0.453 109 999 999 999 957 243 090 875 635 971 315 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.453 109 999 999 999 957 243 090 875 635 971 315 59(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
0.453 109 999 999 999 957 243 090 875 635 971 315 59(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: 0.453 109 999 999 999 957 243 090 875 635 971 315 59.

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) 0.453 109 999 999 999 957 243 090 875 635 971 315 59 × 2 = 0 + 0.906 219 999 999 999 914 486 181 751 271 942 631 18;
  • 2) 0.906 219 999 999 999 914 486 181 751 271 942 631 18 × 2 = 1 + 0.812 439 999 999 999 828 972 363 502 543 885 262 36;
  • 3) 0.812 439 999 999 999 828 972 363 502 543 885 262 36 × 2 = 1 + 0.624 879 999 999 999 657 944 727 005 087 770 524 72;
  • 4) 0.624 879 999 999 999 657 944 727 005 087 770 524 72 × 2 = 1 + 0.249 759 999 999 999 315 889 454 010 175 541 049 44;
  • 5) 0.249 759 999 999 999 315 889 454 010 175 541 049 44 × 2 = 0 + 0.499 519 999 999 998 631 778 908 020 351 082 098 88;
  • 6) 0.499 519 999 999 998 631 778 908 020 351 082 098 88 × 2 = 0 + 0.999 039 999 999 997 263 557 816 040 702 164 197 76;
  • 7) 0.999 039 999 999 997 263 557 816 040 702 164 197 76 × 2 = 1 + 0.998 079 999 999 994 527 115 632 081 404 328 395 52;
  • 8) 0.998 079 999 999 994 527 115 632 081 404 328 395 52 × 2 = 1 + 0.996 159 999 999 989 054 231 264 162 808 656 791 04;
  • 9) 0.996 159 999 999 989 054 231 264 162 808 656 791 04 × 2 = 1 + 0.992 319 999 999 978 108 462 528 325 617 313 582 08;
  • 10) 0.992 319 999 999 978 108 462 528 325 617 313 582 08 × 2 = 1 + 0.984 639 999 999 956 216 925 056 651 234 627 164 16;
  • 11) 0.984 639 999 999 956 216 925 056 651 234 627 164 16 × 2 = 1 + 0.969 279 999 999 912 433 850 113 302 469 254 328 32;
  • 12) 0.969 279 999 999 912 433 850 113 302 469 254 328 32 × 2 = 1 + 0.938 559 999 999 824 867 700 226 604 938 508 656 64;
  • 13) 0.938 559 999 999 824 867 700 226 604 938 508 656 64 × 2 = 1 + 0.877 119 999 999 649 735 400 453 209 877 017 313 28;
  • 14) 0.877 119 999 999 649 735 400 453 209 877 017 313 28 × 2 = 1 + 0.754 239 999 999 299 470 800 906 419 754 034 626 56;
  • 15) 0.754 239 999 999 299 470 800 906 419 754 034 626 56 × 2 = 1 + 0.508 479 999 998 598 941 601 812 839 508 069 253 12;
  • 16) 0.508 479 999 998 598 941 601 812 839 508 069 253 12 × 2 = 1 + 0.016 959 999 997 197 883 203 625 679 016 138 506 24;
  • 17) 0.016 959 999 997 197 883 203 625 679 016 138 506 24 × 2 = 0 + 0.033 919 999 994 395 766 407 251 358 032 277 012 48;
  • 18) 0.033 919 999 994 395 766 407 251 358 032 277 012 48 × 2 = 0 + 0.067 839 999 988 791 532 814 502 716 064 554 024 96;
  • 19) 0.067 839 999 988 791 532 814 502 716 064 554 024 96 × 2 = 0 + 0.135 679 999 977 583 065 629 005 432 129 108 049 92;
  • 20) 0.135 679 999 977 583 065 629 005 432 129 108 049 92 × 2 = 0 + 0.271 359 999 955 166 131 258 010 864 258 216 099 84;
  • 21) 0.271 359 999 955 166 131 258 010 864 258 216 099 84 × 2 = 0 + 0.542 719 999 910 332 262 516 021 728 516 432 199 68;
  • 22) 0.542 719 999 910 332 262 516 021 728 516 432 199 68 × 2 = 1 + 0.085 439 999 820 664 525 032 043 457 032 864 399 36;
  • 23) 0.085 439 999 820 664 525 032 043 457 032 864 399 36 × 2 = 0 + 0.170 879 999 641 329 050 064 086 914 065 728 798 72;
  • 24) 0.170 879 999 641 329 050 064 086 914 065 728 798 72 × 2 = 0 + 0.341 759 999 282 658 100 128 173 828 131 457 597 44;
  • 25) 0.341 759 999 282 658 100 128 173 828 131 457 597 44 × 2 = 0 + 0.683 519 998 565 316 200 256 347 656 262 915 194 88;
  • 26) 0.683 519 998 565 316 200 256 347 656 262 915 194 88 × 2 = 1 + 0.367 039 997 130 632 400 512 695 312 525 830 389 76;
  • 27) 0.367 039 997 130 632 400 512 695 312 525 830 389 76 × 2 = 0 + 0.734 079 994 261 264 801 025 390 625 051 660 779 52;
  • 28) 0.734 079 994 261 264 801 025 390 625 051 660 779 52 × 2 = 1 + 0.468 159 988 522 529 602 050 781 250 103 321 559 04;
  • 29) 0.468 159 988 522 529 602 050 781 250 103 321 559 04 × 2 = 0 + 0.936 319 977 045 059 204 101 562 500 206 643 118 08;
  • 30) 0.936 319 977 045 059 204 101 562 500 206 643 118 08 × 2 = 1 + 0.872 639 954 090 118 408 203 125 000 413 286 236 16;
  • 31) 0.872 639 954 090 118 408 203 125 000 413 286 236 16 × 2 = 1 + 0.745 279 908 180 236 816 406 250 000 826 572 472 32;
  • 32) 0.745 279 908 180 236 816 406 250 000 826 572 472 32 × 2 = 1 + 0.490 559 816 360 473 632 812 500 001 653 144 944 64;
  • 33) 0.490 559 816 360 473 632 812 500 001 653 144 944 64 × 2 = 0 + 0.981 119 632 720 947 265 625 000 003 306 289 889 28;
  • 34) 0.981 119 632 720 947 265 625 000 003 306 289 889 28 × 2 = 1 + 0.962 239 265 441 894 531 250 000 006 612 579 778 56;
  • 35) 0.962 239 265 441 894 531 250 000 006 612 579 778 56 × 2 = 1 + 0.924 478 530 883 789 062 500 000 013 225 159 557 12;
  • 36) 0.924 478 530 883 789 062 500 000 013 225 159 557 12 × 2 = 1 + 0.848 957 061 767 578 125 000 000 026 450 319 114 24;
  • 37) 0.848 957 061 767 578 125 000 000 026 450 319 114 24 × 2 = 1 + 0.697 914 123 535 156 250 000 000 052 900 638 228 48;
  • 38) 0.697 914 123 535 156 250 000 000 052 900 638 228 48 × 2 = 1 + 0.395 828 247 070 312 500 000 000 105 801 276 456 96;
  • 39) 0.395 828 247 070 312 500 000 000 105 801 276 456 96 × 2 = 0 + 0.791 656 494 140 625 000 000 000 211 602 552 913 92;
  • 40) 0.791 656 494 140 625 000 000 000 211 602 552 913 92 × 2 = 1 + 0.583 312 988 281 250 000 000 000 423 205 105 827 84;
  • 41) 0.583 312 988 281 250 000 000 000 423 205 105 827 84 × 2 = 1 + 0.166 625 976 562 500 000 000 000 846 410 211 655 68;
  • 42) 0.166 625 976 562 500 000 000 000 846 410 211 655 68 × 2 = 0 + 0.333 251 953 125 000 000 000 001 692 820 423 311 36;
  • 43) 0.333 251 953 125 000 000 000 001 692 820 423 311 36 × 2 = 0 + 0.666 503 906 250 000 000 000 003 385 640 846 622 72;
  • 44) 0.666 503 906 250 000 000 000 003 385 640 846 622 72 × 2 = 1 + 0.333 007 812 500 000 000 000 006 771 281 693 245 44;
  • 45) 0.333 007 812 500 000 000 000 006 771 281 693 245 44 × 2 = 0 + 0.666 015 625 000 000 000 000 013 542 563 386 490 88;
  • 46) 0.666 015 625 000 000 000 000 013 542 563 386 490 88 × 2 = 1 + 0.332 031 250 000 000 000 000 027 085 126 772 981 76;
  • 47) 0.332 031 250 000 000 000 000 027 085 126 772 981 76 × 2 = 0 + 0.664 062 500 000 000 000 000 054 170 253 545 963 52;
  • 48) 0.664 062 500 000 000 000 000 054 170 253 545 963 52 × 2 = 1 + 0.328 125 000 000 000 000 000 108 340 507 091 927 04;
  • 49) 0.328 125 000 000 000 000 000 108 340 507 091 927 04 × 2 = 0 + 0.656 250 000 000 000 000 000 216 681 014 183 854 08;
  • 50) 0.656 250 000 000 000 000 000 216 681 014 183 854 08 × 2 = 1 + 0.312 500 000 000 000 000 000 433 362 028 367 708 16;
  • 51) 0.312 500 000 000 000 000 000 433 362 028 367 708 16 × 2 = 0 + 0.625 000 000 000 000 000 000 866 724 056 735 416 32;
  • 52) 0.625 000 000 000 000 000 000 866 724 056 735 416 32 × 2 = 1 + 0.250 000 000 000 000 000 001 733 448 113 470 832 64;
  • 53) 0.250 000 000 000 000 000 001 733 448 113 470 832 64 × 2 = 0 + 0.500 000 000 000 000 000 003 466 896 226 941 665 28;
  • 54) 0.500 000 000 000 000 000 003 466 896 226 941 665 28 × 2 = 1 + 0.000 000 000 000 000 000 006 933 792 453 883 330 56;

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:


0.453 109 999 999 999 957 243 090 875 635 971 315 59(10) =

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

5. Positive number before normalization:

0.453 109 999 999 999 957 243 090 875 635 971 315 59(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:


0.453 109 999 999 999 957 243 090 875 635 971 315 59(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 0.453 109 999 999 999 957 243 090 875 635 971 315 59 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

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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