-0.145 067 813 487 901 050 861 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 861 01(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
-0.145 067 813 487 901 050 861 01(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:

|-0.145 067 813 487 901 050 861 01| = 0.145 067 813 487 901 050 861 01


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: 0.145 067 813 487 901 050 861 01.

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) 0.145 067 813 487 901 050 861 01 × 2 = 0 + 0.290 135 626 975 802 101 722 02;
  • 2) 0.290 135 626 975 802 101 722 02 × 2 = 0 + 0.580 271 253 951 604 203 444 04;
  • 3) 0.580 271 253 951 604 203 444 04 × 2 = 1 + 0.160 542 507 903 208 406 888 08;
  • 4) 0.160 542 507 903 208 406 888 08 × 2 = 0 + 0.321 085 015 806 416 813 776 16;
  • 5) 0.321 085 015 806 416 813 776 16 × 2 = 0 + 0.642 170 031 612 833 627 552 32;
  • 6) 0.642 170 031 612 833 627 552 32 × 2 = 1 + 0.284 340 063 225 667 255 104 64;
  • 7) 0.284 340 063 225 667 255 104 64 × 2 = 0 + 0.568 680 126 451 334 510 209 28;
  • 8) 0.568 680 126 451 334 510 209 28 × 2 = 1 + 0.137 360 252 902 669 020 418 56;
  • 9) 0.137 360 252 902 669 020 418 56 × 2 = 0 + 0.274 720 505 805 338 040 837 12;
  • 10) 0.274 720 505 805 338 040 837 12 × 2 = 0 + 0.549 441 011 610 676 081 674 24;
  • 11) 0.549 441 011 610 676 081 674 24 × 2 = 1 + 0.098 882 023 221 352 163 348 48;
  • 12) 0.098 882 023 221 352 163 348 48 × 2 = 0 + 0.197 764 046 442 704 326 696 96;
  • 13) 0.197 764 046 442 704 326 696 96 × 2 = 0 + 0.395 528 092 885 408 653 393 92;
  • 14) 0.395 528 092 885 408 653 393 92 × 2 = 0 + 0.791 056 185 770 817 306 787 84;
  • 15) 0.791 056 185 770 817 306 787 84 × 2 = 1 + 0.582 112 371 541 634 613 575 68;
  • 16) 0.582 112 371 541 634 613 575 68 × 2 = 1 + 0.164 224 743 083 269 227 151 36;
  • 17) 0.164 224 743 083 269 227 151 36 × 2 = 0 + 0.328 449 486 166 538 454 302 72;
  • 18) 0.328 449 486 166 538 454 302 72 × 2 = 0 + 0.656 898 972 333 076 908 605 44;
  • 19) 0.656 898 972 333 076 908 605 44 × 2 = 1 + 0.313 797 944 666 153 817 210 88;
  • 20) 0.313 797 944 666 153 817 210 88 × 2 = 0 + 0.627 595 889 332 307 634 421 76;
  • 21) 0.627 595 889 332 307 634 421 76 × 2 = 1 + 0.255 191 778 664 615 268 843 52;
  • 22) 0.255 191 778 664 615 268 843 52 × 2 = 0 + 0.510 383 557 329 230 537 687 04;
  • 23) 0.510 383 557 329 230 537 687 04 × 2 = 1 + 0.020 767 114 658 461 075 374 08;
  • 24) 0.020 767 114 658 461 075 374 08 × 2 = 0 + 0.041 534 229 316 922 150 748 16;
  • 25) 0.041 534 229 316 922 150 748 16 × 2 = 0 + 0.083 068 458 633 844 301 496 32;
  • 26) 0.083 068 458 633 844 301 496 32 × 2 = 0 + 0.166 136 917 267 688 602 992 64;
  • 27) 0.166 136 917 267 688 602 992 64 × 2 = 0 + 0.332 273 834 535 377 205 985 28;
  • 28) 0.332 273 834 535 377 205 985 28 × 2 = 0 + 0.664 547 669 070 754 411 970 56;
  • 29) 0.664 547 669 070 754 411 970 56 × 2 = 1 + 0.329 095 338 141 508 823 941 12;
  • 30) 0.329 095 338 141 508 823 941 12 × 2 = 0 + 0.658 190 676 283 017 647 882 24;
  • 31) 0.658 190 676 283 017 647 882 24 × 2 = 1 + 0.316 381 352 566 035 295 764 48;
  • 32) 0.316 381 352 566 035 295 764 48 × 2 = 0 + 0.632 762 705 132 070 591 528 96;
  • 33) 0.632 762 705 132 070 591 528 96 × 2 = 1 + 0.265 525 410 264 141 183 057 92;
  • 34) 0.265 525 410 264 141 183 057 92 × 2 = 0 + 0.531 050 820 528 282 366 115 84;
  • 35) 0.531 050 820 528 282 366 115 84 × 2 = 1 + 0.062 101 641 056 564 732 231 68;
  • 36) 0.062 101 641 056 564 732 231 68 × 2 = 0 + 0.124 203 282 113 129 464 463 36;
  • 37) 0.124 203 282 113 129 464 463 36 × 2 = 0 + 0.248 406 564 226 258 928 926 72;
  • 38) 0.248 406 564 226 258 928 926 72 × 2 = 0 + 0.496 813 128 452 517 857 853 44;
  • 39) 0.496 813 128 452 517 857 853 44 × 2 = 0 + 0.993 626 256 905 035 715 706 88;
  • 40) 0.993 626 256 905 035 715 706 88 × 2 = 1 + 0.987 252 513 810 071 431 413 76;
  • 41) 0.987 252 513 810 071 431 413 76 × 2 = 1 + 0.974 505 027 620 142 862 827 52;
  • 42) 0.974 505 027 620 142 862 827 52 × 2 = 1 + 0.949 010 055 240 285 725 655 04;
  • 43) 0.949 010 055 240 285 725 655 04 × 2 = 1 + 0.898 020 110 480 571 451 310 08;
  • 44) 0.898 020 110 480 571 451 310 08 × 2 = 1 + 0.796 040 220 961 142 902 620 16;
  • 45) 0.796 040 220 961 142 902 620 16 × 2 = 1 + 0.592 080 441 922 285 805 240 32;
  • 46) 0.592 080 441 922 285 805 240 32 × 2 = 1 + 0.184 160 883 844 571 610 480 64;
  • 47) 0.184 160 883 844 571 610 480 64 × 2 = 0 + 0.368 321 767 689 143 220 961 28;
  • 48) 0.368 321 767 689 143 220 961 28 × 2 = 0 + 0.736 643 535 378 286 441 922 56;
  • 49) 0.736 643 535 378 286 441 922 56 × 2 = 1 + 0.473 287 070 756 572 883 845 12;
  • 50) 0.473 287 070 756 572 883 845 12 × 2 = 0 + 0.946 574 141 513 145 767 690 24;
  • 51) 0.946 574 141 513 145 767 690 24 × 2 = 1 + 0.893 148 283 026 291 535 380 48;
  • 52) 0.893 148 283 026 291 535 380 48 × 2 = 1 + 0.786 296 566 052 583 070 760 96;
  • 53) 0.786 296 566 052 583 070 760 96 × 2 = 1 + 0.572 593 132 105 166 141 521 92;
  • 54) 0.572 593 132 105 166 141 521 92 × 2 = 1 + 0.145 186 264 210 332 283 043 84;
  • 55) 0.145 186 264 210 332 283 043 84 × 2 = 0 + 0.290 372 528 420 664 566 087 68;

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:


0.145 067 813 487 901 050 861 01(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 861 01(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:


0.145 067 813 487 901 050 861 01(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 -0.145 067 813 487 901 050 861 01 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