1.745 459 324 169 999 826 281 696 186 831 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 831(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
1.745 459 324 169 999 826 281 696 186 831(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: 0.745 459 324 169 999 826 281 696 186 831.

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) 0.745 459 324 169 999 826 281 696 186 831 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 662;
  • 2) 0.490 918 648 339 999 652 563 392 373 662 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 324;
  • 3) 0.981 837 296 679 999 305 126 784 747 324 × 2 = 1 + 0.963 674 593 359 998 610 253 569 494 648;
  • 4) 0.963 674 593 359 998 610 253 569 494 648 × 2 = 1 + 0.927 349 186 719 997 220 507 138 989 296;
  • 5) 0.927 349 186 719 997 220 507 138 989 296 × 2 = 1 + 0.854 698 373 439 994 441 014 277 978 592;
  • 6) 0.854 698 373 439 994 441 014 277 978 592 × 2 = 1 + 0.709 396 746 879 988 882 028 555 957 184;
  • 7) 0.709 396 746 879 988 882 028 555 957 184 × 2 = 1 + 0.418 793 493 759 977 764 057 111 914 368;
  • 8) 0.418 793 493 759 977 764 057 111 914 368 × 2 = 0 + 0.837 586 987 519 955 528 114 223 828 736;
  • 9) 0.837 586 987 519 955 528 114 223 828 736 × 2 = 1 + 0.675 173 975 039 911 056 228 447 657 472;
  • 10) 0.675 173 975 039 911 056 228 447 657 472 × 2 = 1 + 0.350 347 950 079 822 112 456 895 314 944;
  • 11) 0.350 347 950 079 822 112 456 895 314 944 × 2 = 0 + 0.700 695 900 159 644 224 913 790 629 888;
  • 12) 0.700 695 900 159 644 224 913 790 629 888 × 2 = 1 + 0.401 391 800 319 288 449 827 581 259 776;
  • 13) 0.401 391 800 319 288 449 827 581 259 776 × 2 = 0 + 0.802 783 600 638 576 899 655 162 519 552;
  • 14) 0.802 783 600 638 576 899 655 162 519 552 × 2 = 1 + 0.605 567 201 277 153 799 310 325 039 104;
  • 15) 0.605 567 201 277 153 799 310 325 039 104 × 2 = 1 + 0.211 134 402 554 307 598 620 650 078 208;
  • 16) 0.211 134 402 554 307 598 620 650 078 208 × 2 = 0 + 0.422 268 805 108 615 197 241 300 156 416;
  • 17) 0.422 268 805 108 615 197 241 300 156 416 × 2 = 0 + 0.844 537 610 217 230 394 482 600 312 832;
  • 18) 0.844 537 610 217 230 394 482 600 312 832 × 2 = 1 + 0.689 075 220 434 460 788 965 200 625 664;
  • 19) 0.689 075 220 434 460 788 965 200 625 664 × 2 = 1 + 0.378 150 440 868 921 577 930 401 251 328;
  • 20) 0.378 150 440 868 921 577 930 401 251 328 × 2 = 0 + 0.756 300 881 737 843 155 860 802 502 656;
  • 21) 0.756 300 881 737 843 155 860 802 502 656 × 2 = 1 + 0.512 601 763 475 686 311 721 605 005 312;
  • 22) 0.512 601 763 475 686 311 721 605 005 312 × 2 = 1 + 0.025 203 526 951 372 623 443 210 010 624;
  • 23) 0.025 203 526 951 372 623 443 210 010 624 × 2 = 0 + 0.050 407 053 902 745 246 886 420 021 248;
  • 24) 0.050 407 053 902 745 246 886 420 021 248 × 2 = 0 + 0.100 814 107 805 490 493 772 840 042 496;
  • 25) 0.100 814 107 805 490 493 772 840 042 496 × 2 = 0 + 0.201 628 215 610 980 987 545 680 084 992;
  • 26) 0.201 628 215 610 980 987 545 680 084 992 × 2 = 0 + 0.403 256 431 221 961 975 091 360 169 984;
  • 27) 0.403 256 431 221 961 975 091 360 169 984 × 2 = 0 + 0.806 512 862 443 923 950 182 720 339 968;
  • 28) 0.806 512 862 443 923 950 182 720 339 968 × 2 = 1 + 0.613 025 724 887 847 900 365 440 679 936;
  • 29) 0.613 025 724 887 847 900 365 440 679 936 × 2 = 1 + 0.226 051 449 775 695 800 730 881 359 872;
  • 30) 0.226 051 449 775 695 800 730 881 359 872 × 2 = 0 + 0.452 102 899 551 391 601 461 762 719 744;
  • 31) 0.452 102 899 551 391 601 461 762 719 744 × 2 = 0 + 0.904 205 799 102 783 202 923 525 439 488;
  • 32) 0.904 205 799 102 783 202 923 525 439 488 × 2 = 1 + 0.808 411 598 205 566 405 847 050 878 976;
  • 33) 0.808 411 598 205 566 405 847 050 878 976 × 2 = 1 + 0.616 823 196 411 132 811 694 101 757 952;
  • 34) 0.616 823 196 411 132 811 694 101 757 952 × 2 = 1 + 0.233 646 392 822 265 623 388 203 515 904;
  • 35) 0.233 646 392 822 265 623 388 203 515 904 × 2 = 0 + 0.467 292 785 644 531 246 776 407 031 808;
  • 36) 0.467 292 785 644 531 246 776 407 031 808 × 2 = 0 + 0.934 585 571 289 062 493 552 814 063 616;
  • 37) 0.934 585 571 289 062 493 552 814 063 616 × 2 = 1 + 0.869 171 142 578 124 987 105 628 127 232;
  • 38) 0.869 171 142 578 124 987 105 628 127 232 × 2 = 1 + 0.738 342 285 156 249 974 211 256 254 464;
  • 39) 0.738 342 285 156 249 974 211 256 254 464 × 2 = 1 + 0.476 684 570 312 499 948 422 512 508 928;
  • 40) 0.476 684 570 312 499 948 422 512 508 928 × 2 = 0 + 0.953 369 140 624 999 896 845 025 017 856;
  • 41) 0.953 369 140 624 999 896 845 025 017 856 × 2 = 1 + 0.906 738 281 249 999 793 690 050 035 712;
  • 42) 0.906 738 281 249 999 793 690 050 035 712 × 2 = 1 + 0.813 476 562 499 999 587 380 100 071 424;
  • 43) 0.813 476 562 499 999 587 380 100 071 424 × 2 = 1 + 0.626 953 124 999 999 174 760 200 142 848;
  • 44) 0.626 953 124 999 999 174 760 200 142 848 × 2 = 1 + 0.253 906 249 999 998 349 520 400 285 696;
  • 45) 0.253 906 249 999 998 349 520 400 285 696 × 2 = 0 + 0.507 812 499 999 996 699 040 800 571 392;
  • 46) 0.507 812 499 999 996 699 040 800 571 392 × 2 = 1 + 0.015 624 999 999 993 398 081 601 142 784;
  • 47) 0.015 624 999 999 993 398 081 601 142 784 × 2 = 0 + 0.031 249 999 999 986 796 163 202 285 568;
  • 48) 0.031 249 999 999 986 796 163 202 285 568 × 2 = 0 + 0.062 499 999 999 973 592 326 404 571 136;
  • 49) 0.062 499 999 999 973 592 326 404 571 136 × 2 = 0 + 0.124 999 999 999 947 184 652 809 142 272;
  • 50) 0.124 999 999 999 947 184 652 809 142 272 × 2 = 0 + 0.249 999 999 999 894 369 305 618 284 544;
  • 51) 0.249 999 999 999 894 369 305 618 284 544 × 2 = 0 + 0.499 999 999 999 788 738 611 236 569 088;
  • 52) 0.499 999 999 999 788 738 611 236 569 088 × 2 = 0 + 0.999 999 999 999 577 477 222 473 138 176;
  • 53) 0.999 999 999 999 577 477 222 473 138 176 × 2 = 1 + 0.999 999 999 999 154 954 444 946 276 352;

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:


0.745 459 324 169 999 826 281 696 186 831(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 831(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:


1.745 459 324 169 999 826 281 696 186 831(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 1.745 459 324 169 999 826 281 696 186 831 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