Decimal to 64 Bit IEEE 754 Binary: Convert Number 92.299 999 999 999 997 157 829 056 959 599 256 515 638 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 92.299 999 999 999 997 157 829 056 959 599 256 515 638(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 92.
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;
  • 92 ÷ 2 = 46 + 0;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 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.

92(10) =


101 1100(2)


3. Convert to binary (base 2) the fractional part: 0.299 999 999 999 997 157 829 056 959 599 256 515 638.

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) 0.299 999 999 999 997 157 829 056 959 599 256 515 638 × 2 = 0 + 0.599 999 999 999 994 315 658 113 919 198 513 031 276;
  • 2) 0.599 999 999 999 994 315 658 113 919 198 513 031 276 × 2 = 1 + 0.199 999 999 999 988 631 316 227 838 397 026 062 552;
  • 3) 0.199 999 999 999 988 631 316 227 838 397 026 062 552 × 2 = 0 + 0.399 999 999 999 977 262 632 455 676 794 052 125 104;
  • 4) 0.399 999 999 999 977 262 632 455 676 794 052 125 104 × 2 = 0 + 0.799 999 999 999 954 525 264 911 353 588 104 250 208;
  • 5) 0.799 999 999 999 954 525 264 911 353 588 104 250 208 × 2 = 1 + 0.599 999 999 999 909 050 529 822 707 176 208 500 416;
  • 6) 0.599 999 999 999 909 050 529 822 707 176 208 500 416 × 2 = 1 + 0.199 999 999 999 818 101 059 645 414 352 417 000 832;
  • 7) 0.199 999 999 999 818 101 059 645 414 352 417 000 832 × 2 = 0 + 0.399 999 999 999 636 202 119 290 828 704 834 001 664;
  • 8) 0.399 999 999 999 636 202 119 290 828 704 834 001 664 × 2 = 0 + 0.799 999 999 999 272 404 238 581 657 409 668 003 328;
  • 9) 0.799 999 999 999 272 404 238 581 657 409 668 003 328 × 2 = 1 + 0.599 999 999 998 544 808 477 163 314 819 336 006 656;
  • 10) 0.599 999 999 998 544 808 477 163 314 819 336 006 656 × 2 = 1 + 0.199 999 999 997 089 616 954 326 629 638 672 013 312;
  • 11) 0.199 999 999 997 089 616 954 326 629 638 672 013 312 × 2 = 0 + 0.399 999 999 994 179 233 908 653 259 277 344 026 624;
  • 12) 0.399 999 999 994 179 233 908 653 259 277 344 026 624 × 2 = 0 + 0.799 999 999 988 358 467 817 306 518 554 688 053 248;
  • 13) 0.799 999 999 988 358 467 817 306 518 554 688 053 248 × 2 = 1 + 0.599 999 999 976 716 935 634 613 037 109 376 106 496;
  • 14) 0.599 999 999 976 716 935 634 613 037 109 376 106 496 × 2 = 1 + 0.199 999 999 953 433 871 269 226 074 218 752 212 992;
  • 15) 0.199 999 999 953 433 871 269 226 074 218 752 212 992 × 2 = 0 + 0.399 999 999 906 867 742 538 452 148 437 504 425 984;
  • 16) 0.399 999 999 906 867 742 538 452 148 437 504 425 984 × 2 = 0 + 0.799 999 999 813 735 485 076 904 296 875 008 851 968;
  • 17) 0.799 999 999 813 735 485 076 904 296 875 008 851 968 × 2 = 1 + 0.599 999 999 627 470 970 153 808 593 750 017 703 936;
  • 18) 0.599 999 999 627 470 970 153 808 593 750 017 703 936 × 2 = 1 + 0.199 999 999 254 941 940 307 617 187 500 035 407 872;
  • 19) 0.199 999 999 254 941 940 307 617 187 500 035 407 872 × 2 = 0 + 0.399 999 998 509 883 880 615 234 375 000 070 815 744;
  • 20) 0.399 999 998 509 883 880 615 234 375 000 070 815 744 × 2 = 0 + 0.799 999 997 019 767 761 230 468 750 000 141 631 488;
  • 21) 0.799 999 997 019 767 761 230 468 750 000 141 631 488 × 2 = 1 + 0.599 999 994 039 535 522 460 937 500 000 283 262 976;
  • 22) 0.599 999 994 039 535 522 460 937 500 000 283 262 976 × 2 = 1 + 0.199 999 988 079 071 044 921 875 000 000 566 525 952;
  • 23) 0.199 999 988 079 071 044 921 875 000 000 566 525 952 × 2 = 0 + 0.399 999 976 158 142 089 843 750 000 001 133 051 904;
  • 24) 0.399 999 976 158 142 089 843 750 000 001 133 051 904 × 2 = 0 + 0.799 999 952 316 284 179 687 500 000 002 266 103 808;
  • 25) 0.799 999 952 316 284 179 687 500 000 002 266 103 808 × 2 = 1 + 0.599 999 904 632 568 359 375 000 000 004 532 207 616;
  • 26) 0.599 999 904 632 568 359 375 000 000 004 532 207 616 × 2 = 1 + 0.199 999 809 265 136 718 750 000 000 009 064 415 232;
  • 27) 0.199 999 809 265 136 718 750 000 000 009 064 415 232 × 2 = 0 + 0.399 999 618 530 273 437 500 000 000 018 128 830 464;
  • 28) 0.399 999 618 530 273 437 500 000 000 018 128 830 464 × 2 = 0 + 0.799 999 237 060 546 875 000 000 000 036 257 660 928;
  • 29) 0.799 999 237 060 546 875 000 000 000 036 257 660 928 × 2 = 1 + 0.599 998 474 121 093 750 000 000 000 072 515 321 856;
  • 30) 0.599 998 474 121 093 750 000 000 000 072 515 321 856 × 2 = 1 + 0.199 996 948 242 187 500 000 000 000 145 030 643 712;
  • 31) 0.199 996 948 242 187 500 000 000 000 145 030 643 712 × 2 = 0 + 0.399 993 896 484 375 000 000 000 000 290 061 287 424;
  • 32) 0.399 993 896 484 375 000 000 000 000 290 061 287 424 × 2 = 0 + 0.799 987 792 968 750 000 000 000 000 580 122 574 848;
  • 33) 0.799 987 792 968 750 000 000 000 000 580 122 574 848 × 2 = 1 + 0.599 975 585 937 500 000 000 000 001 160 245 149 696;
  • 34) 0.599 975 585 937 500 000 000 000 001 160 245 149 696 × 2 = 1 + 0.199 951 171 875 000 000 000 000 002 320 490 299 392;
  • 35) 0.199 951 171 875 000 000 000 000 002 320 490 299 392 × 2 = 0 + 0.399 902 343 750 000 000 000 000 004 640 980 598 784;
  • 36) 0.399 902 343 750 000 000 000 000 004 640 980 598 784 × 2 = 0 + 0.799 804 687 500 000 000 000 000 009 281 961 197 568;
  • 37) 0.799 804 687 500 000 000 000 000 009 281 961 197 568 × 2 = 1 + 0.599 609 375 000 000 000 000 000 018 563 922 395 136;
  • 38) 0.599 609 375 000 000 000 000 000 018 563 922 395 136 × 2 = 1 + 0.199 218 750 000 000 000 000 000 037 127 844 790 272;
  • 39) 0.199 218 750 000 000 000 000 000 037 127 844 790 272 × 2 = 0 + 0.398 437 500 000 000 000 000 000 074 255 689 580 544;
  • 40) 0.398 437 500 000 000 000 000 000 074 255 689 580 544 × 2 = 0 + 0.796 875 000 000 000 000 000 000 148 511 379 161 088;
  • 41) 0.796 875 000 000 000 000 000 000 148 511 379 161 088 × 2 = 1 + 0.593 750 000 000 000 000 000 000 297 022 758 322 176;
  • 42) 0.593 750 000 000 000 000 000 000 297 022 758 322 176 × 2 = 1 + 0.187 500 000 000 000 000 000 000 594 045 516 644 352;
  • 43) 0.187 500 000 000 000 000 000 000 594 045 516 644 352 × 2 = 0 + 0.375 000 000 000 000 000 000 001 188 091 033 288 704;
  • 44) 0.375 000 000 000 000 000 000 001 188 091 033 288 704 × 2 = 0 + 0.750 000 000 000 000 000 000 002 376 182 066 577 408;
  • 45) 0.750 000 000 000 000 000 000 002 376 182 066 577 408 × 2 = 1 + 0.500 000 000 000 000 000 000 004 752 364 133 154 816;
  • 46) 0.500 000 000 000 000 000 000 004 752 364 133 154 816 × 2 = 1 + 0.000 000 000 000 000 000 000 009 504 728 266 309 632;
  • 47) 0.000 000 000 000 000 000 000 009 504 728 266 309 632 × 2 = 0 + 0.000 000 000 000 000 000 000 019 009 456 532 619 264;
  • 48) 0.000 000 000 000 000 000 000 019 009 456 532 619 264 × 2 = 0 + 0.000 000 000 000 000 000 000 038 018 913 065 238 528;
  • 49) 0.000 000 000 000 000 000 000 038 018 913 065 238 528 × 2 = 0 + 0.000 000 000 000 000 000 000 076 037 826 130 477 056;
  • 50) 0.000 000 000 000 000 000 000 076 037 826 130 477 056 × 2 = 0 + 0.000 000 000 000 000 000 000 152 075 652 260 954 112;
  • 51) 0.000 000 000 000 000 000 000 152 075 652 260 954 112 × 2 = 0 + 0.000 000 000 000 000 000 000 304 151 304 521 908 224;
  • 52) 0.000 000 000 000 000 000 000 304 151 304 521 908 224 × 2 = 0 + 0.000 000 000 000 000 000 000 608 302 609 043 816 448;
  • 53) 0.000 000 000 000 000 000 000 608 302 609 043 816 448 × 2 = 0 + 0.000 000 000 000 000 000 001 216 605 218 087 632 896;

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:


0.299 999 999 999 997 157 829 056 959 599 256 515 638(10) =


0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2)

5. Positive number before normalization:

92.299 999 999 999 997 157 829 056 959 599 256 515 638(10) =


101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 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:


92.299 999 999 999 997 157 829 056 959 599 256 515 638(10) =


101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2) =


101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2) × 20 =


1.0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 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):
1.0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 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. 0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 000 0000 =


0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011


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) =
0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011


The base ten decimal number 92.299 999 999 999 997 157 829 056 959 599 256 515 638 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011

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