1.745 459 324 169 999 826 281 817 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 817(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
1.745 459 324 169 999 826 281 817(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: 0.745 459 324 169 999 826 281 817.

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) 0.745 459 324 169 999 826 281 817 × 2 = 1 + 0.490 918 648 339 999 652 563 634;
  • 2) 0.490 918 648 339 999 652 563 634 × 2 = 0 + 0.981 837 296 679 999 305 127 268;
  • 3) 0.981 837 296 679 999 305 127 268 × 2 = 1 + 0.963 674 593 359 998 610 254 536;
  • 4) 0.963 674 593 359 998 610 254 536 × 2 = 1 + 0.927 349 186 719 997 220 509 072;
  • 5) 0.927 349 186 719 997 220 509 072 × 2 = 1 + 0.854 698 373 439 994 441 018 144;
  • 6) 0.854 698 373 439 994 441 018 144 × 2 = 1 + 0.709 396 746 879 988 882 036 288;
  • 7) 0.709 396 746 879 988 882 036 288 × 2 = 1 + 0.418 793 493 759 977 764 072 576;
  • 8) 0.418 793 493 759 977 764 072 576 × 2 = 0 + 0.837 586 987 519 955 528 145 152;
  • 9) 0.837 586 987 519 955 528 145 152 × 2 = 1 + 0.675 173 975 039 911 056 290 304;
  • 10) 0.675 173 975 039 911 056 290 304 × 2 = 1 + 0.350 347 950 079 822 112 580 608;
  • 11) 0.350 347 950 079 822 112 580 608 × 2 = 0 + 0.700 695 900 159 644 225 161 216;
  • 12) 0.700 695 900 159 644 225 161 216 × 2 = 1 + 0.401 391 800 319 288 450 322 432;
  • 13) 0.401 391 800 319 288 450 322 432 × 2 = 0 + 0.802 783 600 638 576 900 644 864;
  • 14) 0.802 783 600 638 576 900 644 864 × 2 = 1 + 0.605 567 201 277 153 801 289 728;
  • 15) 0.605 567 201 277 153 801 289 728 × 2 = 1 + 0.211 134 402 554 307 602 579 456;
  • 16) 0.211 134 402 554 307 602 579 456 × 2 = 0 + 0.422 268 805 108 615 205 158 912;
  • 17) 0.422 268 805 108 615 205 158 912 × 2 = 0 + 0.844 537 610 217 230 410 317 824;
  • 18) 0.844 537 610 217 230 410 317 824 × 2 = 1 + 0.689 075 220 434 460 820 635 648;
  • 19) 0.689 075 220 434 460 820 635 648 × 2 = 1 + 0.378 150 440 868 921 641 271 296;
  • 20) 0.378 150 440 868 921 641 271 296 × 2 = 0 + 0.756 300 881 737 843 282 542 592;
  • 21) 0.756 300 881 737 843 282 542 592 × 2 = 1 + 0.512 601 763 475 686 565 085 184;
  • 22) 0.512 601 763 475 686 565 085 184 × 2 = 1 + 0.025 203 526 951 373 130 170 368;
  • 23) 0.025 203 526 951 373 130 170 368 × 2 = 0 + 0.050 407 053 902 746 260 340 736;
  • 24) 0.050 407 053 902 746 260 340 736 × 2 = 0 + 0.100 814 107 805 492 520 681 472;
  • 25) 0.100 814 107 805 492 520 681 472 × 2 = 0 + 0.201 628 215 610 985 041 362 944;
  • 26) 0.201 628 215 610 985 041 362 944 × 2 = 0 + 0.403 256 431 221 970 082 725 888;
  • 27) 0.403 256 431 221 970 082 725 888 × 2 = 0 + 0.806 512 862 443 940 165 451 776;
  • 28) 0.806 512 862 443 940 165 451 776 × 2 = 1 + 0.613 025 724 887 880 330 903 552;
  • 29) 0.613 025 724 887 880 330 903 552 × 2 = 1 + 0.226 051 449 775 760 661 807 104;
  • 30) 0.226 051 449 775 760 661 807 104 × 2 = 0 + 0.452 102 899 551 521 323 614 208;
  • 31) 0.452 102 899 551 521 323 614 208 × 2 = 0 + 0.904 205 799 103 042 647 228 416;
  • 32) 0.904 205 799 103 042 647 228 416 × 2 = 1 + 0.808 411 598 206 085 294 456 832;
  • 33) 0.808 411 598 206 085 294 456 832 × 2 = 1 + 0.616 823 196 412 170 588 913 664;
  • 34) 0.616 823 196 412 170 588 913 664 × 2 = 1 + 0.233 646 392 824 341 177 827 328;
  • 35) 0.233 646 392 824 341 177 827 328 × 2 = 0 + 0.467 292 785 648 682 355 654 656;
  • 36) 0.467 292 785 648 682 355 654 656 × 2 = 0 + 0.934 585 571 297 364 711 309 312;
  • 37) 0.934 585 571 297 364 711 309 312 × 2 = 1 + 0.869 171 142 594 729 422 618 624;
  • 38) 0.869 171 142 594 729 422 618 624 × 2 = 1 + 0.738 342 285 189 458 845 237 248;
  • 39) 0.738 342 285 189 458 845 237 248 × 2 = 1 + 0.476 684 570 378 917 690 474 496;
  • 40) 0.476 684 570 378 917 690 474 496 × 2 = 0 + 0.953 369 140 757 835 380 948 992;
  • 41) 0.953 369 140 757 835 380 948 992 × 2 = 1 + 0.906 738 281 515 670 761 897 984;
  • 42) 0.906 738 281 515 670 761 897 984 × 2 = 1 + 0.813 476 563 031 341 523 795 968;
  • 43) 0.813 476 563 031 341 523 795 968 × 2 = 1 + 0.626 953 126 062 683 047 591 936;
  • 44) 0.626 953 126 062 683 047 591 936 × 2 = 1 + 0.253 906 252 125 366 095 183 872;
  • 45) 0.253 906 252 125 366 095 183 872 × 2 = 0 + 0.507 812 504 250 732 190 367 744;
  • 46) 0.507 812 504 250 732 190 367 744 × 2 = 1 + 0.015 625 008 501 464 380 735 488;
  • 47) 0.015 625 008 501 464 380 735 488 × 2 = 0 + 0.031 250 017 002 928 761 470 976;
  • 48) 0.031 250 017 002 928 761 470 976 × 2 = 0 + 0.062 500 034 005 857 522 941 952;
  • 49) 0.062 500 034 005 857 522 941 952 × 2 = 0 + 0.125 000 068 011 715 045 883 904;
  • 50) 0.125 000 068 011 715 045 883 904 × 2 = 0 + 0.250 000 136 023 430 091 767 808;
  • 51) 0.250 000 136 023 430 091 767 808 × 2 = 0 + 0.500 000 272 046 860 183 535 616;
  • 52) 0.500 000 272 046 860 183 535 616 × 2 = 1 + 0.000 000 544 093 720 367 071 232;
  • 53) 0.000 000 544 093 720 367 071 232 × 2 = 0 + 0.000 001 088 187 440 734 142 464;

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:


0.745 459 324 169 999 826 281 817(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 817(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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:


1.745 459 324 169 999 826 281 817(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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 0001 0


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 0001 0 =

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


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 0001


Decimal number 1.745 459 324 169 999 826 281 817 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 0001

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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