0.453 109 999 999 999 957 243 090 875 635 971 316 18 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.453 109 999 999 999 957 243 090 875 635 971 316 18(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
0.453 109 999 999 999 957 243 090 875 635 971 316 18(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.453 109 999 999 999 957 243 090 875 635 971 316 18.

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) 0.453 109 999 999 999 957 243 090 875 635 971 316 18 × 2 = 0 + 0.906 219 999 999 999 914 486 181 751 271 942 632 36;
  • 2) 0.906 219 999 999 999 914 486 181 751 271 942 632 36 × 2 = 1 + 0.812 439 999 999 999 828 972 363 502 543 885 264 72;
  • 3) 0.812 439 999 999 999 828 972 363 502 543 885 264 72 × 2 = 1 + 0.624 879 999 999 999 657 944 727 005 087 770 529 44;
  • 4) 0.624 879 999 999 999 657 944 727 005 087 770 529 44 × 2 = 1 + 0.249 759 999 999 999 315 889 454 010 175 541 058 88;
  • 5) 0.249 759 999 999 999 315 889 454 010 175 541 058 88 × 2 = 0 + 0.499 519 999 999 998 631 778 908 020 351 082 117 76;
  • 6) 0.499 519 999 999 998 631 778 908 020 351 082 117 76 × 2 = 0 + 0.999 039 999 999 997 263 557 816 040 702 164 235 52;
  • 7) 0.999 039 999 999 997 263 557 816 040 702 164 235 52 × 2 = 1 + 0.998 079 999 999 994 527 115 632 081 404 328 471 04;
  • 8) 0.998 079 999 999 994 527 115 632 081 404 328 471 04 × 2 = 1 + 0.996 159 999 999 989 054 231 264 162 808 656 942 08;
  • 9) 0.996 159 999 999 989 054 231 264 162 808 656 942 08 × 2 = 1 + 0.992 319 999 999 978 108 462 528 325 617 313 884 16;
  • 10) 0.992 319 999 999 978 108 462 528 325 617 313 884 16 × 2 = 1 + 0.984 639 999 999 956 216 925 056 651 234 627 768 32;
  • 11) 0.984 639 999 999 956 216 925 056 651 234 627 768 32 × 2 = 1 + 0.969 279 999 999 912 433 850 113 302 469 255 536 64;
  • 12) 0.969 279 999 999 912 433 850 113 302 469 255 536 64 × 2 = 1 + 0.938 559 999 999 824 867 700 226 604 938 511 073 28;
  • 13) 0.938 559 999 999 824 867 700 226 604 938 511 073 28 × 2 = 1 + 0.877 119 999 999 649 735 400 453 209 877 022 146 56;
  • 14) 0.877 119 999 999 649 735 400 453 209 877 022 146 56 × 2 = 1 + 0.754 239 999 999 299 470 800 906 419 754 044 293 12;
  • 15) 0.754 239 999 999 299 470 800 906 419 754 044 293 12 × 2 = 1 + 0.508 479 999 998 598 941 601 812 839 508 088 586 24;
  • 16) 0.508 479 999 998 598 941 601 812 839 508 088 586 24 × 2 = 1 + 0.016 959 999 997 197 883 203 625 679 016 177 172 48;
  • 17) 0.016 959 999 997 197 883 203 625 679 016 177 172 48 × 2 = 0 + 0.033 919 999 994 395 766 407 251 358 032 354 344 96;
  • 18) 0.033 919 999 994 395 766 407 251 358 032 354 344 96 × 2 = 0 + 0.067 839 999 988 791 532 814 502 716 064 708 689 92;
  • 19) 0.067 839 999 988 791 532 814 502 716 064 708 689 92 × 2 = 0 + 0.135 679 999 977 583 065 629 005 432 129 417 379 84;
  • 20) 0.135 679 999 977 583 065 629 005 432 129 417 379 84 × 2 = 0 + 0.271 359 999 955 166 131 258 010 864 258 834 759 68;
  • 21) 0.271 359 999 955 166 131 258 010 864 258 834 759 68 × 2 = 0 + 0.542 719 999 910 332 262 516 021 728 517 669 519 36;
  • 22) 0.542 719 999 910 332 262 516 021 728 517 669 519 36 × 2 = 1 + 0.085 439 999 820 664 525 032 043 457 035 339 038 72;
  • 23) 0.085 439 999 820 664 525 032 043 457 035 339 038 72 × 2 = 0 + 0.170 879 999 641 329 050 064 086 914 070 678 077 44;
  • 24) 0.170 879 999 641 329 050 064 086 914 070 678 077 44 × 2 = 0 + 0.341 759 999 282 658 100 128 173 828 141 356 154 88;
  • 25) 0.341 759 999 282 658 100 128 173 828 141 356 154 88 × 2 = 0 + 0.683 519 998 565 316 200 256 347 656 282 712 309 76;
  • 26) 0.683 519 998 565 316 200 256 347 656 282 712 309 76 × 2 = 1 + 0.367 039 997 130 632 400 512 695 312 565 424 619 52;
  • 27) 0.367 039 997 130 632 400 512 695 312 565 424 619 52 × 2 = 0 + 0.734 079 994 261 264 801 025 390 625 130 849 239 04;
  • 28) 0.734 079 994 261 264 801 025 390 625 130 849 239 04 × 2 = 1 + 0.468 159 988 522 529 602 050 781 250 261 698 478 08;
  • 29) 0.468 159 988 522 529 602 050 781 250 261 698 478 08 × 2 = 0 + 0.936 319 977 045 059 204 101 562 500 523 396 956 16;
  • 30) 0.936 319 977 045 059 204 101 562 500 523 396 956 16 × 2 = 1 + 0.872 639 954 090 118 408 203 125 001 046 793 912 32;
  • 31) 0.872 639 954 090 118 408 203 125 001 046 793 912 32 × 2 = 1 + 0.745 279 908 180 236 816 406 250 002 093 587 824 64;
  • 32) 0.745 279 908 180 236 816 406 250 002 093 587 824 64 × 2 = 1 + 0.490 559 816 360 473 632 812 500 004 187 175 649 28;
  • 33) 0.490 559 816 360 473 632 812 500 004 187 175 649 28 × 2 = 0 + 0.981 119 632 720 947 265 625 000 008 374 351 298 56;
  • 34) 0.981 119 632 720 947 265 625 000 008 374 351 298 56 × 2 = 1 + 0.962 239 265 441 894 531 250 000 016 748 702 597 12;
  • 35) 0.962 239 265 441 894 531 250 000 016 748 702 597 12 × 2 = 1 + 0.924 478 530 883 789 062 500 000 033 497 405 194 24;
  • 36) 0.924 478 530 883 789 062 500 000 033 497 405 194 24 × 2 = 1 + 0.848 957 061 767 578 125 000 000 066 994 810 388 48;
  • 37) 0.848 957 061 767 578 125 000 000 066 994 810 388 48 × 2 = 1 + 0.697 914 123 535 156 250 000 000 133 989 620 776 96;
  • 38) 0.697 914 123 535 156 250 000 000 133 989 620 776 96 × 2 = 1 + 0.395 828 247 070 312 500 000 000 267 979 241 553 92;
  • 39) 0.395 828 247 070 312 500 000 000 267 979 241 553 92 × 2 = 0 + 0.791 656 494 140 625 000 000 000 535 958 483 107 84;
  • 40) 0.791 656 494 140 625 000 000 000 535 958 483 107 84 × 2 = 1 + 0.583 312 988 281 250 000 000 001 071 916 966 215 68;
  • 41) 0.583 312 988 281 250 000 000 001 071 916 966 215 68 × 2 = 1 + 0.166 625 976 562 500 000 000 002 143 833 932 431 36;
  • 42) 0.166 625 976 562 500 000 000 002 143 833 932 431 36 × 2 = 0 + 0.333 251 953 125 000 000 000 004 287 667 864 862 72;
  • 43) 0.333 251 953 125 000 000 000 004 287 667 864 862 72 × 2 = 0 + 0.666 503 906 250 000 000 000 008 575 335 729 725 44;
  • 44) 0.666 503 906 250 000 000 000 008 575 335 729 725 44 × 2 = 1 + 0.333 007 812 500 000 000 000 017 150 671 459 450 88;
  • 45) 0.333 007 812 500 000 000 000 017 150 671 459 450 88 × 2 = 0 + 0.666 015 625 000 000 000 000 034 301 342 918 901 76;
  • 46) 0.666 015 625 000 000 000 000 034 301 342 918 901 76 × 2 = 1 + 0.332 031 250 000 000 000 000 068 602 685 837 803 52;
  • 47) 0.332 031 250 000 000 000 000 068 602 685 837 803 52 × 2 = 0 + 0.664 062 500 000 000 000 000 137 205 371 675 607 04;
  • 48) 0.664 062 500 000 000 000 000 137 205 371 675 607 04 × 2 = 1 + 0.328 125 000 000 000 000 000 274 410 743 351 214 08;
  • 49) 0.328 125 000 000 000 000 000 274 410 743 351 214 08 × 2 = 0 + 0.656 250 000 000 000 000 000 548 821 486 702 428 16;
  • 50) 0.656 250 000 000 000 000 000 548 821 486 702 428 16 × 2 = 1 + 0.312 500 000 000 000 000 001 097 642 973 404 856 32;
  • 51) 0.312 500 000 000 000 000 001 097 642 973 404 856 32 × 2 = 0 + 0.625 000 000 000 000 000 002 195 285 946 809 712 64;
  • 52) 0.625 000 000 000 000 000 002 195 285 946 809 712 64 × 2 = 1 + 0.250 000 000 000 000 000 004 390 571 893 619 425 28;
  • 53) 0.250 000 000 000 000 000 004 390 571 893 619 425 28 × 2 = 0 + 0.500 000 000 000 000 000 008 781 143 787 238 850 56;
  • 54) 0.500 000 000 000 000 000 008 781 143 787 238 850 56 × 2 = 1 + 0.000 000 000 000 000 000 017 562 287 574 477 701 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).


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:


0.453 109 999 999 999 957 243 090 875 635 971 316 18(10) =

0.0111 0011 1111 1111 0000 0100 0101 0111 0111 1101 1001 0101 0101 01(2)

5. Positive number before normalization:

0.453 109 999 999 999 957 243 090 875 635 971 316 18(10) =

0.0111 0011 1111 1111 0000 0100 0101 0111 0111 1101 1001 0101 0101 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.453 109 999 999 999 957 243 090 875 635 971 316 18(10) =

0.0111 0011 1111 1111 0000 0100 0101 0111 0111 1101 1001 0101 0101 01(2) =

0.0111 0011 1111 1111 0000 0100 0101 0111 0111 1101 1001 0101 0101 01(2) × 20 =

1.1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101(2) × 2-2


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): -2

Mantissa (not normalized):
1.1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101 =

1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101


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) =
011 1111 1101

Mantissa (52 bits) =
1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101


Decimal number 0.453 109 999 999 999 957 243 090 875 635 971 316 18 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 1100 1111 1111 1100 0001 0001 0101 1101 1111 0110 0101 0101 0101

Share

» Convert decimal 0.453 109 999 999 999 957 243 090 875 635 971 315 59 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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