-99.452 599 999 999 989 677 235 134 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -99.452 599 999 999 989 677 235 134 33(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
-99.452 599 999 999 989 677 235 134 33(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:

|-99.452 599 999 999 989 677 235 134 33| = 99.452 599 999 999 989 677 235 134 33


2. First, convert to binary (in base 2) the integer part: 99.
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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

99(10) =

110 0011(2)


4. Convert to binary (base 2) the fractional part: 0.452 599 999 999 989 677 235 134 33.

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) 0.452 599 999 999 989 677 235 134 33 × 2 = 0 + 0.905 199 999 999 979 354 470 268 66;
  • 2) 0.905 199 999 999 979 354 470 268 66 × 2 = 1 + 0.810 399 999 999 958 708 940 537 32;
  • 3) 0.810 399 999 999 958 708 940 537 32 × 2 = 1 + 0.620 799 999 999 917 417 881 074 64;
  • 4) 0.620 799 999 999 917 417 881 074 64 × 2 = 1 + 0.241 599 999 999 834 835 762 149 28;
  • 5) 0.241 599 999 999 834 835 762 149 28 × 2 = 0 + 0.483 199 999 999 669 671 524 298 56;
  • 6) 0.483 199 999 999 669 671 524 298 56 × 2 = 0 + 0.966 399 999 999 339 343 048 597 12;
  • 7) 0.966 399 999 999 339 343 048 597 12 × 2 = 1 + 0.932 799 999 998 678 686 097 194 24;
  • 8) 0.932 799 999 998 678 686 097 194 24 × 2 = 1 + 0.865 599 999 997 357 372 194 388 48;
  • 9) 0.865 599 999 997 357 372 194 388 48 × 2 = 1 + 0.731 199 999 994 714 744 388 776 96;
  • 10) 0.731 199 999 994 714 744 388 776 96 × 2 = 1 + 0.462 399 999 989 429 488 777 553 92;
  • 11) 0.462 399 999 989 429 488 777 553 92 × 2 = 0 + 0.924 799 999 978 858 977 555 107 84;
  • 12) 0.924 799 999 978 858 977 555 107 84 × 2 = 1 + 0.849 599 999 957 717 955 110 215 68;
  • 13) 0.849 599 999 957 717 955 110 215 68 × 2 = 1 + 0.699 199 999 915 435 910 220 431 36;
  • 14) 0.699 199 999 915 435 910 220 431 36 × 2 = 1 + 0.398 399 999 830 871 820 440 862 72;
  • 15) 0.398 399 999 830 871 820 440 862 72 × 2 = 0 + 0.796 799 999 661 743 640 881 725 44;
  • 16) 0.796 799 999 661 743 640 881 725 44 × 2 = 1 + 0.593 599 999 323 487 281 763 450 88;
  • 17) 0.593 599 999 323 487 281 763 450 88 × 2 = 1 + 0.187 199 998 646 974 563 526 901 76;
  • 18) 0.187 199 998 646 974 563 526 901 76 × 2 = 0 + 0.374 399 997 293 949 127 053 803 52;
  • 19) 0.374 399 997 293 949 127 053 803 52 × 2 = 0 + 0.748 799 994 587 898 254 107 607 04;
  • 20) 0.748 799 994 587 898 254 107 607 04 × 2 = 1 + 0.497 599 989 175 796 508 215 214 08;
  • 21) 0.497 599 989 175 796 508 215 214 08 × 2 = 0 + 0.995 199 978 351 593 016 430 428 16;
  • 22) 0.995 199 978 351 593 016 430 428 16 × 2 = 1 + 0.990 399 956 703 186 032 860 856 32;
  • 23) 0.990 399 956 703 186 032 860 856 32 × 2 = 1 + 0.980 799 913 406 372 065 721 712 64;
  • 24) 0.980 799 913 406 372 065 721 712 64 × 2 = 1 + 0.961 599 826 812 744 131 443 425 28;
  • 25) 0.961 599 826 812 744 131 443 425 28 × 2 = 1 + 0.923 199 653 625 488 262 886 850 56;
  • 26) 0.923 199 653 625 488 262 886 850 56 × 2 = 1 + 0.846 399 307 250 976 525 773 701 12;
  • 27) 0.846 399 307 250 976 525 773 701 12 × 2 = 1 + 0.692 798 614 501 953 051 547 402 24;
  • 28) 0.692 798 614 501 953 051 547 402 24 × 2 = 1 + 0.385 597 229 003 906 103 094 804 48;
  • 29) 0.385 597 229 003 906 103 094 804 48 × 2 = 0 + 0.771 194 458 007 812 206 189 608 96;
  • 30) 0.771 194 458 007 812 206 189 608 96 × 2 = 1 + 0.542 388 916 015 624 412 379 217 92;
  • 31) 0.542 388 916 015 624 412 379 217 92 × 2 = 1 + 0.084 777 832 031 248 824 758 435 84;
  • 32) 0.084 777 832 031 248 824 758 435 84 × 2 = 0 + 0.169 555 664 062 497 649 516 871 68;
  • 33) 0.169 555 664 062 497 649 516 871 68 × 2 = 0 + 0.339 111 328 124 995 299 033 743 36;
  • 34) 0.339 111 328 124 995 299 033 743 36 × 2 = 0 + 0.678 222 656 249 990 598 067 486 72;
  • 35) 0.678 222 656 249 990 598 067 486 72 × 2 = 1 + 0.356 445 312 499 981 196 134 973 44;
  • 36) 0.356 445 312 499 981 196 134 973 44 × 2 = 0 + 0.712 890 624 999 962 392 269 946 88;
  • 37) 0.712 890 624 999 962 392 269 946 88 × 2 = 1 + 0.425 781 249 999 924 784 539 893 76;
  • 38) 0.425 781 249 999 924 784 539 893 76 × 2 = 0 + 0.851 562 499 999 849 569 079 787 52;
  • 39) 0.851 562 499 999 849 569 079 787 52 × 2 = 1 + 0.703 124 999 999 699 138 159 575 04;
  • 40) 0.703 124 999 999 699 138 159 575 04 × 2 = 1 + 0.406 249 999 999 398 276 319 150 08;
  • 41) 0.406 249 999 999 398 276 319 150 08 × 2 = 0 + 0.812 499 999 998 796 552 638 300 16;
  • 42) 0.812 499 999 998 796 552 638 300 16 × 2 = 1 + 0.624 999 999 997 593 105 276 600 32;
  • 43) 0.624 999 999 997 593 105 276 600 32 × 2 = 1 + 0.249 999 999 995 186 210 553 200 64;
  • 44) 0.249 999 999 995 186 210 553 200 64 × 2 = 0 + 0.499 999 999 990 372 421 106 401 28;
  • 45) 0.499 999 999 990 372 421 106 401 28 × 2 = 0 + 0.999 999 999 980 744 842 212 802 56;
  • 46) 0.999 999 999 980 744 842 212 802 56 × 2 = 1 + 0.999 999 999 961 489 684 425 605 12;
  • 47) 0.999 999 999 961 489 684 425 605 12 × 2 = 1 + 0.999 999 999 922 979 368 851 210 24;
  • 48) 0.999 999 999 922 979 368 851 210 24 × 2 = 1 + 0.999 999 999 845 958 737 702 420 48;
  • 49) 0.999 999 999 845 958 737 702 420 48 × 2 = 1 + 0.999 999 999 691 917 475 404 840 96;
  • 50) 0.999 999 999 691 917 475 404 840 96 × 2 = 1 + 0.999 999 999 383 834 950 809 681 92;
  • 51) 0.999 999 999 383 834 950 809 681 92 × 2 = 1 + 0.999 999 998 767 669 901 619 363 84;
  • 52) 0.999 999 998 767 669 901 619 363 84 × 2 = 1 + 0.999 999 997 535 339 803 238 727 68;
  • 53) 0.999 999 997 535 339 803 238 727 68 × 2 = 1 + 0.999 999 995 070 679 606 477 455 36;

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:


0.452 599 999 999 989 677 235 134 33(10) =

0.0111 0011 1101 1101 1001 0111 1111 0110 0010 1011 0110 0111 1111 1(2)

6. Positive number before normalization:

99.452 599 999 999 989 677 235 134 33(10) =

110 0011.0111 0011 1101 1101 1001 0111 1111 0110 0010 1011 0110 0111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


99.452 599 999 999 989 677 235 134 33(10) =

110 0011.0111 0011 1101 1101 1001 0111 1111 0110 0010 1011 0110 0111 1111 1(2) =

110 0011.0111 0011 1101 1101 1001 0111 1111 0110 0010 1011 0110 0111 1111 1(2) × 20 =

1.1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001 1111 111(2) × 26


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): 6

Mantissa (not normalized):
1.1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001 1111 111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1029(10) =

100 0000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001 111 1111 =

1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001


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) =
100 0000 0101

Mantissa (52 bits) =
1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001


Decimal number -99.452 599 999 999 989 677 235 134 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 1000 1101 1100 1111 0111 0110 0101 1111 1101 1000 1010 1101 1001

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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