-0.000 006 491 636 509 589 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 006 491 636 509 589 22(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
-0.000 006 491 636 509 589 22(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:

|-0.000 006 491 636 509 589 22| = 0.000 006 491 636 509 589 22


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: 0.000 006 491 636 509 589 22.

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) 0.000 006 491 636 509 589 22 × 2 = 0 + 0.000 012 983 273 019 178 44;
  • 2) 0.000 012 983 273 019 178 44 × 2 = 0 + 0.000 025 966 546 038 356 88;
  • 3) 0.000 025 966 546 038 356 88 × 2 = 0 + 0.000 051 933 092 076 713 76;
  • 4) 0.000 051 933 092 076 713 76 × 2 = 0 + 0.000 103 866 184 153 427 52;
  • 5) 0.000 103 866 184 153 427 52 × 2 = 0 + 0.000 207 732 368 306 855 04;
  • 6) 0.000 207 732 368 306 855 04 × 2 = 0 + 0.000 415 464 736 613 710 08;
  • 7) 0.000 415 464 736 613 710 08 × 2 = 0 + 0.000 830 929 473 227 420 16;
  • 8) 0.000 830 929 473 227 420 16 × 2 = 0 + 0.001 661 858 946 454 840 32;
  • 9) 0.001 661 858 946 454 840 32 × 2 = 0 + 0.003 323 717 892 909 680 64;
  • 10) 0.003 323 717 892 909 680 64 × 2 = 0 + 0.006 647 435 785 819 361 28;
  • 11) 0.006 647 435 785 819 361 28 × 2 = 0 + 0.013 294 871 571 638 722 56;
  • 12) 0.013 294 871 571 638 722 56 × 2 = 0 + 0.026 589 743 143 277 445 12;
  • 13) 0.026 589 743 143 277 445 12 × 2 = 0 + 0.053 179 486 286 554 890 24;
  • 14) 0.053 179 486 286 554 890 24 × 2 = 0 + 0.106 358 972 573 109 780 48;
  • 15) 0.106 358 972 573 109 780 48 × 2 = 0 + 0.212 717 945 146 219 560 96;
  • 16) 0.212 717 945 146 219 560 96 × 2 = 0 + 0.425 435 890 292 439 121 92;
  • 17) 0.425 435 890 292 439 121 92 × 2 = 0 + 0.850 871 780 584 878 243 84;
  • 18) 0.850 871 780 584 878 243 84 × 2 = 1 + 0.701 743 561 169 756 487 68;
  • 19) 0.701 743 561 169 756 487 68 × 2 = 1 + 0.403 487 122 339 512 975 36;
  • 20) 0.403 487 122 339 512 975 36 × 2 = 0 + 0.806 974 244 679 025 950 72;
  • 21) 0.806 974 244 679 025 950 72 × 2 = 1 + 0.613 948 489 358 051 901 44;
  • 22) 0.613 948 489 358 051 901 44 × 2 = 1 + 0.227 896 978 716 103 802 88;
  • 23) 0.227 896 978 716 103 802 88 × 2 = 0 + 0.455 793 957 432 207 605 76;
  • 24) 0.455 793 957 432 207 605 76 × 2 = 0 + 0.911 587 914 864 415 211 52;
  • 25) 0.911 587 914 864 415 211 52 × 2 = 1 + 0.823 175 829 728 830 423 04;
  • 26) 0.823 175 829 728 830 423 04 × 2 = 1 + 0.646 351 659 457 660 846 08;
  • 27) 0.646 351 659 457 660 846 08 × 2 = 1 + 0.292 703 318 915 321 692 16;
  • 28) 0.292 703 318 915 321 692 16 × 2 = 0 + 0.585 406 637 830 643 384 32;
  • 29) 0.585 406 637 830 643 384 32 × 2 = 1 + 0.170 813 275 661 286 768 64;
  • 30) 0.170 813 275 661 286 768 64 × 2 = 0 + 0.341 626 551 322 573 537 28;
  • 31) 0.341 626 551 322 573 537 28 × 2 = 0 + 0.683 253 102 645 147 074 56;
  • 32) 0.683 253 102 645 147 074 56 × 2 = 1 + 0.366 506 205 290 294 149 12;
  • 33) 0.366 506 205 290 294 149 12 × 2 = 0 + 0.733 012 410 580 588 298 24;
  • 34) 0.733 012 410 580 588 298 24 × 2 = 1 + 0.466 024 821 161 176 596 48;
  • 35) 0.466 024 821 161 176 596 48 × 2 = 0 + 0.932 049 642 322 353 192 96;
  • 36) 0.932 049 642 322 353 192 96 × 2 = 1 + 0.864 099 284 644 706 385 92;
  • 37) 0.864 099 284 644 706 385 92 × 2 = 1 + 0.728 198 569 289 412 771 84;
  • 38) 0.728 198 569 289 412 771 84 × 2 = 1 + 0.456 397 138 578 825 543 68;
  • 39) 0.456 397 138 578 825 543 68 × 2 = 0 + 0.912 794 277 157 651 087 36;
  • 40) 0.912 794 277 157 651 087 36 × 2 = 1 + 0.825 588 554 315 302 174 72;
  • 41) 0.825 588 554 315 302 174 72 × 2 = 1 + 0.651 177 108 630 604 349 44;
  • 42) 0.651 177 108 630 604 349 44 × 2 = 1 + 0.302 354 217 261 208 698 88;
  • 43) 0.302 354 217 261 208 698 88 × 2 = 0 + 0.604 708 434 522 417 397 76;
  • 44) 0.604 708 434 522 417 397 76 × 2 = 1 + 0.209 416 869 044 834 795 52;
  • 45) 0.209 416 869 044 834 795 52 × 2 = 0 + 0.418 833 738 089 669 591 04;
  • 46) 0.418 833 738 089 669 591 04 × 2 = 0 + 0.837 667 476 179 339 182 08;
  • 47) 0.837 667 476 179 339 182 08 × 2 = 1 + 0.675 334 952 358 678 364 16;
  • 48) 0.675 334 952 358 678 364 16 × 2 = 1 + 0.350 669 904 717 356 728 32;
  • 49) 0.350 669 904 717 356 728 32 × 2 = 0 + 0.701 339 809 434 713 456 64;
  • 50) 0.701 339 809 434 713 456 64 × 2 = 1 + 0.402 679 618 869 426 913 28;
  • 51) 0.402 679 618 869 426 913 28 × 2 = 0 + 0.805 359 237 738 853 826 56;
  • 52) 0.805 359 237 738 853 826 56 × 2 = 1 + 0.610 718 475 477 707 653 12;
  • 53) 0.610 718 475 477 707 653 12 × 2 = 1 + 0.221 436 950 955 415 306 24;
  • 54) 0.221 436 950 955 415 306 24 × 2 = 0 + 0.442 873 901 910 830 612 48;
  • 55) 0.442 873 901 910 830 612 48 × 2 = 0 + 0.885 747 803 821 661 224 96;
  • 56) 0.885 747 803 821 661 224 96 × 2 = 1 + 0.771 495 607 643 322 449 92;
  • 57) 0.771 495 607 643 322 449 92 × 2 = 1 + 0.542 991 215 286 644 899 84;
  • 58) 0.542 991 215 286 644 899 84 × 2 = 1 + 0.085 982 430 573 289 799 68;
  • 59) 0.085 982 430 573 289 799 68 × 2 = 0 + 0.171 964 861 146 579 599 36;
  • 60) 0.171 964 861 146 579 599 36 × 2 = 0 + 0.343 929 722 293 159 198 72;
  • 61) 0.343 929 722 293 159 198 72 × 2 = 0 + 0.687 859 444 586 318 397 44;
  • 62) 0.687 859 444 586 318 397 44 × 2 = 1 + 0.375 718 889 172 636 794 88;
  • 63) 0.375 718 889 172 636 794 88 × 2 = 0 + 0.751 437 778 345 273 589 76;
  • 64) 0.751 437 778 345 273 589 76 × 2 = 1 + 0.502 875 556 690 547 179 52;
  • 65) 0.502 875 556 690 547 179 52 × 2 = 1 + 0.005 751 113 381 094 359 04;
  • 66) 0.005 751 113 381 094 359 04 × 2 = 0 + 0.011 502 226 762 188 718 08;
  • 67) 0.011 502 226 762 188 718 08 × 2 = 0 + 0.023 004 453 524 377 436 16;
  • 68) 0.023 004 453 524 377 436 16 × 2 = 0 + 0.046 008 907 048 754 872 32;
  • 69) 0.046 008 907 048 754 872 32 × 2 = 0 + 0.092 017 814 097 509 744 64;
  • 70) 0.092 017 814 097 509 744 64 × 2 = 0 + 0.184 035 628 195 019 489 28;

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:


0.000 006 491 636 509 589 22(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1100 0101 1000 00(2)

6. Positive number before normalization:

0.000 006 491 636 509 589 22(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1100 0101 1000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 491 636 509 589 22(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1100 0101 1000 00(2) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1100 0101 1000 00(2) × 20 =

1.1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000(2) × 2-18


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): -18

Mantissa (not normalized):
1.1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(11-1) - 1 =

(-18 + 1 023)(10) =

1 005(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;
  • 1 005 ÷ 2 = 502 + 1;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 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) =

1005(10) =

011 1110 1101(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. 1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000 =

1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000


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 1101

Mantissa (52 bits) =
1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000


Decimal number -0.000 006 491 636 509 589 22 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1101 - 1011 0011 1010 0101 0111 0111 0100 1101 0110 0111 0001 0110 0000

» Convert decimal -0.000 006 491 636 509 589 75 to 64 bit double precision IEEE 754 binary floating point representation standard

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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