1.745 459 324 169 999 826 281 696 181 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 181 6(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
1.745 459 324 169 999 826 281 696 181 6(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: 0.745 459 324 169 999 826 281 696 181 6.

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) 0.745 459 324 169 999 826 281 696 181 6 × 2 = 1 + 0.490 918 648 339 999 652 563 392 363 2;
  • 2) 0.490 918 648 339 999 652 563 392 363 2 × 2 = 0 + 0.981 837 296 679 999 305 126 784 726 4;
  • 3) 0.981 837 296 679 999 305 126 784 726 4 × 2 = 1 + 0.963 674 593 359 998 610 253 569 452 8;
  • 4) 0.963 674 593 359 998 610 253 569 452 8 × 2 = 1 + 0.927 349 186 719 997 220 507 138 905 6;
  • 5) 0.927 349 186 719 997 220 507 138 905 6 × 2 = 1 + 0.854 698 373 439 994 441 014 277 811 2;
  • 6) 0.854 698 373 439 994 441 014 277 811 2 × 2 = 1 + 0.709 396 746 879 988 882 028 555 622 4;
  • 7) 0.709 396 746 879 988 882 028 555 622 4 × 2 = 1 + 0.418 793 493 759 977 764 057 111 244 8;
  • 8) 0.418 793 493 759 977 764 057 111 244 8 × 2 = 0 + 0.837 586 987 519 955 528 114 222 489 6;
  • 9) 0.837 586 987 519 955 528 114 222 489 6 × 2 = 1 + 0.675 173 975 039 911 056 228 444 979 2;
  • 10) 0.675 173 975 039 911 056 228 444 979 2 × 2 = 1 + 0.350 347 950 079 822 112 456 889 958 4;
  • 11) 0.350 347 950 079 822 112 456 889 958 4 × 2 = 0 + 0.700 695 900 159 644 224 913 779 916 8;
  • 12) 0.700 695 900 159 644 224 913 779 916 8 × 2 = 1 + 0.401 391 800 319 288 449 827 559 833 6;
  • 13) 0.401 391 800 319 288 449 827 559 833 6 × 2 = 0 + 0.802 783 600 638 576 899 655 119 667 2;
  • 14) 0.802 783 600 638 576 899 655 119 667 2 × 2 = 1 + 0.605 567 201 277 153 799 310 239 334 4;
  • 15) 0.605 567 201 277 153 799 310 239 334 4 × 2 = 1 + 0.211 134 402 554 307 598 620 478 668 8;
  • 16) 0.211 134 402 554 307 598 620 478 668 8 × 2 = 0 + 0.422 268 805 108 615 197 240 957 337 6;
  • 17) 0.422 268 805 108 615 197 240 957 337 6 × 2 = 0 + 0.844 537 610 217 230 394 481 914 675 2;
  • 18) 0.844 537 610 217 230 394 481 914 675 2 × 2 = 1 + 0.689 075 220 434 460 788 963 829 350 4;
  • 19) 0.689 075 220 434 460 788 963 829 350 4 × 2 = 1 + 0.378 150 440 868 921 577 927 658 700 8;
  • 20) 0.378 150 440 868 921 577 927 658 700 8 × 2 = 0 + 0.756 300 881 737 843 155 855 317 401 6;
  • 21) 0.756 300 881 737 843 155 855 317 401 6 × 2 = 1 + 0.512 601 763 475 686 311 710 634 803 2;
  • 22) 0.512 601 763 475 686 311 710 634 803 2 × 2 = 1 + 0.025 203 526 951 372 623 421 269 606 4;
  • 23) 0.025 203 526 951 372 623 421 269 606 4 × 2 = 0 + 0.050 407 053 902 745 246 842 539 212 8;
  • 24) 0.050 407 053 902 745 246 842 539 212 8 × 2 = 0 + 0.100 814 107 805 490 493 685 078 425 6;
  • 25) 0.100 814 107 805 490 493 685 078 425 6 × 2 = 0 + 0.201 628 215 610 980 987 370 156 851 2;
  • 26) 0.201 628 215 610 980 987 370 156 851 2 × 2 = 0 + 0.403 256 431 221 961 974 740 313 702 4;
  • 27) 0.403 256 431 221 961 974 740 313 702 4 × 2 = 0 + 0.806 512 862 443 923 949 480 627 404 8;
  • 28) 0.806 512 862 443 923 949 480 627 404 8 × 2 = 1 + 0.613 025 724 887 847 898 961 254 809 6;
  • 29) 0.613 025 724 887 847 898 961 254 809 6 × 2 = 1 + 0.226 051 449 775 695 797 922 509 619 2;
  • 30) 0.226 051 449 775 695 797 922 509 619 2 × 2 = 0 + 0.452 102 899 551 391 595 845 019 238 4;
  • 31) 0.452 102 899 551 391 595 845 019 238 4 × 2 = 0 + 0.904 205 799 102 783 191 690 038 476 8;
  • 32) 0.904 205 799 102 783 191 690 038 476 8 × 2 = 1 + 0.808 411 598 205 566 383 380 076 953 6;
  • 33) 0.808 411 598 205 566 383 380 076 953 6 × 2 = 1 + 0.616 823 196 411 132 766 760 153 907 2;
  • 34) 0.616 823 196 411 132 766 760 153 907 2 × 2 = 1 + 0.233 646 392 822 265 533 520 307 814 4;
  • 35) 0.233 646 392 822 265 533 520 307 814 4 × 2 = 0 + 0.467 292 785 644 531 067 040 615 628 8;
  • 36) 0.467 292 785 644 531 067 040 615 628 8 × 2 = 0 + 0.934 585 571 289 062 134 081 231 257 6;
  • 37) 0.934 585 571 289 062 134 081 231 257 6 × 2 = 1 + 0.869 171 142 578 124 268 162 462 515 2;
  • 38) 0.869 171 142 578 124 268 162 462 515 2 × 2 = 1 + 0.738 342 285 156 248 536 324 925 030 4;
  • 39) 0.738 342 285 156 248 536 324 925 030 4 × 2 = 1 + 0.476 684 570 312 497 072 649 850 060 8;
  • 40) 0.476 684 570 312 497 072 649 850 060 8 × 2 = 0 + 0.953 369 140 624 994 145 299 700 121 6;
  • 41) 0.953 369 140 624 994 145 299 700 121 6 × 2 = 1 + 0.906 738 281 249 988 290 599 400 243 2;
  • 42) 0.906 738 281 249 988 290 599 400 243 2 × 2 = 1 + 0.813 476 562 499 976 581 198 800 486 4;
  • 43) 0.813 476 562 499 976 581 198 800 486 4 × 2 = 1 + 0.626 953 124 999 953 162 397 600 972 8;
  • 44) 0.626 953 124 999 953 162 397 600 972 8 × 2 = 1 + 0.253 906 249 999 906 324 795 201 945 6;
  • 45) 0.253 906 249 999 906 324 795 201 945 6 × 2 = 0 + 0.507 812 499 999 812 649 590 403 891 2;
  • 46) 0.507 812 499 999 812 649 590 403 891 2 × 2 = 1 + 0.015 624 999 999 625 299 180 807 782 4;
  • 47) 0.015 624 999 999 625 299 180 807 782 4 × 2 = 0 + 0.031 249 999 999 250 598 361 615 564 8;
  • 48) 0.031 249 999 999 250 598 361 615 564 8 × 2 = 0 + 0.062 499 999 998 501 196 723 231 129 6;
  • 49) 0.062 499 999 998 501 196 723 231 129 6 × 2 = 0 + 0.124 999 999 997 002 393 446 462 259 2;
  • 50) 0.124 999 999 997 002 393 446 462 259 2 × 2 = 0 + 0.249 999 999 994 004 786 892 924 518 4;
  • 51) 0.249 999 999 994 004 786 892 924 518 4 × 2 = 0 + 0.499 999 999 988 009 573 785 849 036 8;
  • 52) 0.499 999 999 988 009 573 785 849 036 8 × 2 = 0 + 0.999 999 999 976 019 147 571 698 073 6;
  • 53) 0.999 999 999 976 019 147 571 698 073 6 × 2 = 1 + 0.999 999 999 952 038 295 143 396 147 2;

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:


0.745 459 324 169 999 826 281 696 181 6(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 181 6(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:


1.745 459 324 169 999 826 281 696 181 6(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 1.745 459 324 169 999 826 281 696 181 6 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