-0.145 067 813 487 901 050 713 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 713(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
-0.145 067 813 487 901 050 713(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:

|-0.145 067 813 487 901 050 713| = 0.145 067 813 487 901 050 713


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: 0.145 067 813 487 901 050 713.

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) 0.145 067 813 487 901 050 713 × 2 = 0 + 0.290 135 626 975 802 101 426;
  • 2) 0.290 135 626 975 802 101 426 × 2 = 0 + 0.580 271 253 951 604 202 852;
  • 3) 0.580 271 253 951 604 202 852 × 2 = 1 + 0.160 542 507 903 208 405 704;
  • 4) 0.160 542 507 903 208 405 704 × 2 = 0 + 0.321 085 015 806 416 811 408;
  • 5) 0.321 085 015 806 416 811 408 × 2 = 0 + 0.642 170 031 612 833 622 816;
  • 6) 0.642 170 031 612 833 622 816 × 2 = 1 + 0.284 340 063 225 667 245 632;
  • 7) 0.284 340 063 225 667 245 632 × 2 = 0 + 0.568 680 126 451 334 491 264;
  • 8) 0.568 680 126 451 334 491 264 × 2 = 1 + 0.137 360 252 902 668 982 528;
  • 9) 0.137 360 252 902 668 982 528 × 2 = 0 + 0.274 720 505 805 337 965 056;
  • 10) 0.274 720 505 805 337 965 056 × 2 = 0 + 0.549 441 011 610 675 930 112;
  • 11) 0.549 441 011 610 675 930 112 × 2 = 1 + 0.098 882 023 221 351 860 224;
  • 12) 0.098 882 023 221 351 860 224 × 2 = 0 + 0.197 764 046 442 703 720 448;
  • 13) 0.197 764 046 442 703 720 448 × 2 = 0 + 0.395 528 092 885 407 440 896;
  • 14) 0.395 528 092 885 407 440 896 × 2 = 0 + 0.791 056 185 770 814 881 792;
  • 15) 0.791 056 185 770 814 881 792 × 2 = 1 + 0.582 112 371 541 629 763 584;
  • 16) 0.582 112 371 541 629 763 584 × 2 = 1 + 0.164 224 743 083 259 527 168;
  • 17) 0.164 224 743 083 259 527 168 × 2 = 0 + 0.328 449 486 166 519 054 336;
  • 18) 0.328 449 486 166 519 054 336 × 2 = 0 + 0.656 898 972 333 038 108 672;
  • 19) 0.656 898 972 333 038 108 672 × 2 = 1 + 0.313 797 944 666 076 217 344;
  • 20) 0.313 797 944 666 076 217 344 × 2 = 0 + 0.627 595 889 332 152 434 688;
  • 21) 0.627 595 889 332 152 434 688 × 2 = 1 + 0.255 191 778 664 304 869 376;
  • 22) 0.255 191 778 664 304 869 376 × 2 = 0 + 0.510 383 557 328 609 738 752;
  • 23) 0.510 383 557 328 609 738 752 × 2 = 1 + 0.020 767 114 657 219 477 504;
  • 24) 0.020 767 114 657 219 477 504 × 2 = 0 + 0.041 534 229 314 438 955 008;
  • 25) 0.041 534 229 314 438 955 008 × 2 = 0 + 0.083 068 458 628 877 910 016;
  • 26) 0.083 068 458 628 877 910 016 × 2 = 0 + 0.166 136 917 257 755 820 032;
  • 27) 0.166 136 917 257 755 820 032 × 2 = 0 + 0.332 273 834 515 511 640 064;
  • 28) 0.332 273 834 515 511 640 064 × 2 = 0 + 0.664 547 669 031 023 280 128;
  • 29) 0.664 547 669 031 023 280 128 × 2 = 1 + 0.329 095 338 062 046 560 256;
  • 30) 0.329 095 338 062 046 560 256 × 2 = 0 + 0.658 190 676 124 093 120 512;
  • 31) 0.658 190 676 124 093 120 512 × 2 = 1 + 0.316 381 352 248 186 241 024;
  • 32) 0.316 381 352 248 186 241 024 × 2 = 0 + 0.632 762 704 496 372 482 048;
  • 33) 0.632 762 704 496 372 482 048 × 2 = 1 + 0.265 525 408 992 744 964 096;
  • 34) 0.265 525 408 992 744 964 096 × 2 = 0 + 0.531 050 817 985 489 928 192;
  • 35) 0.531 050 817 985 489 928 192 × 2 = 1 + 0.062 101 635 970 979 856 384;
  • 36) 0.062 101 635 970 979 856 384 × 2 = 0 + 0.124 203 271 941 959 712 768;
  • 37) 0.124 203 271 941 959 712 768 × 2 = 0 + 0.248 406 543 883 919 425 536;
  • 38) 0.248 406 543 883 919 425 536 × 2 = 0 + 0.496 813 087 767 838 851 072;
  • 39) 0.496 813 087 767 838 851 072 × 2 = 0 + 0.993 626 175 535 677 702 144;
  • 40) 0.993 626 175 535 677 702 144 × 2 = 1 + 0.987 252 351 071 355 404 288;
  • 41) 0.987 252 351 071 355 404 288 × 2 = 1 + 0.974 504 702 142 710 808 576;
  • 42) 0.974 504 702 142 710 808 576 × 2 = 1 + 0.949 009 404 285 421 617 152;
  • 43) 0.949 009 404 285 421 617 152 × 2 = 1 + 0.898 018 808 570 843 234 304;
  • 44) 0.898 018 808 570 843 234 304 × 2 = 1 + 0.796 037 617 141 686 468 608;
  • 45) 0.796 037 617 141 686 468 608 × 2 = 1 + 0.592 075 234 283 372 937 216;
  • 46) 0.592 075 234 283 372 937 216 × 2 = 1 + 0.184 150 468 566 745 874 432;
  • 47) 0.184 150 468 566 745 874 432 × 2 = 0 + 0.368 300 937 133 491 748 864;
  • 48) 0.368 300 937 133 491 748 864 × 2 = 0 + 0.736 601 874 266 983 497 728;
  • 49) 0.736 601 874 266 983 497 728 × 2 = 1 + 0.473 203 748 533 966 995 456;
  • 50) 0.473 203 748 533 966 995 456 × 2 = 0 + 0.946 407 497 067 933 990 912;
  • 51) 0.946 407 497 067 933 990 912 × 2 = 1 + 0.892 814 994 135 867 981 824;
  • 52) 0.892 814 994 135 867 981 824 × 2 = 1 + 0.785 629 988 271 735 963 648;
  • 53) 0.785 629 988 271 735 963 648 × 2 = 1 + 0.571 259 976 543 471 927 296;
  • 54) 0.571 259 976 543 471 927 296 × 2 = 1 + 0.142 519 953 086 943 854 592;
  • 55) 0.142 519 953 086 943 854 592 × 2 = 0 + 0.285 039 906 173 887 709 184;

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:


0.145 067 813 487 901 050 713(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 713(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:


0.145 067 813 487 901 050 713(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 -0.145 067 813 487 901 050 713 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