-0.145 067 813 487 901 050 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 89(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
-0.145 067 813 487 901 050 89(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:

|-0.145 067 813 487 901 050 89| = 0.145 067 813 487 901 050 89


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: 0.145 067 813 487 901 050 89.

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) 0.145 067 813 487 901 050 89 × 2 = 0 + 0.290 135 626 975 802 101 78;
  • 2) 0.290 135 626 975 802 101 78 × 2 = 0 + 0.580 271 253 951 604 203 56;
  • 3) 0.580 271 253 951 604 203 56 × 2 = 1 + 0.160 542 507 903 208 407 12;
  • 4) 0.160 542 507 903 208 407 12 × 2 = 0 + 0.321 085 015 806 416 814 24;
  • 5) 0.321 085 015 806 416 814 24 × 2 = 0 + 0.642 170 031 612 833 628 48;
  • 6) 0.642 170 031 612 833 628 48 × 2 = 1 + 0.284 340 063 225 667 256 96;
  • 7) 0.284 340 063 225 667 256 96 × 2 = 0 + 0.568 680 126 451 334 513 92;
  • 8) 0.568 680 126 451 334 513 92 × 2 = 1 + 0.137 360 252 902 669 027 84;
  • 9) 0.137 360 252 902 669 027 84 × 2 = 0 + 0.274 720 505 805 338 055 68;
  • 10) 0.274 720 505 805 338 055 68 × 2 = 0 + 0.549 441 011 610 676 111 36;
  • 11) 0.549 441 011 610 676 111 36 × 2 = 1 + 0.098 882 023 221 352 222 72;
  • 12) 0.098 882 023 221 352 222 72 × 2 = 0 + 0.197 764 046 442 704 445 44;
  • 13) 0.197 764 046 442 704 445 44 × 2 = 0 + 0.395 528 092 885 408 890 88;
  • 14) 0.395 528 092 885 408 890 88 × 2 = 0 + 0.791 056 185 770 817 781 76;
  • 15) 0.791 056 185 770 817 781 76 × 2 = 1 + 0.582 112 371 541 635 563 52;
  • 16) 0.582 112 371 541 635 563 52 × 2 = 1 + 0.164 224 743 083 271 127 04;
  • 17) 0.164 224 743 083 271 127 04 × 2 = 0 + 0.328 449 486 166 542 254 08;
  • 18) 0.328 449 486 166 542 254 08 × 2 = 0 + 0.656 898 972 333 084 508 16;
  • 19) 0.656 898 972 333 084 508 16 × 2 = 1 + 0.313 797 944 666 169 016 32;
  • 20) 0.313 797 944 666 169 016 32 × 2 = 0 + 0.627 595 889 332 338 032 64;
  • 21) 0.627 595 889 332 338 032 64 × 2 = 1 + 0.255 191 778 664 676 065 28;
  • 22) 0.255 191 778 664 676 065 28 × 2 = 0 + 0.510 383 557 329 352 130 56;
  • 23) 0.510 383 557 329 352 130 56 × 2 = 1 + 0.020 767 114 658 704 261 12;
  • 24) 0.020 767 114 658 704 261 12 × 2 = 0 + 0.041 534 229 317 408 522 24;
  • 25) 0.041 534 229 317 408 522 24 × 2 = 0 + 0.083 068 458 634 817 044 48;
  • 26) 0.083 068 458 634 817 044 48 × 2 = 0 + 0.166 136 917 269 634 088 96;
  • 27) 0.166 136 917 269 634 088 96 × 2 = 0 + 0.332 273 834 539 268 177 92;
  • 28) 0.332 273 834 539 268 177 92 × 2 = 0 + 0.664 547 669 078 536 355 84;
  • 29) 0.664 547 669 078 536 355 84 × 2 = 1 + 0.329 095 338 157 072 711 68;
  • 30) 0.329 095 338 157 072 711 68 × 2 = 0 + 0.658 190 676 314 145 423 36;
  • 31) 0.658 190 676 314 145 423 36 × 2 = 1 + 0.316 381 352 628 290 846 72;
  • 32) 0.316 381 352 628 290 846 72 × 2 = 0 + 0.632 762 705 256 581 693 44;
  • 33) 0.632 762 705 256 581 693 44 × 2 = 1 + 0.265 525 410 513 163 386 88;
  • 34) 0.265 525 410 513 163 386 88 × 2 = 0 + 0.531 050 821 026 326 773 76;
  • 35) 0.531 050 821 026 326 773 76 × 2 = 1 + 0.062 101 642 052 653 547 52;
  • 36) 0.062 101 642 052 653 547 52 × 2 = 0 + 0.124 203 284 105 307 095 04;
  • 37) 0.124 203 284 105 307 095 04 × 2 = 0 + 0.248 406 568 210 614 190 08;
  • 38) 0.248 406 568 210 614 190 08 × 2 = 0 + 0.496 813 136 421 228 380 16;
  • 39) 0.496 813 136 421 228 380 16 × 2 = 0 + 0.993 626 272 842 456 760 32;
  • 40) 0.993 626 272 842 456 760 32 × 2 = 1 + 0.987 252 545 684 913 520 64;
  • 41) 0.987 252 545 684 913 520 64 × 2 = 1 + 0.974 505 091 369 827 041 28;
  • 42) 0.974 505 091 369 827 041 28 × 2 = 1 + 0.949 010 182 739 654 082 56;
  • 43) 0.949 010 182 739 654 082 56 × 2 = 1 + 0.898 020 365 479 308 165 12;
  • 44) 0.898 020 365 479 308 165 12 × 2 = 1 + 0.796 040 730 958 616 330 24;
  • 45) 0.796 040 730 958 616 330 24 × 2 = 1 + 0.592 081 461 917 232 660 48;
  • 46) 0.592 081 461 917 232 660 48 × 2 = 1 + 0.184 162 923 834 465 320 96;
  • 47) 0.184 162 923 834 465 320 96 × 2 = 0 + 0.368 325 847 668 930 641 92;
  • 48) 0.368 325 847 668 930 641 92 × 2 = 0 + 0.736 651 695 337 861 283 84;
  • 49) 0.736 651 695 337 861 283 84 × 2 = 1 + 0.473 303 390 675 722 567 68;
  • 50) 0.473 303 390 675 722 567 68 × 2 = 0 + 0.946 606 781 351 445 135 36;
  • 51) 0.946 606 781 351 445 135 36 × 2 = 1 + 0.893 213 562 702 890 270 72;
  • 52) 0.893 213 562 702 890 270 72 × 2 = 1 + 0.786 427 125 405 780 541 44;
  • 53) 0.786 427 125 405 780 541 44 × 2 = 1 + 0.572 854 250 811 561 082 88;
  • 54) 0.572 854 250 811 561 082 88 × 2 = 1 + 0.145 708 501 623 122 165 76;
  • 55) 0.145 708 501 623 122 165 76 × 2 = 0 + 0.291 417 003 246 244 331 52;

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:


0.145 067 813 487 901 050 89(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

6. Positive number before normalization:

0.145 067 813 487 901 050 89(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(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:


0.145 067 813 487 901 050 89(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) × 20 =

1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110(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 1111 1110 0101 1110


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 1111 1110 0101 1110 =

0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


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 1111 1110 0101 1110


Decimal number -0.145 067 813 487 901 050 89 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 1111 1110 0101 1110

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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