1.745 459 324 169 999 826 281 696 186 677 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 677(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
1.745 459 324 169 999 826 281 696 186 677(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: 0.745 459 324 169 999 826 281 696 186 677.

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) 0.745 459 324 169 999 826 281 696 186 677 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 354;
  • 2) 0.490 918 648 339 999 652 563 392 373 354 × 2 = 0 + 0.981 837 296 679 999 305 126 784 746 708;
  • 3) 0.981 837 296 679 999 305 126 784 746 708 × 2 = 1 + 0.963 674 593 359 998 610 253 569 493 416;
  • 4) 0.963 674 593 359 998 610 253 569 493 416 × 2 = 1 + 0.927 349 186 719 997 220 507 138 986 832;
  • 5) 0.927 349 186 719 997 220 507 138 986 832 × 2 = 1 + 0.854 698 373 439 994 441 014 277 973 664;
  • 6) 0.854 698 373 439 994 441 014 277 973 664 × 2 = 1 + 0.709 396 746 879 988 882 028 555 947 328;
  • 7) 0.709 396 746 879 988 882 028 555 947 328 × 2 = 1 + 0.418 793 493 759 977 764 057 111 894 656;
  • 8) 0.418 793 493 759 977 764 057 111 894 656 × 2 = 0 + 0.837 586 987 519 955 528 114 223 789 312;
  • 9) 0.837 586 987 519 955 528 114 223 789 312 × 2 = 1 + 0.675 173 975 039 911 056 228 447 578 624;
  • 10) 0.675 173 975 039 911 056 228 447 578 624 × 2 = 1 + 0.350 347 950 079 822 112 456 895 157 248;
  • 11) 0.350 347 950 079 822 112 456 895 157 248 × 2 = 0 + 0.700 695 900 159 644 224 913 790 314 496;
  • 12) 0.700 695 900 159 644 224 913 790 314 496 × 2 = 1 + 0.401 391 800 319 288 449 827 580 628 992;
  • 13) 0.401 391 800 319 288 449 827 580 628 992 × 2 = 0 + 0.802 783 600 638 576 899 655 161 257 984;
  • 14) 0.802 783 600 638 576 899 655 161 257 984 × 2 = 1 + 0.605 567 201 277 153 799 310 322 515 968;
  • 15) 0.605 567 201 277 153 799 310 322 515 968 × 2 = 1 + 0.211 134 402 554 307 598 620 645 031 936;
  • 16) 0.211 134 402 554 307 598 620 645 031 936 × 2 = 0 + 0.422 268 805 108 615 197 241 290 063 872;
  • 17) 0.422 268 805 108 615 197 241 290 063 872 × 2 = 0 + 0.844 537 610 217 230 394 482 580 127 744;
  • 18) 0.844 537 610 217 230 394 482 580 127 744 × 2 = 1 + 0.689 075 220 434 460 788 965 160 255 488;
  • 19) 0.689 075 220 434 460 788 965 160 255 488 × 2 = 1 + 0.378 150 440 868 921 577 930 320 510 976;
  • 20) 0.378 150 440 868 921 577 930 320 510 976 × 2 = 0 + 0.756 300 881 737 843 155 860 641 021 952;
  • 21) 0.756 300 881 737 843 155 860 641 021 952 × 2 = 1 + 0.512 601 763 475 686 311 721 282 043 904;
  • 22) 0.512 601 763 475 686 311 721 282 043 904 × 2 = 1 + 0.025 203 526 951 372 623 442 564 087 808;
  • 23) 0.025 203 526 951 372 623 442 564 087 808 × 2 = 0 + 0.050 407 053 902 745 246 885 128 175 616;
  • 24) 0.050 407 053 902 745 246 885 128 175 616 × 2 = 0 + 0.100 814 107 805 490 493 770 256 351 232;
  • 25) 0.100 814 107 805 490 493 770 256 351 232 × 2 = 0 + 0.201 628 215 610 980 987 540 512 702 464;
  • 26) 0.201 628 215 610 980 987 540 512 702 464 × 2 = 0 + 0.403 256 431 221 961 975 081 025 404 928;
  • 27) 0.403 256 431 221 961 975 081 025 404 928 × 2 = 0 + 0.806 512 862 443 923 950 162 050 809 856;
  • 28) 0.806 512 862 443 923 950 162 050 809 856 × 2 = 1 + 0.613 025 724 887 847 900 324 101 619 712;
  • 29) 0.613 025 724 887 847 900 324 101 619 712 × 2 = 1 + 0.226 051 449 775 695 800 648 203 239 424;
  • 30) 0.226 051 449 775 695 800 648 203 239 424 × 2 = 0 + 0.452 102 899 551 391 601 296 406 478 848;
  • 31) 0.452 102 899 551 391 601 296 406 478 848 × 2 = 0 + 0.904 205 799 102 783 202 592 812 957 696;
  • 32) 0.904 205 799 102 783 202 592 812 957 696 × 2 = 1 + 0.808 411 598 205 566 405 185 625 915 392;
  • 33) 0.808 411 598 205 566 405 185 625 915 392 × 2 = 1 + 0.616 823 196 411 132 810 371 251 830 784;
  • 34) 0.616 823 196 411 132 810 371 251 830 784 × 2 = 1 + 0.233 646 392 822 265 620 742 503 661 568;
  • 35) 0.233 646 392 822 265 620 742 503 661 568 × 2 = 0 + 0.467 292 785 644 531 241 485 007 323 136;
  • 36) 0.467 292 785 644 531 241 485 007 323 136 × 2 = 0 + 0.934 585 571 289 062 482 970 014 646 272;
  • 37) 0.934 585 571 289 062 482 970 014 646 272 × 2 = 1 + 0.869 171 142 578 124 965 940 029 292 544;
  • 38) 0.869 171 142 578 124 965 940 029 292 544 × 2 = 1 + 0.738 342 285 156 249 931 880 058 585 088;
  • 39) 0.738 342 285 156 249 931 880 058 585 088 × 2 = 1 + 0.476 684 570 312 499 863 760 117 170 176;
  • 40) 0.476 684 570 312 499 863 760 117 170 176 × 2 = 0 + 0.953 369 140 624 999 727 520 234 340 352;
  • 41) 0.953 369 140 624 999 727 520 234 340 352 × 2 = 1 + 0.906 738 281 249 999 455 040 468 680 704;
  • 42) 0.906 738 281 249 999 455 040 468 680 704 × 2 = 1 + 0.813 476 562 499 998 910 080 937 361 408;
  • 43) 0.813 476 562 499 998 910 080 937 361 408 × 2 = 1 + 0.626 953 124 999 997 820 161 874 722 816;
  • 44) 0.626 953 124 999 997 820 161 874 722 816 × 2 = 1 + 0.253 906 249 999 995 640 323 749 445 632;
  • 45) 0.253 906 249 999 995 640 323 749 445 632 × 2 = 0 + 0.507 812 499 999 991 280 647 498 891 264;
  • 46) 0.507 812 499 999 991 280 647 498 891 264 × 2 = 1 + 0.015 624 999 999 982 561 294 997 782 528;
  • 47) 0.015 624 999 999 982 561 294 997 782 528 × 2 = 0 + 0.031 249 999 999 965 122 589 995 565 056;
  • 48) 0.031 249 999 999 965 122 589 995 565 056 × 2 = 0 + 0.062 499 999 999 930 245 179 991 130 112;
  • 49) 0.062 499 999 999 930 245 179 991 130 112 × 2 = 0 + 0.124 999 999 999 860 490 359 982 260 224;
  • 50) 0.124 999 999 999 860 490 359 982 260 224 × 2 = 0 + 0.249 999 999 999 720 980 719 964 520 448;
  • 51) 0.249 999 999 999 720 980 719 964 520 448 × 2 = 0 + 0.499 999 999 999 441 961 439 929 040 896;
  • 52) 0.499 999 999 999 441 961 439 929 040 896 × 2 = 0 + 0.999 999 999 998 883 922 879 858 081 792;
  • 53) 0.999 999 999 998 883 922 879 858 081 792 × 2 = 1 + 0.999 999 999 997 767 845 759 716 163 584;

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:


0.745 459 324 169 999 826 281 696 186 677(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 677(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:


1.745 459 324 169 999 826 281 696 186 677(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 1.745 459 324 169 999 826 281 696 186 677 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