1.745 459 324 169 999 826 281 696 183 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 183 62(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
1.745 459 324 169 999 826 281 696 183 62(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.745 459 324 169 999 826 281 696 183 62.

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) 0.745 459 324 169 999 826 281 696 183 62 × 2 = 1 + 0.490 918 648 339 999 652 563 392 367 24;
  • 2) 0.490 918 648 339 999 652 563 392 367 24 × 2 = 0 + 0.981 837 296 679 999 305 126 784 734 48;
  • 3) 0.981 837 296 679 999 305 126 784 734 48 × 2 = 1 + 0.963 674 593 359 998 610 253 569 468 96;
  • 4) 0.963 674 593 359 998 610 253 569 468 96 × 2 = 1 + 0.927 349 186 719 997 220 507 138 937 92;
  • 5) 0.927 349 186 719 997 220 507 138 937 92 × 2 = 1 + 0.854 698 373 439 994 441 014 277 875 84;
  • 6) 0.854 698 373 439 994 441 014 277 875 84 × 2 = 1 + 0.709 396 746 879 988 882 028 555 751 68;
  • 7) 0.709 396 746 879 988 882 028 555 751 68 × 2 = 1 + 0.418 793 493 759 977 764 057 111 503 36;
  • 8) 0.418 793 493 759 977 764 057 111 503 36 × 2 = 0 + 0.837 586 987 519 955 528 114 223 006 72;
  • 9) 0.837 586 987 519 955 528 114 223 006 72 × 2 = 1 + 0.675 173 975 039 911 056 228 446 013 44;
  • 10) 0.675 173 975 039 911 056 228 446 013 44 × 2 = 1 + 0.350 347 950 079 822 112 456 892 026 88;
  • 11) 0.350 347 950 079 822 112 456 892 026 88 × 2 = 0 + 0.700 695 900 159 644 224 913 784 053 76;
  • 12) 0.700 695 900 159 644 224 913 784 053 76 × 2 = 1 + 0.401 391 800 319 288 449 827 568 107 52;
  • 13) 0.401 391 800 319 288 449 827 568 107 52 × 2 = 0 + 0.802 783 600 638 576 899 655 136 215 04;
  • 14) 0.802 783 600 638 576 899 655 136 215 04 × 2 = 1 + 0.605 567 201 277 153 799 310 272 430 08;
  • 15) 0.605 567 201 277 153 799 310 272 430 08 × 2 = 1 + 0.211 134 402 554 307 598 620 544 860 16;
  • 16) 0.211 134 402 554 307 598 620 544 860 16 × 2 = 0 + 0.422 268 805 108 615 197 241 089 720 32;
  • 17) 0.422 268 805 108 615 197 241 089 720 32 × 2 = 0 + 0.844 537 610 217 230 394 482 179 440 64;
  • 18) 0.844 537 610 217 230 394 482 179 440 64 × 2 = 1 + 0.689 075 220 434 460 788 964 358 881 28;
  • 19) 0.689 075 220 434 460 788 964 358 881 28 × 2 = 1 + 0.378 150 440 868 921 577 928 717 762 56;
  • 20) 0.378 150 440 868 921 577 928 717 762 56 × 2 = 0 + 0.756 300 881 737 843 155 857 435 525 12;
  • 21) 0.756 300 881 737 843 155 857 435 525 12 × 2 = 1 + 0.512 601 763 475 686 311 714 871 050 24;
  • 22) 0.512 601 763 475 686 311 714 871 050 24 × 2 = 1 + 0.025 203 526 951 372 623 429 742 100 48;
  • 23) 0.025 203 526 951 372 623 429 742 100 48 × 2 = 0 + 0.050 407 053 902 745 246 859 484 200 96;
  • 24) 0.050 407 053 902 745 246 859 484 200 96 × 2 = 0 + 0.100 814 107 805 490 493 718 968 401 92;
  • 25) 0.100 814 107 805 490 493 718 968 401 92 × 2 = 0 + 0.201 628 215 610 980 987 437 936 803 84;
  • 26) 0.201 628 215 610 980 987 437 936 803 84 × 2 = 0 + 0.403 256 431 221 961 974 875 873 607 68;
  • 27) 0.403 256 431 221 961 974 875 873 607 68 × 2 = 0 + 0.806 512 862 443 923 949 751 747 215 36;
  • 28) 0.806 512 862 443 923 949 751 747 215 36 × 2 = 1 + 0.613 025 724 887 847 899 503 494 430 72;
  • 29) 0.613 025 724 887 847 899 503 494 430 72 × 2 = 1 + 0.226 051 449 775 695 799 006 988 861 44;
  • 30) 0.226 051 449 775 695 799 006 988 861 44 × 2 = 0 + 0.452 102 899 551 391 598 013 977 722 88;
  • 31) 0.452 102 899 551 391 598 013 977 722 88 × 2 = 0 + 0.904 205 799 102 783 196 027 955 445 76;
  • 32) 0.904 205 799 102 783 196 027 955 445 76 × 2 = 1 + 0.808 411 598 205 566 392 055 910 891 52;
  • 33) 0.808 411 598 205 566 392 055 910 891 52 × 2 = 1 + 0.616 823 196 411 132 784 111 821 783 04;
  • 34) 0.616 823 196 411 132 784 111 821 783 04 × 2 = 1 + 0.233 646 392 822 265 568 223 643 566 08;
  • 35) 0.233 646 392 822 265 568 223 643 566 08 × 2 = 0 + 0.467 292 785 644 531 136 447 287 132 16;
  • 36) 0.467 292 785 644 531 136 447 287 132 16 × 2 = 0 + 0.934 585 571 289 062 272 894 574 264 32;
  • 37) 0.934 585 571 289 062 272 894 574 264 32 × 2 = 1 + 0.869 171 142 578 124 545 789 148 528 64;
  • 38) 0.869 171 142 578 124 545 789 148 528 64 × 2 = 1 + 0.738 342 285 156 249 091 578 297 057 28;
  • 39) 0.738 342 285 156 249 091 578 297 057 28 × 2 = 1 + 0.476 684 570 312 498 183 156 594 114 56;
  • 40) 0.476 684 570 312 498 183 156 594 114 56 × 2 = 0 + 0.953 369 140 624 996 366 313 188 229 12;
  • 41) 0.953 369 140 624 996 366 313 188 229 12 × 2 = 1 + 0.906 738 281 249 992 732 626 376 458 24;
  • 42) 0.906 738 281 249 992 732 626 376 458 24 × 2 = 1 + 0.813 476 562 499 985 465 252 752 916 48;
  • 43) 0.813 476 562 499 985 465 252 752 916 48 × 2 = 1 + 0.626 953 124 999 970 930 505 505 832 96;
  • 44) 0.626 953 124 999 970 930 505 505 832 96 × 2 = 1 + 0.253 906 249 999 941 861 011 011 665 92;
  • 45) 0.253 906 249 999 941 861 011 011 665 92 × 2 = 0 + 0.507 812 499 999 883 722 022 023 331 84;
  • 46) 0.507 812 499 999 883 722 022 023 331 84 × 2 = 1 + 0.015 624 999 999 767 444 044 046 663 68;
  • 47) 0.015 624 999 999 767 444 044 046 663 68 × 2 = 0 + 0.031 249 999 999 534 888 088 093 327 36;
  • 48) 0.031 249 999 999 534 888 088 093 327 36 × 2 = 0 + 0.062 499 999 999 069 776 176 186 654 72;
  • 49) 0.062 499 999 999 069 776 176 186 654 72 × 2 = 0 + 0.124 999 999 998 139 552 352 373 309 44;
  • 50) 0.124 999 999 998 139 552 352 373 309 44 × 2 = 0 + 0.249 999 999 996 279 104 704 746 618 88;
  • 51) 0.249 999 999 996 279 104 704 746 618 88 × 2 = 0 + 0.499 999 999 992 558 209 409 493 237 76;
  • 52) 0.499 999 999 992 558 209 409 493 237 76 × 2 = 0 + 0.999 999 999 985 116 418 818 986 475 52;
  • 53) 0.999 999 999 985 116 418 818 986 475 52 × 2 = 1 + 0.999 999 999 970 232 837 637 972 951 04;

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:


0.745 459 324 169 999 826 281 696 183 62(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 183 62(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.745 459 324 169 999 826 281 696 183 62(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) × 20


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): 0

Mantissa (not normalized):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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 1111

Mantissa (52 bits) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 696 183 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

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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