1.745 459 324 169 999 826 281 694 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 694 99(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
1.745 459 324 169 999 826 281 694 99(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: 0.745 459 324 169 999 826 281 694 99.

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) 0.745 459 324 169 999 826 281 694 99 × 2 = 1 + 0.490 918 648 339 999 652 563 389 98;
  • 2) 0.490 918 648 339 999 652 563 389 98 × 2 = 0 + 0.981 837 296 679 999 305 126 779 96;
  • 3) 0.981 837 296 679 999 305 126 779 96 × 2 = 1 + 0.963 674 593 359 998 610 253 559 92;
  • 4) 0.963 674 593 359 998 610 253 559 92 × 2 = 1 + 0.927 349 186 719 997 220 507 119 84;
  • 5) 0.927 349 186 719 997 220 507 119 84 × 2 = 1 + 0.854 698 373 439 994 441 014 239 68;
  • 6) 0.854 698 373 439 994 441 014 239 68 × 2 = 1 + 0.709 396 746 879 988 882 028 479 36;
  • 7) 0.709 396 746 879 988 882 028 479 36 × 2 = 1 + 0.418 793 493 759 977 764 056 958 72;
  • 8) 0.418 793 493 759 977 764 056 958 72 × 2 = 0 + 0.837 586 987 519 955 528 113 917 44;
  • 9) 0.837 586 987 519 955 528 113 917 44 × 2 = 1 + 0.675 173 975 039 911 056 227 834 88;
  • 10) 0.675 173 975 039 911 056 227 834 88 × 2 = 1 + 0.350 347 950 079 822 112 455 669 76;
  • 11) 0.350 347 950 079 822 112 455 669 76 × 2 = 0 + 0.700 695 900 159 644 224 911 339 52;
  • 12) 0.700 695 900 159 644 224 911 339 52 × 2 = 1 + 0.401 391 800 319 288 449 822 679 04;
  • 13) 0.401 391 800 319 288 449 822 679 04 × 2 = 0 + 0.802 783 600 638 576 899 645 358 08;
  • 14) 0.802 783 600 638 576 899 645 358 08 × 2 = 1 + 0.605 567 201 277 153 799 290 716 16;
  • 15) 0.605 567 201 277 153 799 290 716 16 × 2 = 1 + 0.211 134 402 554 307 598 581 432 32;
  • 16) 0.211 134 402 554 307 598 581 432 32 × 2 = 0 + 0.422 268 805 108 615 197 162 864 64;
  • 17) 0.422 268 805 108 615 197 162 864 64 × 2 = 0 + 0.844 537 610 217 230 394 325 729 28;
  • 18) 0.844 537 610 217 230 394 325 729 28 × 2 = 1 + 0.689 075 220 434 460 788 651 458 56;
  • 19) 0.689 075 220 434 460 788 651 458 56 × 2 = 1 + 0.378 150 440 868 921 577 302 917 12;
  • 20) 0.378 150 440 868 921 577 302 917 12 × 2 = 0 + 0.756 300 881 737 843 154 605 834 24;
  • 21) 0.756 300 881 737 843 154 605 834 24 × 2 = 1 + 0.512 601 763 475 686 309 211 668 48;
  • 22) 0.512 601 763 475 686 309 211 668 48 × 2 = 1 + 0.025 203 526 951 372 618 423 336 96;
  • 23) 0.025 203 526 951 372 618 423 336 96 × 2 = 0 + 0.050 407 053 902 745 236 846 673 92;
  • 24) 0.050 407 053 902 745 236 846 673 92 × 2 = 0 + 0.100 814 107 805 490 473 693 347 84;
  • 25) 0.100 814 107 805 490 473 693 347 84 × 2 = 0 + 0.201 628 215 610 980 947 386 695 68;
  • 26) 0.201 628 215 610 980 947 386 695 68 × 2 = 0 + 0.403 256 431 221 961 894 773 391 36;
  • 27) 0.403 256 431 221 961 894 773 391 36 × 2 = 0 + 0.806 512 862 443 923 789 546 782 72;
  • 28) 0.806 512 862 443 923 789 546 782 72 × 2 = 1 + 0.613 025 724 887 847 579 093 565 44;
  • 29) 0.613 025 724 887 847 579 093 565 44 × 2 = 1 + 0.226 051 449 775 695 158 187 130 88;
  • 30) 0.226 051 449 775 695 158 187 130 88 × 2 = 0 + 0.452 102 899 551 390 316 374 261 76;
  • 31) 0.452 102 899 551 390 316 374 261 76 × 2 = 0 + 0.904 205 799 102 780 632 748 523 52;
  • 32) 0.904 205 799 102 780 632 748 523 52 × 2 = 1 + 0.808 411 598 205 561 265 497 047 04;
  • 33) 0.808 411 598 205 561 265 497 047 04 × 2 = 1 + 0.616 823 196 411 122 530 994 094 08;
  • 34) 0.616 823 196 411 122 530 994 094 08 × 2 = 1 + 0.233 646 392 822 245 061 988 188 16;
  • 35) 0.233 646 392 822 245 061 988 188 16 × 2 = 0 + 0.467 292 785 644 490 123 976 376 32;
  • 36) 0.467 292 785 644 490 123 976 376 32 × 2 = 0 + 0.934 585 571 288 980 247 952 752 64;
  • 37) 0.934 585 571 288 980 247 952 752 64 × 2 = 1 + 0.869 171 142 577 960 495 905 505 28;
  • 38) 0.869 171 142 577 960 495 905 505 28 × 2 = 1 + 0.738 342 285 155 920 991 811 010 56;
  • 39) 0.738 342 285 155 920 991 811 010 56 × 2 = 1 + 0.476 684 570 311 841 983 622 021 12;
  • 40) 0.476 684 570 311 841 983 622 021 12 × 2 = 0 + 0.953 369 140 623 683 967 244 042 24;
  • 41) 0.953 369 140 623 683 967 244 042 24 × 2 = 1 + 0.906 738 281 247 367 934 488 084 48;
  • 42) 0.906 738 281 247 367 934 488 084 48 × 2 = 1 + 0.813 476 562 494 735 868 976 168 96;
  • 43) 0.813 476 562 494 735 868 976 168 96 × 2 = 1 + 0.626 953 124 989 471 737 952 337 92;
  • 44) 0.626 953 124 989 471 737 952 337 92 × 2 = 1 + 0.253 906 249 978 943 475 904 675 84;
  • 45) 0.253 906 249 978 943 475 904 675 84 × 2 = 0 + 0.507 812 499 957 886 951 809 351 68;
  • 46) 0.507 812 499 957 886 951 809 351 68 × 2 = 1 + 0.015 624 999 915 773 903 618 703 36;
  • 47) 0.015 624 999 915 773 903 618 703 36 × 2 = 0 + 0.031 249 999 831 547 807 237 406 72;
  • 48) 0.031 249 999 831 547 807 237 406 72 × 2 = 0 + 0.062 499 999 663 095 614 474 813 44;
  • 49) 0.062 499 999 663 095 614 474 813 44 × 2 = 0 + 0.124 999 999 326 191 228 949 626 88;
  • 50) 0.124 999 999 326 191 228 949 626 88 × 2 = 0 + 0.249 999 998 652 382 457 899 253 76;
  • 51) 0.249 999 998 652 382 457 899 253 76 × 2 = 0 + 0.499 999 997 304 764 915 798 507 52;
  • 52) 0.499 999 997 304 764 915 798 507 52 × 2 = 0 + 0.999 999 994 609 529 831 597 015 04;
  • 53) 0.999 999 994 609 529 831 597 015 04 × 2 = 1 + 0.999 999 989 219 059 663 194 030 08;

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:


0.745 459 324 169 999 826 281 694 99(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 694 99(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:


1.745 459 324 169 999 826 281 694 99(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 1.745 459 324 169 999 826 281 694 99 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