-0.024 189 051 235 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.024 189 051 235 4(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
-0.024 189 051 235 4(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:

|-0.024 189 051 235 4| = 0.024 189 051 235 4


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: 0.024 189 051 235 4.

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) 0.024 189 051 235 4 × 2 = 0 + 0.048 378 102 470 8;
  • 2) 0.048 378 102 470 8 × 2 = 0 + 0.096 756 204 941 6;
  • 3) 0.096 756 204 941 6 × 2 = 0 + 0.193 512 409 883 2;
  • 4) 0.193 512 409 883 2 × 2 = 0 + 0.387 024 819 766 4;
  • 5) 0.387 024 819 766 4 × 2 = 0 + 0.774 049 639 532 8;
  • 6) 0.774 049 639 532 8 × 2 = 1 + 0.548 099 279 065 6;
  • 7) 0.548 099 279 065 6 × 2 = 1 + 0.096 198 558 131 2;
  • 8) 0.096 198 558 131 2 × 2 = 0 + 0.192 397 116 262 4;
  • 9) 0.192 397 116 262 4 × 2 = 0 + 0.384 794 232 524 8;
  • 10) 0.384 794 232 524 8 × 2 = 0 + 0.769 588 465 049 6;
  • 11) 0.769 588 465 049 6 × 2 = 1 + 0.539 176 930 099 2;
  • 12) 0.539 176 930 099 2 × 2 = 1 + 0.078 353 860 198 4;
  • 13) 0.078 353 860 198 4 × 2 = 0 + 0.156 707 720 396 8;
  • 14) 0.156 707 720 396 8 × 2 = 0 + 0.313 415 440 793 6;
  • 15) 0.313 415 440 793 6 × 2 = 0 + 0.626 830 881 587 2;
  • 16) 0.626 830 881 587 2 × 2 = 1 + 0.253 661 763 174 4;
  • 17) 0.253 661 763 174 4 × 2 = 0 + 0.507 323 526 348 8;
  • 18) 0.507 323 526 348 8 × 2 = 1 + 0.014 647 052 697 6;
  • 19) 0.014 647 052 697 6 × 2 = 0 + 0.029 294 105 395 2;
  • 20) 0.029 294 105 395 2 × 2 = 0 + 0.058 588 210 790 4;
  • 21) 0.058 588 210 790 4 × 2 = 0 + 0.117 176 421 580 8;
  • 22) 0.117 176 421 580 8 × 2 = 0 + 0.234 352 843 161 6;
  • 23) 0.234 352 843 161 6 × 2 = 0 + 0.468 705 686 323 2;
  • 24) 0.468 705 686 323 2 × 2 = 0 + 0.937 411 372 646 4;
  • 25) 0.937 411 372 646 4 × 2 = 1 + 0.874 822 745 292 8;
  • 26) 0.874 822 745 292 8 × 2 = 1 + 0.749 645 490 585 6;
  • 27) 0.749 645 490 585 6 × 2 = 1 + 0.499 290 981 171 2;
  • 28) 0.499 290 981 171 2 × 2 = 0 + 0.998 581 962 342 4;
  • 29) 0.998 581 962 342 4 × 2 = 1 + 0.997 163 924 684 8;
  • 30) 0.997 163 924 684 8 × 2 = 1 + 0.994 327 849 369 6;
  • 31) 0.994 327 849 369 6 × 2 = 1 + 0.988 655 698 739 2;
  • 32) 0.988 655 698 739 2 × 2 = 1 + 0.977 311 397 478 4;
  • 33) 0.977 311 397 478 4 × 2 = 1 + 0.954 622 794 956 8;
  • 34) 0.954 622 794 956 8 × 2 = 1 + 0.909 245 589 913 6;
  • 35) 0.909 245 589 913 6 × 2 = 1 + 0.818 491 179 827 2;
  • 36) 0.818 491 179 827 2 × 2 = 1 + 0.636 982 359 654 4;
  • 37) 0.636 982 359 654 4 × 2 = 1 + 0.273 964 719 308 8;
  • 38) 0.273 964 719 308 8 × 2 = 0 + 0.547 929 438 617 6;
  • 39) 0.547 929 438 617 6 × 2 = 1 + 0.095 858 877 235 2;
  • 40) 0.095 858 877 235 2 × 2 = 0 + 0.191 717 754 470 4;
  • 41) 0.191 717 754 470 4 × 2 = 0 + 0.383 435 508 940 8;
  • 42) 0.383 435 508 940 8 × 2 = 0 + 0.766 871 017 881 6;
  • 43) 0.766 871 017 881 6 × 2 = 1 + 0.533 742 035 763 2;
  • 44) 0.533 742 035 763 2 × 2 = 1 + 0.067 484 071 526 4;
  • 45) 0.067 484 071 526 4 × 2 = 0 + 0.134 968 143 052 8;
  • 46) 0.134 968 143 052 8 × 2 = 0 + 0.269 936 286 105 6;
  • 47) 0.269 936 286 105 6 × 2 = 0 + 0.539 872 572 211 2;
  • 48) 0.539 872 572 211 2 × 2 = 1 + 0.079 745 144 422 4;
  • 49) 0.079 745 144 422 4 × 2 = 0 + 0.159 490 288 844 8;
  • 50) 0.159 490 288 844 8 × 2 = 0 + 0.318 980 577 689 6;
  • 51) 0.318 980 577 689 6 × 2 = 0 + 0.637 961 155 379 2;
  • 52) 0.637 961 155 379 2 × 2 = 1 + 0.275 922 310 758 4;
  • 53) 0.275 922 310 758 4 × 2 = 0 + 0.551 844 621 516 8;
  • 54) 0.551 844 621 516 8 × 2 = 1 + 0.103 689 243 033 6;
  • 55) 0.103 689 243 033 6 × 2 = 0 + 0.207 378 486 067 2;
  • 56) 0.207 378 486 067 2 × 2 = 0 + 0.414 756 972 134 4;
  • 57) 0.414 756 972 134 4 × 2 = 0 + 0.829 513 944 268 8;
  • 58) 0.829 513 944 268 8 × 2 = 1 + 0.659 027 888 537 6;

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:


0.024 189 051 235 4(10) =

0.0000 0110 0011 0001 0100 0000 1110 1111 1111 1010 0011 0001 0001 0100 01(2)

6. Positive number before normalization:

0.024 189 051 235 4(10) =

0.0000 0110 0011 0001 0100 0000 1110 1111 1111 1010 0011 0001 0001 0100 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.024 189 051 235 4(10) =

0.0000 0110 0011 0001 0100 0000 1110 1111 1111 1010 0011 0001 0001 0100 01(2) =

0.0000 0110 0011 0001 0100 0000 1110 1111 1111 1010 0011 0001 0001 0100 01(2) × 20 =

1.1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 0101 0001(2) × 2-6


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): -6

Mantissa (not normalized):
1.1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 0101 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 0101 0001 =

1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 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 1001

Mantissa (52 bits) =
1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 0101 0001


Decimal number -0.024 189 051 235 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 1000 1100 0101 0000 0011 1011 1111 1110 1000 1100 0100 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