90.899 999 999 999 991 473 487 171 103 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 171 103(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
90.899 999 999 999 991 473 487 171 103(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. First, convert to binary (in base 2) the integer part: 90.
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;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

90(10) =

101 1010(2)


3. Convert to binary (base 2) the fractional part: 0.899 999 999 999 991 473 487 171 103.

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) 0.899 999 999 999 991 473 487 171 103 × 2 = 1 + 0.799 999 999 999 982 946 974 342 206;
  • 2) 0.799 999 999 999 982 946 974 342 206 × 2 = 1 + 0.599 999 999 999 965 893 948 684 412;
  • 3) 0.599 999 999 999 965 893 948 684 412 × 2 = 1 + 0.199 999 999 999 931 787 897 368 824;
  • 4) 0.199 999 999 999 931 787 897 368 824 × 2 = 0 + 0.399 999 999 999 863 575 794 737 648;
  • 5) 0.399 999 999 999 863 575 794 737 648 × 2 = 0 + 0.799 999 999 999 727 151 589 475 296;
  • 6) 0.799 999 999 999 727 151 589 475 296 × 2 = 1 + 0.599 999 999 999 454 303 178 950 592;
  • 7) 0.599 999 999 999 454 303 178 950 592 × 2 = 1 + 0.199 999 999 998 908 606 357 901 184;
  • 8) 0.199 999 999 998 908 606 357 901 184 × 2 = 0 + 0.399 999 999 997 817 212 715 802 368;
  • 9) 0.399 999 999 997 817 212 715 802 368 × 2 = 0 + 0.799 999 999 995 634 425 431 604 736;
  • 10) 0.799 999 999 995 634 425 431 604 736 × 2 = 1 + 0.599 999 999 991 268 850 863 209 472;
  • 11) 0.599 999 999 991 268 850 863 209 472 × 2 = 1 + 0.199 999 999 982 537 701 726 418 944;
  • 12) 0.199 999 999 982 537 701 726 418 944 × 2 = 0 + 0.399 999 999 965 075 403 452 837 888;
  • 13) 0.399 999 999 965 075 403 452 837 888 × 2 = 0 + 0.799 999 999 930 150 806 905 675 776;
  • 14) 0.799 999 999 930 150 806 905 675 776 × 2 = 1 + 0.599 999 999 860 301 613 811 351 552;
  • 15) 0.599 999 999 860 301 613 811 351 552 × 2 = 1 + 0.199 999 999 720 603 227 622 703 104;
  • 16) 0.199 999 999 720 603 227 622 703 104 × 2 = 0 + 0.399 999 999 441 206 455 245 406 208;
  • 17) 0.399 999 999 441 206 455 245 406 208 × 2 = 0 + 0.799 999 998 882 412 910 490 812 416;
  • 18) 0.799 999 998 882 412 910 490 812 416 × 2 = 1 + 0.599 999 997 764 825 820 981 624 832;
  • 19) 0.599 999 997 764 825 820 981 624 832 × 2 = 1 + 0.199 999 995 529 651 641 963 249 664;
  • 20) 0.199 999 995 529 651 641 963 249 664 × 2 = 0 + 0.399 999 991 059 303 283 926 499 328;
  • 21) 0.399 999 991 059 303 283 926 499 328 × 2 = 0 + 0.799 999 982 118 606 567 852 998 656;
  • 22) 0.799 999 982 118 606 567 852 998 656 × 2 = 1 + 0.599 999 964 237 213 135 705 997 312;
  • 23) 0.599 999 964 237 213 135 705 997 312 × 2 = 1 + 0.199 999 928 474 426 271 411 994 624;
  • 24) 0.199 999 928 474 426 271 411 994 624 × 2 = 0 + 0.399 999 856 948 852 542 823 989 248;
  • 25) 0.399 999 856 948 852 542 823 989 248 × 2 = 0 + 0.799 999 713 897 705 085 647 978 496;
  • 26) 0.799 999 713 897 705 085 647 978 496 × 2 = 1 + 0.599 999 427 795 410 171 295 956 992;
  • 27) 0.599 999 427 795 410 171 295 956 992 × 2 = 1 + 0.199 998 855 590 820 342 591 913 984;
  • 28) 0.199 998 855 590 820 342 591 913 984 × 2 = 0 + 0.399 997 711 181 640 685 183 827 968;
  • 29) 0.399 997 711 181 640 685 183 827 968 × 2 = 0 + 0.799 995 422 363 281 370 367 655 936;
  • 30) 0.799 995 422 363 281 370 367 655 936 × 2 = 1 + 0.599 990 844 726 562 740 735 311 872;
  • 31) 0.599 990 844 726 562 740 735 311 872 × 2 = 1 + 0.199 981 689 453 125 481 470 623 744;
  • 32) 0.199 981 689 453 125 481 470 623 744 × 2 = 0 + 0.399 963 378 906 250 962 941 247 488;
  • 33) 0.399 963 378 906 250 962 941 247 488 × 2 = 0 + 0.799 926 757 812 501 925 882 494 976;
  • 34) 0.799 926 757 812 501 925 882 494 976 × 2 = 1 + 0.599 853 515 625 003 851 764 989 952;
  • 35) 0.599 853 515 625 003 851 764 989 952 × 2 = 1 + 0.199 707 031 250 007 703 529 979 904;
  • 36) 0.199 707 031 250 007 703 529 979 904 × 2 = 0 + 0.399 414 062 500 015 407 059 959 808;
  • 37) 0.399 414 062 500 015 407 059 959 808 × 2 = 0 + 0.798 828 125 000 030 814 119 919 616;
  • 38) 0.798 828 125 000 030 814 119 919 616 × 2 = 1 + 0.597 656 250 000 061 628 239 839 232;
  • 39) 0.597 656 250 000 061 628 239 839 232 × 2 = 1 + 0.195 312 500 000 123 256 479 678 464;
  • 40) 0.195 312 500 000 123 256 479 678 464 × 2 = 0 + 0.390 625 000 000 246 512 959 356 928;
  • 41) 0.390 625 000 000 246 512 959 356 928 × 2 = 0 + 0.781 250 000 000 493 025 918 713 856;
  • 42) 0.781 250 000 000 493 025 918 713 856 × 2 = 1 + 0.562 500 000 000 986 051 837 427 712;
  • 43) 0.562 500 000 000 986 051 837 427 712 × 2 = 1 + 0.125 000 000 001 972 103 674 855 424;
  • 44) 0.125 000 000 001 972 103 674 855 424 × 2 = 0 + 0.250 000 000 003 944 207 349 710 848;
  • 45) 0.250 000 000 003 944 207 349 710 848 × 2 = 0 + 0.500 000 000 007 888 414 699 421 696;
  • 46) 0.500 000 000 007 888 414 699 421 696 × 2 = 1 + 0.000 000 000 015 776 829 398 843 392;
  • 47) 0.000 000 000 015 776 829 398 843 392 × 2 = 0 + 0.000 000 000 031 553 658 797 686 784;
  • 48) 0.000 000 000 031 553 658 797 686 784 × 2 = 0 + 0.000 000 000 063 107 317 595 373 568;
  • 49) 0.000 000 000 063 107 317 595 373 568 × 2 = 0 + 0.000 000 000 126 214 635 190 747 136;
  • 50) 0.000 000 000 126 214 635 190 747 136 × 2 = 0 + 0.000 000 000 252 429 270 381 494 272;
  • 51) 0.000 000 000 252 429 270 381 494 272 × 2 = 0 + 0.000 000 000 504 858 540 762 988 544;
  • 52) 0.000 000 000 504 858 540 762 988 544 × 2 = 0 + 0.000 000 001 009 717 081 525 977 088;
  • 53) 0.000 000 001 009 717 081 525 977 088 × 2 = 0 + 0.000 000 002 019 434 163 051 954 176;

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).


4. 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:


0.899 999 999 999 991 473 487 171 103(10) =

0.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2)

5. Positive number before normalization:

90.899 999 999 999 991 473 487 171 103(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2)

6. 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:


90.899 999 999 999 991 473 487 171 103(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) × 20 =

1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 000(2) × 26


7. 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 0 (a positive number)

Exponent (unadjusted): 6

Mantissa (not normalized):
1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 000


8. 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)


9. 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;

10. 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)


11. 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. 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 000 0000 =

0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0101

Mantissa (52 bits) =
0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


Decimal number 90.899 999 999 999 991 473 487 171 103 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001

» Convert decimal 90.899 999 999 999 991 473 487 171 072 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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