1.745 459 324 169 999 826 281 679 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 679 3(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
1.745 459 324 169 999 826 281 679 3(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: 0.745 459 324 169 999 826 281 679 3.

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) 0.745 459 324 169 999 826 281 679 3 × 2 = 1 + 0.490 918 648 339 999 652 563 358 6;
  • 2) 0.490 918 648 339 999 652 563 358 6 × 2 = 0 + 0.981 837 296 679 999 305 126 717 2;
  • 3) 0.981 837 296 679 999 305 126 717 2 × 2 = 1 + 0.963 674 593 359 998 610 253 434 4;
  • 4) 0.963 674 593 359 998 610 253 434 4 × 2 = 1 + 0.927 349 186 719 997 220 506 868 8;
  • 5) 0.927 349 186 719 997 220 506 868 8 × 2 = 1 + 0.854 698 373 439 994 441 013 737 6;
  • 6) 0.854 698 373 439 994 441 013 737 6 × 2 = 1 + 0.709 396 746 879 988 882 027 475 2;
  • 7) 0.709 396 746 879 988 882 027 475 2 × 2 = 1 + 0.418 793 493 759 977 764 054 950 4;
  • 8) 0.418 793 493 759 977 764 054 950 4 × 2 = 0 + 0.837 586 987 519 955 528 109 900 8;
  • 9) 0.837 586 987 519 955 528 109 900 8 × 2 = 1 + 0.675 173 975 039 911 056 219 801 6;
  • 10) 0.675 173 975 039 911 056 219 801 6 × 2 = 1 + 0.350 347 950 079 822 112 439 603 2;
  • 11) 0.350 347 950 079 822 112 439 603 2 × 2 = 0 + 0.700 695 900 159 644 224 879 206 4;
  • 12) 0.700 695 900 159 644 224 879 206 4 × 2 = 1 + 0.401 391 800 319 288 449 758 412 8;
  • 13) 0.401 391 800 319 288 449 758 412 8 × 2 = 0 + 0.802 783 600 638 576 899 516 825 6;
  • 14) 0.802 783 600 638 576 899 516 825 6 × 2 = 1 + 0.605 567 201 277 153 799 033 651 2;
  • 15) 0.605 567 201 277 153 799 033 651 2 × 2 = 1 + 0.211 134 402 554 307 598 067 302 4;
  • 16) 0.211 134 402 554 307 598 067 302 4 × 2 = 0 + 0.422 268 805 108 615 196 134 604 8;
  • 17) 0.422 268 805 108 615 196 134 604 8 × 2 = 0 + 0.844 537 610 217 230 392 269 209 6;
  • 18) 0.844 537 610 217 230 392 269 209 6 × 2 = 1 + 0.689 075 220 434 460 784 538 419 2;
  • 19) 0.689 075 220 434 460 784 538 419 2 × 2 = 1 + 0.378 150 440 868 921 569 076 838 4;
  • 20) 0.378 150 440 868 921 569 076 838 4 × 2 = 0 + 0.756 300 881 737 843 138 153 676 8;
  • 21) 0.756 300 881 737 843 138 153 676 8 × 2 = 1 + 0.512 601 763 475 686 276 307 353 6;
  • 22) 0.512 601 763 475 686 276 307 353 6 × 2 = 1 + 0.025 203 526 951 372 552 614 707 2;
  • 23) 0.025 203 526 951 372 552 614 707 2 × 2 = 0 + 0.050 407 053 902 745 105 229 414 4;
  • 24) 0.050 407 053 902 745 105 229 414 4 × 2 = 0 + 0.100 814 107 805 490 210 458 828 8;
  • 25) 0.100 814 107 805 490 210 458 828 8 × 2 = 0 + 0.201 628 215 610 980 420 917 657 6;
  • 26) 0.201 628 215 610 980 420 917 657 6 × 2 = 0 + 0.403 256 431 221 960 841 835 315 2;
  • 27) 0.403 256 431 221 960 841 835 315 2 × 2 = 0 + 0.806 512 862 443 921 683 670 630 4;
  • 28) 0.806 512 862 443 921 683 670 630 4 × 2 = 1 + 0.613 025 724 887 843 367 341 260 8;
  • 29) 0.613 025 724 887 843 367 341 260 8 × 2 = 1 + 0.226 051 449 775 686 734 682 521 6;
  • 30) 0.226 051 449 775 686 734 682 521 6 × 2 = 0 + 0.452 102 899 551 373 469 365 043 2;
  • 31) 0.452 102 899 551 373 469 365 043 2 × 2 = 0 + 0.904 205 799 102 746 938 730 086 4;
  • 32) 0.904 205 799 102 746 938 730 086 4 × 2 = 1 + 0.808 411 598 205 493 877 460 172 8;
  • 33) 0.808 411 598 205 493 877 460 172 8 × 2 = 1 + 0.616 823 196 410 987 754 920 345 6;
  • 34) 0.616 823 196 410 987 754 920 345 6 × 2 = 1 + 0.233 646 392 821 975 509 840 691 2;
  • 35) 0.233 646 392 821 975 509 840 691 2 × 2 = 0 + 0.467 292 785 643 951 019 681 382 4;
  • 36) 0.467 292 785 643 951 019 681 382 4 × 2 = 0 + 0.934 585 571 287 902 039 362 764 8;
  • 37) 0.934 585 571 287 902 039 362 764 8 × 2 = 1 + 0.869 171 142 575 804 078 725 529 6;
  • 38) 0.869 171 142 575 804 078 725 529 6 × 2 = 1 + 0.738 342 285 151 608 157 451 059 2;
  • 39) 0.738 342 285 151 608 157 451 059 2 × 2 = 1 + 0.476 684 570 303 216 314 902 118 4;
  • 40) 0.476 684 570 303 216 314 902 118 4 × 2 = 0 + 0.953 369 140 606 432 629 804 236 8;
  • 41) 0.953 369 140 606 432 629 804 236 8 × 2 = 1 + 0.906 738 281 212 865 259 608 473 6;
  • 42) 0.906 738 281 212 865 259 608 473 6 × 2 = 1 + 0.813 476 562 425 730 519 216 947 2;
  • 43) 0.813 476 562 425 730 519 216 947 2 × 2 = 1 + 0.626 953 124 851 461 038 433 894 4;
  • 44) 0.626 953 124 851 461 038 433 894 4 × 2 = 1 + 0.253 906 249 702 922 076 867 788 8;
  • 45) 0.253 906 249 702 922 076 867 788 8 × 2 = 0 + 0.507 812 499 405 844 153 735 577 6;
  • 46) 0.507 812 499 405 844 153 735 577 6 × 2 = 1 + 0.015 624 998 811 688 307 471 155 2;
  • 47) 0.015 624 998 811 688 307 471 155 2 × 2 = 0 + 0.031 249 997 623 376 614 942 310 4;
  • 48) 0.031 249 997 623 376 614 942 310 4 × 2 = 0 + 0.062 499 995 246 753 229 884 620 8;
  • 49) 0.062 499 995 246 753 229 884 620 8 × 2 = 0 + 0.124 999 990 493 506 459 769 241 6;
  • 50) 0.124 999 990 493 506 459 769 241 6 × 2 = 0 + 0.249 999 980 987 012 919 538 483 2;
  • 51) 0.249 999 980 987 012 919 538 483 2 × 2 = 0 + 0.499 999 961 974 025 839 076 966 4;
  • 52) 0.499 999 961 974 025 839 076 966 4 × 2 = 0 + 0.999 999 923 948 051 678 153 932 8;
  • 53) 0.999 999 923 948 051 678 153 932 8 × 2 = 1 + 0.999 999 847 896 103 356 307 865 6;

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:


0.745 459 324 169 999 826 281 679 3(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 679 3(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:


1.745 459 324 169 999 826 281 679 3(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 1.745 459 324 169 999 826 281 679 3 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