-0.000 001 324 478 226 374 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 324 478 226 374 11(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
-0.000 001 324 478 226 374 11(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:

|-0.000 001 324 478 226 374 11| = 0.000 001 324 478 226 374 11


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: 0.000 001 324 478 226 374 11.

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) 0.000 001 324 478 226 374 11 × 2 = 0 + 0.000 002 648 956 452 748 22;
  • 2) 0.000 002 648 956 452 748 22 × 2 = 0 + 0.000 005 297 912 905 496 44;
  • 3) 0.000 005 297 912 905 496 44 × 2 = 0 + 0.000 010 595 825 810 992 88;
  • 4) 0.000 010 595 825 810 992 88 × 2 = 0 + 0.000 021 191 651 621 985 76;
  • 5) 0.000 021 191 651 621 985 76 × 2 = 0 + 0.000 042 383 303 243 971 52;
  • 6) 0.000 042 383 303 243 971 52 × 2 = 0 + 0.000 084 766 606 487 943 04;
  • 7) 0.000 084 766 606 487 943 04 × 2 = 0 + 0.000 169 533 212 975 886 08;
  • 8) 0.000 169 533 212 975 886 08 × 2 = 0 + 0.000 339 066 425 951 772 16;
  • 9) 0.000 339 066 425 951 772 16 × 2 = 0 + 0.000 678 132 851 903 544 32;
  • 10) 0.000 678 132 851 903 544 32 × 2 = 0 + 0.001 356 265 703 807 088 64;
  • 11) 0.001 356 265 703 807 088 64 × 2 = 0 + 0.002 712 531 407 614 177 28;
  • 12) 0.002 712 531 407 614 177 28 × 2 = 0 + 0.005 425 062 815 228 354 56;
  • 13) 0.005 425 062 815 228 354 56 × 2 = 0 + 0.010 850 125 630 456 709 12;
  • 14) 0.010 850 125 630 456 709 12 × 2 = 0 + 0.021 700 251 260 913 418 24;
  • 15) 0.021 700 251 260 913 418 24 × 2 = 0 + 0.043 400 502 521 826 836 48;
  • 16) 0.043 400 502 521 826 836 48 × 2 = 0 + 0.086 801 005 043 653 672 96;
  • 17) 0.086 801 005 043 653 672 96 × 2 = 0 + 0.173 602 010 087 307 345 92;
  • 18) 0.173 602 010 087 307 345 92 × 2 = 0 + 0.347 204 020 174 614 691 84;
  • 19) 0.347 204 020 174 614 691 84 × 2 = 0 + 0.694 408 040 349 229 383 68;
  • 20) 0.694 408 040 349 229 383 68 × 2 = 1 + 0.388 816 080 698 458 767 36;
  • 21) 0.388 816 080 698 458 767 36 × 2 = 0 + 0.777 632 161 396 917 534 72;
  • 22) 0.777 632 161 396 917 534 72 × 2 = 1 + 0.555 264 322 793 835 069 44;
  • 23) 0.555 264 322 793 835 069 44 × 2 = 1 + 0.110 528 645 587 670 138 88;
  • 24) 0.110 528 645 587 670 138 88 × 2 = 0 + 0.221 057 291 175 340 277 76;
  • 25) 0.221 057 291 175 340 277 76 × 2 = 0 + 0.442 114 582 350 680 555 52;
  • 26) 0.442 114 582 350 680 555 52 × 2 = 0 + 0.884 229 164 701 361 111 04;
  • 27) 0.884 229 164 701 361 111 04 × 2 = 1 + 0.768 458 329 402 722 222 08;
  • 28) 0.768 458 329 402 722 222 08 × 2 = 1 + 0.536 916 658 805 444 444 16;
  • 29) 0.536 916 658 805 444 444 16 × 2 = 1 + 0.073 833 317 610 888 888 32;
  • 30) 0.073 833 317 610 888 888 32 × 2 = 0 + 0.147 666 635 221 777 776 64;
  • 31) 0.147 666 635 221 777 776 64 × 2 = 0 + 0.295 333 270 443 555 553 28;
  • 32) 0.295 333 270 443 555 553 28 × 2 = 0 + 0.590 666 540 887 111 106 56;
  • 33) 0.590 666 540 887 111 106 56 × 2 = 1 + 0.181 333 081 774 222 213 12;
  • 34) 0.181 333 081 774 222 213 12 × 2 = 0 + 0.362 666 163 548 444 426 24;
  • 35) 0.362 666 163 548 444 426 24 × 2 = 0 + 0.725 332 327 096 888 852 48;
  • 36) 0.725 332 327 096 888 852 48 × 2 = 1 + 0.450 664 654 193 777 704 96;
  • 37) 0.450 664 654 193 777 704 96 × 2 = 0 + 0.901 329 308 387 555 409 92;
  • 38) 0.901 329 308 387 555 409 92 × 2 = 1 + 0.802 658 616 775 110 819 84;
  • 39) 0.802 658 616 775 110 819 84 × 2 = 1 + 0.605 317 233 550 221 639 68;
  • 40) 0.605 317 233 550 221 639 68 × 2 = 1 + 0.210 634 467 100 443 279 36;
  • 41) 0.210 634 467 100 443 279 36 × 2 = 0 + 0.421 268 934 200 886 558 72;
  • 42) 0.421 268 934 200 886 558 72 × 2 = 0 + 0.842 537 868 401 773 117 44;
  • 43) 0.842 537 868 401 773 117 44 × 2 = 1 + 0.685 075 736 803 546 234 88;
  • 44) 0.685 075 736 803 546 234 88 × 2 = 1 + 0.370 151 473 607 092 469 76;
  • 45) 0.370 151 473 607 092 469 76 × 2 = 0 + 0.740 302 947 214 184 939 52;
  • 46) 0.740 302 947 214 184 939 52 × 2 = 1 + 0.480 605 894 428 369 879 04;
  • 47) 0.480 605 894 428 369 879 04 × 2 = 0 + 0.961 211 788 856 739 758 08;
  • 48) 0.961 211 788 856 739 758 08 × 2 = 1 + 0.922 423 577 713 479 516 16;
  • 49) 0.922 423 577 713 479 516 16 × 2 = 1 + 0.844 847 155 426 959 032 32;
  • 50) 0.844 847 155 426 959 032 32 × 2 = 1 + 0.689 694 310 853 918 064 64;
  • 51) 0.689 694 310 853 918 064 64 × 2 = 1 + 0.379 388 621 707 836 129 28;
  • 52) 0.379 388 621 707 836 129 28 × 2 = 0 + 0.758 777 243 415 672 258 56;
  • 53) 0.758 777 243 415 672 258 56 × 2 = 1 + 0.517 554 486 831 344 517 12;
  • 54) 0.517 554 486 831 344 517 12 × 2 = 1 + 0.035 108 973 662 689 034 24;
  • 55) 0.035 108 973 662 689 034 24 × 2 = 0 + 0.070 217 947 325 378 068 48;
  • 56) 0.070 217 947 325 378 068 48 × 2 = 0 + 0.140 435 894 650 756 136 96;
  • 57) 0.140 435 894 650 756 136 96 × 2 = 0 + 0.280 871 789 301 512 273 92;
  • 58) 0.280 871 789 301 512 273 92 × 2 = 0 + 0.561 743 578 603 024 547 84;
  • 59) 0.561 743 578 603 024 547 84 × 2 = 1 + 0.123 487 157 206 049 095 68;
  • 60) 0.123 487 157 206 049 095 68 × 2 = 0 + 0.246 974 314 412 098 191 36;
  • 61) 0.246 974 314 412 098 191 36 × 2 = 0 + 0.493 948 628 824 196 382 72;
  • 62) 0.493 948 628 824 196 382 72 × 2 = 0 + 0.987 897 257 648 392 765 44;
  • 63) 0.987 897 257 648 392 765 44 × 2 = 1 + 0.975 794 515 296 785 530 88;
  • 64) 0.975 794 515 296 785 530 88 × 2 = 1 + 0.951 589 030 593 571 061 76;
  • 65) 0.951 589 030 593 571 061 76 × 2 = 1 + 0.903 178 061 187 142 123 52;
  • 66) 0.903 178 061 187 142 123 52 × 2 = 1 + 0.806 356 122 374 284 247 04;
  • 67) 0.806 356 122 374 284 247 04 × 2 = 1 + 0.612 712 244 748 568 494 08;
  • 68) 0.612 712 244 748 568 494 08 × 2 = 1 + 0.225 424 489 497 136 988 16;
  • 69) 0.225 424 489 497 136 988 16 × 2 = 0 + 0.450 848 978 994 273 976 32;
  • 70) 0.450 848 978 994 273 976 32 × 2 = 0 + 0.901 697 957 988 547 952 64;
  • 71) 0.901 697 957 988 547 952 64 × 2 = 1 + 0.803 395 915 977 095 905 28;
  • 72) 0.803 395 915 977 095 905 28 × 2 = 1 + 0.606 791 831 954 191 810 56;

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:


0.000 001 324 478 226 374 11(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011(2)

6. Positive number before normalization:

0.000 001 324 478 226 374 11(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 324 478 226 374 11(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011(2) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011(2) × 20 =

1.0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011(2) × 2-20


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): -20

Mantissa (not normalized):
1.0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1003(10) =

011 1110 1011(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. 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011 =

0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011


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 1110 1011

Mantissa (52 bits) =
0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011


Decimal number -0.000 001 324 478 226 374 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 0011 1111 0011

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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