-0.145 067 813 487 901 050 789 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 789(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
-0.145 067 813 487 901 050 789(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:

|-0.145 067 813 487 901 050 789| = 0.145 067 813 487 901 050 789


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: 0.145 067 813 487 901 050 789.

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) 0.145 067 813 487 901 050 789 × 2 = 0 + 0.290 135 626 975 802 101 578;
  • 2) 0.290 135 626 975 802 101 578 × 2 = 0 + 0.580 271 253 951 604 203 156;
  • 3) 0.580 271 253 951 604 203 156 × 2 = 1 + 0.160 542 507 903 208 406 312;
  • 4) 0.160 542 507 903 208 406 312 × 2 = 0 + 0.321 085 015 806 416 812 624;
  • 5) 0.321 085 015 806 416 812 624 × 2 = 0 + 0.642 170 031 612 833 625 248;
  • 6) 0.642 170 031 612 833 625 248 × 2 = 1 + 0.284 340 063 225 667 250 496;
  • 7) 0.284 340 063 225 667 250 496 × 2 = 0 + 0.568 680 126 451 334 500 992;
  • 8) 0.568 680 126 451 334 500 992 × 2 = 1 + 0.137 360 252 902 669 001 984;
  • 9) 0.137 360 252 902 669 001 984 × 2 = 0 + 0.274 720 505 805 338 003 968;
  • 10) 0.274 720 505 805 338 003 968 × 2 = 0 + 0.549 441 011 610 676 007 936;
  • 11) 0.549 441 011 610 676 007 936 × 2 = 1 + 0.098 882 023 221 352 015 872;
  • 12) 0.098 882 023 221 352 015 872 × 2 = 0 + 0.197 764 046 442 704 031 744;
  • 13) 0.197 764 046 442 704 031 744 × 2 = 0 + 0.395 528 092 885 408 063 488;
  • 14) 0.395 528 092 885 408 063 488 × 2 = 0 + 0.791 056 185 770 816 126 976;
  • 15) 0.791 056 185 770 816 126 976 × 2 = 1 + 0.582 112 371 541 632 253 952;
  • 16) 0.582 112 371 541 632 253 952 × 2 = 1 + 0.164 224 743 083 264 507 904;
  • 17) 0.164 224 743 083 264 507 904 × 2 = 0 + 0.328 449 486 166 529 015 808;
  • 18) 0.328 449 486 166 529 015 808 × 2 = 0 + 0.656 898 972 333 058 031 616;
  • 19) 0.656 898 972 333 058 031 616 × 2 = 1 + 0.313 797 944 666 116 063 232;
  • 20) 0.313 797 944 666 116 063 232 × 2 = 0 + 0.627 595 889 332 232 126 464;
  • 21) 0.627 595 889 332 232 126 464 × 2 = 1 + 0.255 191 778 664 464 252 928;
  • 22) 0.255 191 778 664 464 252 928 × 2 = 0 + 0.510 383 557 328 928 505 856;
  • 23) 0.510 383 557 328 928 505 856 × 2 = 1 + 0.020 767 114 657 857 011 712;
  • 24) 0.020 767 114 657 857 011 712 × 2 = 0 + 0.041 534 229 315 714 023 424;
  • 25) 0.041 534 229 315 714 023 424 × 2 = 0 + 0.083 068 458 631 428 046 848;
  • 26) 0.083 068 458 631 428 046 848 × 2 = 0 + 0.166 136 917 262 856 093 696;
  • 27) 0.166 136 917 262 856 093 696 × 2 = 0 + 0.332 273 834 525 712 187 392;
  • 28) 0.332 273 834 525 712 187 392 × 2 = 0 + 0.664 547 669 051 424 374 784;
  • 29) 0.664 547 669 051 424 374 784 × 2 = 1 + 0.329 095 338 102 848 749 568;
  • 30) 0.329 095 338 102 848 749 568 × 2 = 0 + 0.658 190 676 205 697 499 136;
  • 31) 0.658 190 676 205 697 499 136 × 2 = 1 + 0.316 381 352 411 394 998 272;
  • 32) 0.316 381 352 411 394 998 272 × 2 = 0 + 0.632 762 704 822 789 996 544;
  • 33) 0.632 762 704 822 789 996 544 × 2 = 1 + 0.265 525 409 645 579 993 088;
  • 34) 0.265 525 409 645 579 993 088 × 2 = 0 + 0.531 050 819 291 159 986 176;
  • 35) 0.531 050 819 291 159 986 176 × 2 = 1 + 0.062 101 638 582 319 972 352;
  • 36) 0.062 101 638 582 319 972 352 × 2 = 0 + 0.124 203 277 164 639 944 704;
  • 37) 0.124 203 277 164 639 944 704 × 2 = 0 + 0.248 406 554 329 279 889 408;
  • 38) 0.248 406 554 329 279 889 408 × 2 = 0 + 0.496 813 108 658 559 778 816;
  • 39) 0.496 813 108 658 559 778 816 × 2 = 0 + 0.993 626 217 317 119 557 632;
  • 40) 0.993 626 217 317 119 557 632 × 2 = 1 + 0.987 252 434 634 239 115 264;
  • 41) 0.987 252 434 634 239 115 264 × 2 = 1 + 0.974 504 869 268 478 230 528;
  • 42) 0.974 504 869 268 478 230 528 × 2 = 1 + 0.949 009 738 536 956 461 056;
  • 43) 0.949 009 738 536 956 461 056 × 2 = 1 + 0.898 019 477 073 912 922 112;
  • 44) 0.898 019 477 073 912 922 112 × 2 = 1 + 0.796 038 954 147 825 844 224;
  • 45) 0.796 038 954 147 825 844 224 × 2 = 1 + 0.592 077 908 295 651 688 448;
  • 46) 0.592 077 908 295 651 688 448 × 2 = 1 + 0.184 155 816 591 303 376 896;
  • 47) 0.184 155 816 591 303 376 896 × 2 = 0 + 0.368 311 633 182 606 753 792;
  • 48) 0.368 311 633 182 606 753 792 × 2 = 0 + 0.736 623 266 365 213 507 584;
  • 49) 0.736 623 266 365 213 507 584 × 2 = 1 + 0.473 246 532 730 427 015 168;
  • 50) 0.473 246 532 730 427 015 168 × 2 = 0 + 0.946 493 065 460 854 030 336;
  • 51) 0.946 493 065 460 854 030 336 × 2 = 1 + 0.892 986 130 921 708 060 672;
  • 52) 0.892 986 130 921 708 060 672 × 2 = 1 + 0.785 972 261 843 416 121 344;
  • 53) 0.785 972 261 843 416 121 344 × 2 = 1 + 0.571 944 523 686 832 242 688;
  • 54) 0.571 944 523 686 832 242 688 × 2 = 1 + 0.143 889 047 373 664 485 376;
  • 55) 0.143 889 047 373 664 485 376 × 2 = 0 + 0.287 778 094 747 328 970 752;

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:


0.145 067 813 487 901 050 789(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 789(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:


0.145 067 813 487 901 050 789(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 -0.145 067 813 487 901 050 789 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