-0.145 067 813 487 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 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
-0.145 067 813 487 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:

|-0.145 067 813 487 4| = 0.145 067 813 487 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: 0.145 067 813 487 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) 0.145 067 813 487 4 × 2 = 0 + 0.290 135 626 974 8;
  • 2) 0.290 135 626 974 8 × 2 = 0 + 0.580 271 253 949 6;
  • 3) 0.580 271 253 949 6 × 2 = 1 + 0.160 542 507 899 2;
  • 4) 0.160 542 507 899 2 × 2 = 0 + 0.321 085 015 798 4;
  • 5) 0.321 085 015 798 4 × 2 = 0 + 0.642 170 031 596 8;
  • 6) 0.642 170 031 596 8 × 2 = 1 + 0.284 340 063 193 6;
  • 7) 0.284 340 063 193 6 × 2 = 0 + 0.568 680 126 387 2;
  • 8) 0.568 680 126 387 2 × 2 = 1 + 0.137 360 252 774 4;
  • 9) 0.137 360 252 774 4 × 2 = 0 + 0.274 720 505 548 8;
  • 10) 0.274 720 505 548 8 × 2 = 0 + 0.549 441 011 097 6;
  • 11) 0.549 441 011 097 6 × 2 = 1 + 0.098 882 022 195 2;
  • 12) 0.098 882 022 195 2 × 2 = 0 + 0.197 764 044 390 4;
  • 13) 0.197 764 044 390 4 × 2 = 0 + 0.395 528 088 780 8;
  • 14) 0.395 528 088 780 8 × 2 = 0 + 0.791 056 177 561 6;
  • 15) 0.791 056 177 561 6 × 2 = 1 + 0.582 112 355 123 2;
  • 16) 0.582 112 355 123 2 × 2 = 1 + 0.164 224 710 246 4;
  • 17) 0.164 224 710 246 4 × 2 = 0 + 0.328 449 420 492 8;
  • 18) 0.328 449 420 492 8 × 2 = 0 + 0.656 898 840 985 6;
  • 19) 0.656 898 840 985 6 × 2 = 1 + 0.313 797 681 971 2;
  • 20) 0.313 797 681 971 2 × 2 = 0 + 0.627 595 363 942 4;
  • 21) 0.627 595 363 942 4 × 2 = 1 + 0.255 190 727 884 8;
  • 22) 0.255 190 727 884 8 × 2 = 0 + 0.510 381 455 769 6;
  • 23) 0.510 381 455 769 6 × 2 = 1 + 0.020 762 911 539 2;
  • 24) 0.020 762 911 539 2 × 2 = 0 + 0.041 525 823 078 4;
  • 25) 0.041 525 823 078 4 × 2 = 0 + 0.083 051 646 156 8;
  • 26) 0.083 051 646 156 8 × 2 = 0 + 0.166 103 292 313 6;
  • 27) 0.166 103 292 313 6 × 2 = 0 + 0.332 206 584 627 2;
  • 28) 0.332 206 584 627 2 × 2 = 0 + 0.664 413 169 254 4;
  • 29) 0.664 413 169 254 4 × 2 = 1 + 0.328 826 338 508 8;
  • 30) 0.328 826 338 508 8 × 2 = 0 + 0.657 652 677 017 6;
  • 31) 0.657 652 677 017 6 × 2 = 1 + 0.315 305 354 035 2;
  • 32) 0.315 305 354 035 2 × 2 = 0 + 0.630 610 708 070 4;
  • 33) 0.630 610 708 070 4 × 2 = 1 + 0.261 221 416 140 8;
  • 34) 0.261 221 416 140 8 × 2 = 0 + 0.522 442 832 281 6;
  • 35) 0.522 442 832 281 6 × 2 = 1 + 0.044 885 664 563 2;
  • 36) 0.044 885 664 563 2 × 2 = 0 + 0.089 771 329 126 4;
  • 37) 0.089 771 329 126 4 × 2 = 0 + 0.179 542 658 252 8;
  • 38) 0.179 542 658 252 8 × 2 = 0 + 0.359 085 316 505 6;
  • 39) 0.359 085 316 505 6 × 2 = 0 + 0.718 170 633 011 2;
  • 40) 0.718 170 633 011 2 × 2 = 1 + 0.436 341 266 022 4;
  • 41) 0.436 341 266 022 4 × 2 = 0 + 0.872 682 532 044 8;
  • 42) 0.872 682 532 044 8 × 2 = 1 + 0.745 365 064 089 6;
  • 43) 0.745 365 064 089 6 × 2 = 1 + 0.490 730 128 179 2;
  • 44) 0.490 730 128 179 2 × 2 = 0 + 0.981 460 256 358 4;
  • 45) 0.981 460 256 358 4 × 2 = 1 + 0.962 920 512 716 8;
  • 46) 0.962 920 512 716 8 × 2 = 1 + 0.925 841 025 433 6;
  • 47) 0.925 841 025 433 6 × 2 = 1 + 0.851 682 050 867 2;
  • 48) 0.851 682 050 867 2 × 2 = 1 + 0.703 364 101 734 4;
  • 49) 0.703 364 101 734 4 × 2 = 1 + 0.406 728 203 468 8;
  • 50) 0.406 728 203 468 8 × 2 = 0 + 0.813 456 406 937 6;
  • 51) 0.813 456 406 937 6 × 2 = 1 + 0.626 912 813 875 2;
  • 52) 0.626 912 813 875 2 × 2 = 1 + 0.253 825 627 750 4;
  • 53) 0.253 825 627 750 4 × 2 = 0 + 0.507 651 255 500 8;
  • 54) 0.507 651 255 500 8 × 2 = 1 + 0.015 302 511 001 6;
  • 55) 0.015 302 511 001 6 × 2 = 0 + 0.030 605 022 003 2;

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:


0.145 067 813 487 4(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 0110 1111 1011 010(2)

6. Positive number before normalization:

0.145 067 813 487 4(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 0110 1111 1011 010(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:


0.145 067 813 487 4(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 0110 1111 1011 010(2) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 0110 1111 1011 010(2) × 20 =

1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010(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):
1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010


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. 0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010 =

0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010


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) =
0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010


Decimal number -0.145 067 813 487 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0010 1001 0001 1001 0101 0000 0101 0101 0000 1011 0111 1101 1010

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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