-190.296 199 409 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -190.296 199 409(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
-190.296 199 409(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:

|-190.296 199 409| = 190.296 199 409


2. First, convert to binary (in base 2) the integer part: 190.
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;
  • 190 ÷ 2 = 95 + 0;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

190(10) =

1011 1110(2)


4. Convert to binary (base 2) the fractional part: 0.296 199 409.

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) 0.296 199 409 × 2 = 0 + 0.592 398 818;
  • 2) 0.592 398 818 × 2 = 1 + 0.184 797 636;
  • 3) 0.184 797 636 × 2 = 0 + 0.369 595 272;
  • 4) 0.369 595 272 × 2 = 0 + 0.739 190 544;
  • 5) 0.739 190 544 × 2 = 1 + 0.478 381 088;
  • 6) 0.478 381 088 × 2 = 0 + 0.956 762 176;
  • 7) 0.956 762 176 × 2 = 1 + 0.913 524 352;
  • 8) 0.913 524 352 × 2 = 1 + 0.827 048 704;
  • 9) 0.827 048 704 × 2 = 1 + 0.654 097 408;
  • 10) 0.654 097 408 × 2 = 1 + 0.308 194 816;
  • 11) 0.308 194 816 × 2 = 0 + 0.616 389 632;
  • 12) 0.616 389 632 × 2 = 1 + 0.232 779 264;
  • 13) 0.232 779 264 × 2 = 0 + 0.465 558 528;
  • 14) 0.465 558 528 × 2 = 0 + 0.931 117 056;
  • 15) 0.931 117 056 × 2 = 1 + 0.862 234 112;
  • 16) 0.862 234 112 × 2 = 1 + 0.724 468 224;
  • 17) 0.724 468 224 × 2 = 1 + 0.448 936 448;
  • 18) 0.448 936 448 × 2 = 0 + 0.897 872 896;
  • 19) 0.897 872 896 × 2 = 1 + 0.795 745 792;
  • 20) 0.795 745 792 × 2 = 1 + 0.591 491 584;
  • 21) 0.591 491 584 × 2 = 1 + 0.182 983 168;
  • 22) 0.182 983 168 × 2 = 0 + 0.365 966 336;
  • 23) 0.365 966 336 × 2 = 0 + 0.731 932 672;
  • 24) 0.731 932 672 × 2 = 1 + 0.463 865 344;
  • 25) 0.463 865 344 × 2 = 0 + 0.927 730 688;
  • 26) 0.927 730 688 × 2 = 1 + 0.855 461 376;
  • 27) 0.855 461 376 × 2 = 1 + 0.710 922 752;
  • 28) 0.710 922 752 × 2 = 1 + 0.421 845 504;
  • 29) 0.421 845 504 × 2 = 0 + 0.843 691 008;
  • 30) 0.843 691 008 × 2 = 1 + 0.687 382 016;
  • 31) 0.687 382 016 × 2 = 1 + 0.374 764 032;
  • 32) 0.374 764 032 × 2 = 0 + 0.749 528 064;
  • 33) 0.749 528 064 × 2 = 1 + 0.499 056 128;
  • 34) 0.499 056 128 × 2 = 0 + 0.998 112 256;
  • 35) 0.998 112 256 × 2 = 1 + 0.996 224 512;
  • 36) 0.996 224 512 × 2 = 1 + 0.992 449 024;
  • 37) 0.992 449 024 × 2 = 1 + 0.984 898 048;
  • 38) 0.984 898 048 × 2 = 1 + 0.969 796 096;
  • 39) 0.969 796 096 × 2 = 1 + 0.939 592 192;
  • 40) 0.939 592 192 × 2 = 1 + 0.879 184 384;
  • 41) 0.879 184 384 × 2 = 1 + 0.758 368 768;
  • 42) 0.758 368 768 × 2 = 1 + 0.516 737 536;
  • 43) 0.516 737 536 × 2 = 1 + 0.033 475 072;
  • 44) 0.033 475 072 × 2 = 0 + 0.066 950 144;
  • 45) 0.066 950 144 × 2 = 0 + 0.133 900 288;
  • 46) 0.133 900 288 × 2 = 0 + 0.267 800 576;
  • 47) 0.267 800 576 × 2 = 0 + 0.535 601 152;
  • 48) 0.535 601 152 × 2 = 1 + 0.071 202 304;
  • 49) 0.071 202 304 × 2 = 0 + 0.142 404 608;
  • 50) 0.142 404 608 × 2 = 0 + 0.284 809 216;
  • 51) 0.284 809 216 × 2 = 0 + 0.569 618 432;
  • 52) 0.569 618 432 × 2 = 1 + 0.139 236 864;
  • 53) 0.139 236 864 × 2 = 0 + 0.278 473 728;

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:


0.296 199 409(10) =

0.0100 1011 1101 0011 1011 1001 0111 0110 1011 1111 1110 0001 0001 0(2)

6. Positive number before normalization:

190.296 199 409(10) =

1011 1110.0100 1011 1101 0011 1011 1001 0111 0110 1011 1111 1110 0001 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


190.296 199 409(10) =

1011 1110.0100 1011 1101 0011 1011 1001 0111 0110 1011 1111 1110 0001 0001 0(2) =

1011 1110.0100 1011 1101 0011 1011 1001 0111 0110 1011 1111 1110 0001 0001 0(2) × 20 =

1.0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 1100 0010 0010(2) × 27


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): 7

Mantissa (not normalized):
1.0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 1100 0010 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1030(10) =

100 0000 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 1100 0010 0010 =

0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 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) =
100 0000 0110

Mantissa (52 bits) =
0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 1100


Decimal number -190.296 199 409 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0110 - 0111 1100 1001 0111 1010 0111 0111 0010 1110 1101 0111 1111 1100

» Convert decimal -190.296 199 395 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