1.745 459 324 169 999 826 281 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 57(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 57(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 57.

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 57 × 2 = 1 + 0.490 918 648 339 999 652 563 14;
  • 2) 0.490 918 648 339 999 652 563 14 × 2 = 0 + 0.981 837 296 679 999 305 126 28;
  • 3) 0.981 837 296 679 999 305 126 28 × 2 = 1 + 0.963 674 593 359 998 610 252 56;
  • 4) 0.963 674 593 359 998 610 252 56 × 2 = 1 + 0.927 349 186 719 997 220 505 12;
  • 5) 0.927 349 186 719 997 220 505 12 × 2 = 1 + 0.854 698 373 439 994 441 010 24;
  • 6) 0.854 698 373 439 994 441 010 24 × 2 = 1 + 0.709 396 746 879 988 882 020 48;
  • 7) 0.709 396 746 879 988 882 020 48 × 2 = 1 + 0.418 793 493 759 977 764 040 96;
  • 8) 0.418 793 493 759 977 764 040 96 × 2 = 0 + 0.837 586 987 519 955 528 081 92;
  • 9) 0.837 586 987 519 955 528 081 92 × 2 = 1 + 0.675 173 975 039 911 056 163 84;
  • 10) 0.675 173 975 039 911 056 163 84 × 2 = 1 + 0.350 347 950 079 822 112 327 68;
  • 11) 0.350 347 950 079 822 112 327 68 × 2 = 0 + 0.700 695 900 159 644 224 655 36;
  • 12) 0.700 695 900 159 644 224 655 36 × 2 = 1 + 0.401 391 800 319 288 449 310 72;
  • 13) 0.401 391 800 319 288 449 310 72 × 2 = 0 + 0.802 783 600 638 576 898 621 44;
  • 14) 0.802 783 600 638 576 898 621 44 × 2 = 1 + 0.605 567 201 277 153 797 242 88;
  • 15) 0.605 567 201 277 153 797 242 88 × 2 = 1 + 0.211 134 402 554 307 594 485 76;
  • 16) 0.211 134 402 554 307 594 485 76 × 2 = 0 + 0.422 268 805 108 615 188 971 52;
  • 17) 0.422 268 805 108 615 188 971 52 × 2 = 0 + 0.844 537 610 217 230 377 943 04;
  • 18) 0.844 537 610 217 230 377 943 04 × 2 = 1 + 0.689 075 220 434 460 755 886 08;
  • 19) 0.689 075 220 434 460 755 886 08 × 2 = 1 + 0.378 150 440 868 921 511 772 16;
  • 20) 0.378 150 440 868 921 511 772 16 × 2 = 0 + 0.756 300 881 737 843 023 544 32;
  • 21) 0.756 300 881 737 843 023 544 32 × 2 = 1 + 0.512 601 763 475 686 047 088 64;
  • 22) 0.512 601 763 475 686 047 088 64 × 2 = 1 + 0.025 203 526 951 372 094 177 28;
  • 23) 0.025 203 526 951 372 094 177 28 × 2 = 0 + 0.050 407 053 902 744 188 354 56;
  • 24) 0.050 407 053 902 744 188 354 56 × 2 = 0 + 0.100 814 107 805 488 376 709 12;
  • 25) 0.100 814 107 805 488 376 709 12 × 2 = 0 + 0.201 628 215 610 976 753 418 24;
  • 26) 0.201 628 215 610 976 753 418 24 × 2 = 0 + 0.403 256 431 221 953 506 836 48;
  • 27) 0.403 256 431 221 953 506 836 48 × 2 = 0 + 0.806 512 862 443 907 013 672 96;
  • 28) 0.806 512 862 443 907 013 672 96 × 2 = 1 + 0.613 025 724 887 814 027 345 92;
  • 29) 0.613 025 724 887 814 027 345 92 × 2 = 1 + 0.226 051 449 775 628 054 691 84;
  • 30) 0.226 051 449 775 628 054 691 84 × 2 = 0 + 0.452 102 899 551 256 109 383 68;
  • 31) 0.452 102 899 551 256 109 383 68 × 2 = 0 + 0.904 205 799 102 512 218 767 36;
  • 32) 0.904 205 799 102 512 218 767 36 × 2 = 1 + 0.808 411 598 205 024 437 534 72;
  • 33) 0.808 411 598 205 024 437 534 72 × 2 = 1 + 0.616 823 196 410 048 875 069 44;
  • 34) 0.616 823 196 410 048 875 069 44 × 2 = 1 + 0.233 646 392 820 097 750 138 88;
  • 35) 0.233 646 392 820 097 750 138 88 × 2 = 0 + 0.467 292 785 640 195 500 277 76;
  • 36) 0.467 292 785 640 195 500 277 76 × 2 = 0 + 0.934 585 571 280 391 000 555 52;
  • 37) 0.934 585 571 280 391 000 555 52 × 2 = 1 + 0.869 171 142 560 782 001 111 04;
  • 38) 0.869 171 142 560 782 001 111 04 × 2 = 1 + 0.738 342 285 121 564 002 222 08;
  • 39) 0.738 342 285 121 564 002 222 08 × 2 = 1 + 0.476 684 570 243 128 004 444 16;
  • 40) 0.476 684 570 243 128 004 444 16 × 2 = 0 + 0.953 369 140 486 256 008 888 32;
  • 41) 0.953 369 140 486 256 008 888 32 × 2 = 1 + 0.906 738 280 972 512 017 776 64;
  • 42) 0.906 738 280 972 512 017 776 64 × 2 = 1 + 0.813 476 561 945 024 035 553 28;
  • 43) 0.813 476 561 945 024 035 553 28 × 2 = 1 + 0.626 953 123 890 048 071 106 56;
  • 44) 0.626 953 123 890 048 071 106 56 × 2 = 1 + 0.253 906 247 780 096 142 213 12;
  • 45) 0.253 906 247 780 096 142 213 12 × 2 = 0 + 0.507 812 495 560 192 284 426 24;
  • 46) 0.507 812 495 560 192 284 426 24 × 2 = 1 + 0.015 624 991 120 384 568 852 48;
  • 47) 0.015 624 991 120 384 568 852 48 × 2 = 0 + 0.031 249 982 240 769 137 704 96;
  • 48) 0.031 249 982 240 769 137 704 96 × 2 = 0 + 0.062 499 964 481 538 275 409 92;
  • 49) 0.062 499 964 481 538 275 409 92 × 2 = 0 + 0.124 999 928 963 076 550 819 84;
  • 50) 0.124 999 928 963 076 550 819 84 × 2 = 0 + 0.249 999 857 926 153 101 639 68;
  • 51) 0.249 999 857 926 153 101 639 68 × 2 = 0 + 0.499 999 715 852 306 203 279 36;
  • 52) 0.499 999 715 852 306 203 279 36 × 2 = 0 + 0.999 999 431 704 612 406 558 72;
  • 53) 0.999 999 431 704 612 406 558 72 × 2 = 1 + 0.999 998 863 409 224 813 117 44;

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 57(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 57(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 57(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 57 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