-0.000 001 324 478 226 374 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 324 478 226 374 47(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 47(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 47| = 0.000 001 324 478 226 374 47


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 47.

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 47 × 2 = 0 + 0.000 002 648 956 452 748 94;
  • 2) 0.000 002 648 956 452 748 94 × 2 = 0 + 0.000 005 297 912 905 497 88;
  • 3) 0.000 005 297 912 905 497 88 × 2 = 0 + 0.000 010 595 825 810 995 76;
  • 4) 0.000 010 595 825 810 995 76 × 2 = 0 + 0.000 021 191 651 621 991 52;
  • 5) 0.000 021 191 651 621 991 52 × 2 = 0 + 0.000 042 383 303 243 983 04;
  • 6) 0.000 042 383 303 243 983 04 × 2 = 0 + 0.000 084 766 606 487 966 08;
  • 7) 0.000 084 766 606 487 966 08 × 2 = 0 + 0.000 169 533 212 975 932 16;
  • 8) 0.000 169 533 212 975 932 16 × 2 = 0 + 0.000 339 066 425 951 864 32;
  • 9) 0.000 339 066 425 951 864 32 × 2 = 0 + 0.000 678 132 851 903 728 64;
  • 10) 0.000 678 132 851 903 728 64 × 2 = 0 + 0.001 356 265 703 807 457 28;
  • 11) 0.001 356 265 703 807 457 28 × 2 = 0 + 0.002 712 531 407 614 914 56;
  • 12) 0.002 712 531 407 614 914 56 × 2 = 0 + 0.005 425 062 815 229 829 12;
  • 13) 0.005 425 062 815 229 829 12 × 2 = 0 + 0.010 850 125 630 459 658 24;
  • 14) 0.010 850 125 630 459 658 24 × 2 = 0 + 0.021 700 251 260 919 316 48;
  • 15) 0.021 700 251 260 919 316 48 × 2 = 0 + 0.043 400 502 521 838 632 96;
  • 16) 0.043 400 502 521 838 632 96 × 2 = 0 + 0.086 801 005 043 677 265 92;
  • 17) 0.086 801 005 043 677 265 92 × 2 = 0 + 0.173 602 010 087 354 531 84;
  • 18) 0.173 602 010 087 354 531 84 × 2 = 0 + 0.347 204 020 174 709 063 68;
  • 19) 0.347 204 020 174 709 063 68 × 2 = 0 + 0.694 408 040 349 418 127 36;
  • 20) 0.694 408 040 349 418 127 36 × 2 = 1 + 0.388 816 080 698 836 254 72;
  • 21) 0.388 816 080 698 836 254 72 × 2 = 0 + 0.777 632 161 397 672 509 44;
  • 22) 0.777 632 161 397 672 509 44 × 2 = 1 + 0.555 264 322 795 345 018 88;
  • 23) 0.555 264 322 795 345 018 88 × 2 = 1 + 0.110 528 645 590 690 037 76;
  • 24) 0.110 528 645 590 690 037 76 × 2 = 0 + 0.221 057 291 181 380 075 52;
  • 25) 0.221 057 291 181 380 075 52 × 2 = 0 + 0.442 114 582 362 760 151 04;
  • 26) 0.442 114 582 362 760 151 04 × 2 = 0 + 0.884 229 164 725 520 302 08;
  • 27) 0.884 229 164 725 520 302 08 × 2 = 1 + 0.768 458 329 451 040 604 16;
  • 28) 0.768 458 329 451 040 604 16 × 2 = 1 + 0.536 916 658 902 081 208 32;
  • 29) 0.536 916 658 902 081 208 32 × 2 = 1 + 0.073 833 317 804 162 416 64;
  • 30) 0.073 833 317 804 162 416 64 × 2 = 0 + 0.147 666 635 608 324 833 28;
  • 31) 0.147 666 635 608 324 833 28 × 2 = 0 + 0.295 333 271 216 649 666 56;
  • 32) 0.295 333 271 216 649 666 56 × 2 = 0 + 0.590 666 542 433 299 333 12;
  • 33) 0.590 666 542 433 299 333 12 × 2 = 1 + 0.181 333 084 866 598 666 24;
  • 34) 0.181 333 084 866 598 666 24 × 2 = 0 + 0.362 666 169 733 197 332 48;
  • 35) 0.362 666 169 733 197 332 48 × 2 = 0 + 0.725 332 339 466 394 664 96;
  • 36) 0.725 332 339 466 394 664 96 × 2 = 1 + 0.450 664 678 932 789 329 92;
  • 37) 0.450 664 678 932 789 329 92 × 2 = 0 + 0.901 329 357 865 578 659 84;
  • 38) 0.901 329 357 865 578 659 84 × 2 = 1 + 0.802 658 715 731 157 319 68;
  • 39) 0.802 658 715 731 157 319 68 × 2 = 1 + 0.605 317 431 462 314 639 36;
  • 40) 0.605 317 431 462 314 639 36 × 2 = 1 + 0.210 634 862 924 629 278 72;
  • 41) 0.210 634 862 924 629 278 72 × 2 = 0 + 0.421 269 725 849 258 557 44;
  • 42) 0.421 269 725 849 258 557 44 × 2 = 0 + 0.842 539 451 698 517 114 88;
  • 43) 0.842 539 451 698 517 114 88 × 2 = 1 + 0.685 078 903 397 034 229 76;
  • 44) 0.685 078 903 397 034 229 76 × 2 = 1 + 0.370 157 806 794 068 459 52;
  • 45) 0.370 157 806 794 068 459 52 × 2 = 0 + 0.740 315 613 588 136 919 04;
  • 46) 0.740 315 613 588 136 919 04 × 2 = 1 + 0.480 631 227 176 273 838 08;
  • 47) 0.480 631 227 176 273 838 08 × 2 = 0 + 0.961 262 454 352 547 676 16;
  • 48) 0.961 262 454 352 547 676 16 × 2 = 1 + 0.922 524 908 705 095 352 32;
  • 49) 0.922 524 908 705 095 352 32 × 2 = 1 + 0.845 049 817 410 190 704 64;
  • 50) 0.845 049 817 410 190 704 64 × 2 = 1 + 0.690 099 634 820 381 409 28;
  • 51) 0.690 099 634 820 381 409 28 × 2 = 1 + 0.380 199 269 640 762 818 56;
  • 52) 0.380 199 269 640 762 818 56 × 2 = 0 + 0.760 398 539 281 525 637 12;
  • 53) 0.760 398 539 281 525 637 12 × 2 = 1 + 0.520 797 078 563 051 274 24;
  • 54) 0.520 797 078 563 051 274 24 × 2 = 1 + 0.041 594 157 126 102 548 48;
  • 55) 0.041 594 157 126 102 548 48 × 2 = 0 + 0.083 188 314 252 205 096 96;
  • 56) 0.083 188 314 252 205 096 96 × 2 = 0 + 0.166 376 628 504 410 193 92;
  • 57) 0.166 376 628 504 410 193 92 × 2 = 0 + 0.332 753 257 008 820 387 84;
  • 58) 0.332 753 257 008 820 387 84 × 2 = 0 + 0.665 506 514 017 640 775 68;
  • 59) 0.665 506 514 017 640 775 68 × 2 = 1 + 0.331 013 028 035 281 551 36;
  • 60) 0.331 013 028 035 281 551 36 × 2 = 0 + 0.662 026 056 070 563 102 72;
  • 61) 0.662 026 056 070 563 102 72 × 2 = 1 + 0.324 052 112 141 126 205 44;
  • 62) 0.324 052 112 141 126 205 44 × 2 = 0 + 0.648 104 224 282 252 410 88;
  • 63) 0.648 104 224 282 252 410 88 × 2 = 1 + 0.296 208 448 564 504 821 76;
  • 64) 0.296 208 448 564 504 821 76 × 2 = 0 + 0.592 416 897 129 009 643 52;
  • 65) 0.592 416 897 129 009 643 52 × 2 = 1 + 0.184 833 794 258 019 287 04;
  • 66) 0.184 833 794 258 019 287 04 × 2 = 0 + 0.369 667 588 516 038 574 08;
  • 67) 0.369 667 588 516 038 574 08 × 2 = 0 + 0.739 335 177 032 077 148 16;
  • 68) 0.739 335 177 032 077 148 16 × 2 = 1 + 0.478 670 354 064 154 296 32;
  • 69) 0.478 670 354 064 154 296 32 × 2 = 0 + 0.957 340 708 128 308 592 64;
  • 70) 0.957 340 708 128 308 592 64 × 2 = 1 + 0.914 681 416 256 617 185 28;
  • 71) 0.914 681 416 256 617 185 28 × 2 = 1 + 0.829 362 832 513 234 370 56;
  • 72) 0.829 362 832 513 234 370 56 × 2 = 1 + 0.658 725 665 026 468 741 12;

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 47(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111(2)

6. Positive number before normalization:

0.000 001 324 478 226 374 47(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111(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 47(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111(2) =

0.0000 0000 0000 0000 0001 0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111(2) × 20 =

1.0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111(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 1010 1001 0111


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 1010 1001 0111 =

0110 0011 1000 1001 0111 0011 0101 1110 1100 0010 1010 1001 0111


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 1010 1001 0111


Decimal number -0.000 001 324 478 226 374 47 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 1010 1001 0111

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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