1.745 459 324 169 999 826 281 696 186 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 82(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
1.745 459 324 169 999 826 281 696 186 82(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: 0.745 459 324 169 999 826 281 696 186 82.

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) 0.745 459 324 169 999 826 281 696 186 82 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 64;
  • 2) 0.490 918 648 339 999 652 563 392 373 64 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 28;
  • 3) 0.981 837 296 679 999 305 126 784 747 28 × 2 = 1 + 0.963 674 593 359 998 610 253 569 494 56;
  • 4) 0.963 674 593 359 998 610 253 569 494 56 × 2 = 1 + 0.927 349 186 719 997 220 507 138 989 12;
  • 5) 0.927 349 186 719 997 220 507 138 989 12 × 2 = 1 + 0.854 698 373 439 994 441 014 277 978 24;
  • 6) 0.854 698 373 439 994 441 014 277 978 24 × 2 = 1 + 0.709 396 746 879 988 882 028 555 956 48;
  • 7) 0.709 396 746 879 988 882 028 555 956 48 × 2 = 1 + 0.418 793 493 759 977 764 057 111 912 96;
  • 8) 0.418 793 493 759 977 764 057 111 912 96 × 2 = 0 + 0.837 586 987 519 955 528 114 223 825 92;
  • 9) 0.837 586 987 519 955 528 114 223 825 92 × 2 = 1 + 0.675 173 975 039 911 056 228 447 651 84;
  • 10) 0.675 173 975 039 911 056 228 447 651 84 × 2 = 1 + 0.350 347 950 079 822 112 456 895 303 68;
  • 11) 0.350 347 950 079 822 112 456 895 303 68 × 2 = 0 + 0.700 695 900 159 644 224 913 790 607 36;
  • 12) 0.700 695 900 159 644 224 913 790 607 36 × 2 = 1 + 0.401 391 800 319 288 449 827 581 214 72;
  • 13) 0.401 391 800 319 288 449 827 581 214 72 × 2 = 0 + 0.802 783 600 638 576 899 655 162 429 44;
  • 14) 0.802 783 600 638 576 899 655 162 429 44 × 2 = 1 + 0.605 567 201 277 153 799 310 324 858 88;
  • 15) 0.605 567 201 277 153 799 310 324 858 88 × 2 = 1 + 0.211 134 402 554 307 598 620 649 717 76;
  • 16) 0.211 134 402 554 307 598 620 649 717 76 × 2 = 0 + 0.422 268 805 108 615 197 241 299 435 52;
  • 17) 0.422 268 805 108 615 197 241 299 435 52 × 2 = 0 + 0.844 537 610 217 230 394 482 598 871 04;
  • 18) 0.844 537 610 217 230 394 482 598 871 04 × 2 = 1 + 0.689 075 220 434 460 788 965 197 742 08;
  • 19) 0.689 075 220 434 460 788 965 197 742 08 × 2 = 1 + 0.378 150 440 868 921 577 930 395 484 16;
  • 20) 0.378 150 440 868 921 577 930 395 484 16 × 2 = 0 + 0.756 300 881 737 843 155 860 790 968 32;
  • 21) 0.756 300 881 737 843 155 860 790 968 32 × 2 = 1 + 0.512 601 763 475 686 311 721 581 936 64;
  • 22) 0.512 601 763 475 686 311 721 581 936 64 × 2 = 1 + 0.025 203 526 951 372 623 443 163 873 28;
  • 23) 0.025 203 526 951 372 623 443 163 873 28 × 2 = 0 + 0.050 407 053 902 745 246 886 327 746 56;
  • 24) 0.050 407 053 902 745 246 886 327 746 56 × 2 = 0 + 0.100 814 107 805 490 493 772 655 493 12;
  • 25) 0.100 814 107 805 490 493 772 655 493 12 × 2 = 0 + 0.201 628 215 610 980 987 545 310 986 24;
  • 26) 0.201 628 215 610 980 987 545 310 986 24 × 2 = 0 + 0.403 256 431 221 961 975 090 621 972 48;
  • 27) 0.403 256 431 221 961 975 090 621 972 48 × 2 = 0 + 0.806 512 862 443 923 950 181 243 944 96;
  • 28) 0.806 512 862 443 923 950 181 243 944 96 × 2 = 1 + 0.613 025 724 887 847 900 362 487 889 92;
  • 29) 0.613 025 724 887 847 900 362 487 889 92 × 2 = 1 + 0.226 051 449 775 695 800 724 975 779 84;
  • 30) 0.226 051 449 775 695 800 724 975 779 84 × 2 = 0 + 0.452 102 899 551 391 601 449 951 559 68;
  • 31) 0.452 102 899 551 391 601 449 951 559 68 × 2 = 0 + 0.904 205 799 102 783 202 899 903 119 36;
  • 32) 0.904 205 799 102 783 202 899 903 119 36 × 2 = 1 + 0.808 411 598 205 566 405 799 806 238 72;
  • 33) 0.808 411 598 205 566 405 799 806 238 72 × 2 = 1 + 0.616 823 196 411 132 811 599 612 477 44;
  • 34) 0.616 823 196 411 132 811 599 612 477 44 × 2 = 1 + 0.233 646 392 822 265 623 199 224 954 88;
  • 35) 0.233 646 392 822 265 623 199 224 954 88 × 2 = 0 + 0.467 292 785 644 531 246 398 449 909 76;
  • 36) 0.467 292 785 644 531 246 398 449 909 76 × 2 = 0 + 0.934 585 571 289 062 492 796 899 819 52;
  • 37) 0.934 585 571 289 062 492 796 899 819 52 × 2 = 1 + 0.869 171 142 578 124 985 593 799 639 04;
  • 38) 0.869 171 142 578 124 985 593 799 639 04 × 2 = 1 + 0.738 342 285 156 249 971 187 599 278 08;
  • 39) 0.738 342 285 156 249 971 187 599 278 08 × 2 = 1 + 0.476 684 570 312 499 942 375 198 556 16;
  • 40) 0.476 684 570 312 499 942 375 198 556 16 × 2 = 0 + 0.953 369 140 624 999 884 750 397 112 32;
  • 41) 0.953 369 140 624 999 884 750 397 112 32 × 2 = 1 + 0.906 738 281 249 999 769 500 794 224 64;
  • 42) 0.906 738 281 249 999 769 500 794 224 64 × 2 = 1 + 0.813 476 562 499 999 539 001 588 449 28;
  • 43) 0.813 476 562 499 999 539 001 588 449 28 × 2 = 1 + 0.626 953 124 999 999 078 003 176 898 56;
  • 44) 0.626 953 124 999 999 078 003 176 898 56 × 2 = 1 + 0.253 906 249 999 998 156 006 353 797 12;
  • 45) 0.253 906 249 999 998 156 006 353 797 12 × 2 = 0 + 0.507 812 499 999 996 312 012 707 594 24;
  • 46) 0.507 812 499 999 996 312 012 707 594 24 × 2 = 1 + 0.015 624 999 999 992 624 025 415 188 48;
  • 47) 0.015 624 999 999 992 624 025 415 188 48 × 2 = 0 + 0.031 249 999 999 985 248 050 830 376 96;
  • 48) 0.031 249 999 999 985 248 050 830 376 96 × 2 = 0 + 0.062 499 999 999 970 496 101 660 753 92;
  • 49) 0.062 499 999 999 970 496 101 660 753 92 × 2 = 0 + 0.124 999 999 999 940 992 203 321 507 84;
  • 50) 0.124 999 999 999 940 992 203 321 507 84 × 2 = 0 + 0.249 999 999 999 881 984 406 643 015 68;
  • 51) 0.249 999 999 999 881 984 406 643 015 68 × 2 = 0 + 0.499 999 999 999 763 968 813 286 031 36;
  • 52) 0.499 999 999 999 763 968 813 286 031 36 × 2 = 0 + 0.999 999 999 999 527 937 626 572 062 72;
  • 53) 0.999 999 999 999 527 937 626 572 062 72 × 2 = 1 + 0.999 999 999 999 055 875 253 144 125 44;

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:


0.745 459 324 169 999 826 281 696 186 82(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 82(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:


1.745 459 324 169 999 826 281 696 186 82(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 1.745 459 324 169 999 826 281 696 186 82 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