296;
  • 43) 0.248 697 655 296 × 2 = 0 + 0.497 395 310 592;
  • 44) 0.497 395 310 592 × 2 = 0 + 0.994 790 621 184;
  • 45) 0.994 790 621 184 × 2 = 1 + 0.989 581 242 368;
  • 46) 0.989 581 242 368 × 2 = 1 + 0.979 162 484 736;
  • 47) 0.979 162 484 736 × 2 = 1 + 0.958 324 969 472;
  • 48) 0.958 324 969 472 × 2 = 1 + 0.916 649 938 944;
  • 49) 0.916 649 938 944 × 2 = 1 + 0.833 299 877 888;
  • 50) 0.833 299 877 888 × 2 = 1 + 0.666 599 755 776;
  • 51) 0.666 599 755 776 × 2 = 1 + 0.333 199 511 552;
  • 52) 0.333 199 511 552 × 2 = 0 + 0.666 399 023 104;
  • 53) 0.666 399 023 104 × 2 = 1 + 0.332 798 046 208;
  • 54) 0.332 798 046 208 × 2 = 0 + 0.665 596 092 416;
  • 55) 0.665 596 092 416 × 2 = 1 + 0.331 192 184 832;

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:


0.180 000 000 036 5(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1111 1100 1111 1110 101(2)

6. Positive number before normalization:

0.180 000 000 036 5(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1111 1100 1111 1110 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.180 000 000 036 5(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1111 1100 1111 1110 101(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1111 1100 1111 1110 101(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101(2) × 2-3


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): -3

Mantissa (not normalized):
1.0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101 =

0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101


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 1100

Mantissa (52 bits) =
0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101


Decimal number -0.180 000 000 036 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 0111 1110 0111 1111 0101

» Convert decimal -0.180 000 000 036 2 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