1.745 459 324 169 999 826 281 698 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 698 37(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
1.745 459 324 169 999 826 281 698 37(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: 0.745 459 324 169 999 826 281 698 37.

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) 0.745 459 324 169 999 826 281 698 37 × 2 = 1 + 0.490 918 648 339 999 652 563 396 74;
  • 2) 0.490 918 648 339 999 652 563 396 74 × 2 = 0 + 0.981 837 296 679 999 305 126 793 48;
  • 3) 0.981 837 296 679 999 305 126 793 48 × 2 = 1 + 0.963 674 593 359 998 610 253 586 96;
  • 4) 0.963 674 593 359 998 610 253 586 96 × 2 = 1 + 0.927 349 186 719 997 220 507 173 92;
  • 5) 0.927 349 186 719 997 220 507 173 92 × 2 = 1 + 0.854 698 373 439 994 441 014 347 84;
  • 6) 0.854 698 373 439 994 441 014 347 84 × 2 = 1 + 0.709 396 746 879 988 882 028 695 68;
  • 7) 0.709 396 746 879 988 882 028 695 68 × 2 = 1 + 0.418 793 493 759 977 764 057 391 36;
  • 8) 0.418 793 493 759 977 764 057 391 36 × 2 = 0 + 0.837 586 987 519 955 528 114 782 72;
  • 9) 0.837 586 987 519 955 528 114 782 72 × 2 = 1 + 0.675 173 975 039 911 056 229 565 44;
  • 10) 0.675 173 975 039 911 056 229 565 44 × 2 = 1 + 0.350 347 950 079 822 112 459 130 88;
  • 11) 0.350 347 950 079 822 112 459 130 88 × 2 = 0 + 0.700 695 900 159 644 224 918 261 76;
  • 12) 0.700 695 900 159 644 224 918 261 76 × 2 = 1 + 0.401 391 800 319 288 449 836 523 52;
  • 13) 0.401 391 800 319 288 449 836 523 52 × 2 = 0 + 0.802 783 600 638 576 899 673 047 04;
  • 14) 0.802 783 600 638 576 899 673 047 04 × 2 = 1 + 0.605 567 201 277 153 799 346 094 08;
  • 15) 0.605 567 201 277 153 799 346 094 08 × 2 = 1 + 0.211 134 402 554 307 598 692 188 16;
  • 16) 0.211 134 402 554 307 598 692 188 16 × 2 = 0 + 0.422 268 805 108 615 197 384 376 32;
  • 17) 0.422 268 805 108 615 197 384 376 32 × 2 = 0 + 0.844 537 610 217 230 394 768 752 64;
  • 18) 0.844 537 610 217 230 394 768 752 64 × 2 = 1 + 0.689 075 220 434 460 789 537 505 28;
  • 19) 0.689 075 220 434 460 789 537 505 28 × 2 = 1 + 0.378 150 440 868 921 579 075 010 56;
  • 20) 0.378 150 440 868 921 579 075 010 56 × 2 = 0 + 0.756 300 881 737 843 158 150 021 12;
  • 21) 0.756 300 881 737 843 158 150 021 12 × 2 = 1 + 0.512 601 763 475 686 316 300 042 24;
  • 22) 0.512 601 763 475 686 316 300 042 24 × 2 = 1 + 0.025 203 526 951 372 632 600 084 48;
  • 23) 0.025 203 526 951 372 632 600 084 48 × 2 = 0 + 0.050 407 053 902 745 265 200 168 96;
  • 24) 0.050 407 053 902 745 265 200 168 96 × 2 = 0 + 0.100 814 107 805 490 530 400 337 92;
  • 25) 0.100 814 107 805 490 530 400 337 92 × 2 = 0 + 0.201 628 215 610 981 060 800 675 84;
  • 26) 0.201 628 215 610 981 060 800 675 84 × 2 = 0 + 0.403 256 431 221 962 121 601 351 68;
  • 27) 0.403 256 431 221 962 121 601 351 68 × 2 = 0 + 0.806 512 862 443 924 243 202 703 36;
  • 28) 0.806 512 862 443 924 243 202 703 36 × 2 = 1 + 0.613 025 724 887 848 486 405 406 72;
  • 29) 0.613 025 724 887 848 486 405 406 72 × 2 = 1 + 0.226 051 449 775 696 972 810 813 44;
  • 30) 0.226 051 449 775 696 972 810 813 44 × 2 = 0 + 0.452 102 899 551 393 945 621 626 88;
  • 31) 0.452 102 899 551 393 945 621 626 88 × 2 = 0 + 0.904 205 799 102 787 891 243 253 76;
  • 32) 0.904 205 799 102 787 891 243 253 76 × 2 = 1 + 0.808 411 598 205 575 782 486 507 52;
  • 33) 0.808 411 598 205 575 782 486 507 52 × 2 = 1 + 0.616 823 196 411 151 564 973 015 04;
  • 34) 0.616 823 196 411 151 564 973 015 04 × 2 = 1 + 0.233 646 392 822 303 129 946 030 08;
  • 35) 0.233 646 392 822 303 129 946 030 08 × 2 = 0 + 0.467 292 785 644 606 259 892 060 16;
  • 36) 0.467 292 785 644 606 259 892 060 16 × 2 = 0 + 0.934 585 571 289 212 519 784 120 32;
  • 37) 0.934 585 571 289 212 519 784 120 32 × 2 = 1 + 0.869 171 142 578 425 039 568 240 64;
  • 38) 0.869 171 142 578 425 039 568 240 64 × 2 = 1 + 0.738 342 285 156 850 079 136 481 28;
  • 39) 0.738 342 285 156 850 079 136 481 28 × 2 = 1 + 0.476 684 570 313 700 158 272 962 56;
  • 40) 0.476 684 570 313 700 158 272 962 56 × 2 = 0 + 0.953 369 140 627 400 316 545 925 12;
  • 41) 0.953 369 140 627 400 316 545 925 12 × 2 = 1 + 0.906 738 281 254 800 633 091 850 24;
  • 42) 0.906 738 281 254 800 633 091 850 24 × 2 = 1 + 0.813 476 562 509 601 266 183 700 48;
  • 43) 0.813 476 562 509 601 266 183 700 48 × 2 = 1 + 0.626 953 125 019 202 532 367 400 96;
  • 44) 0.626 953 125 019 202 532 367 400 96 × 2 = 1 + 0.253 906 250 038 405 064 734 801 92;
  • 45) 0.253 906 250 038 405 064 734 801 92 × 2 = 0 + 0.507 812 500 076 810 129 469 603 84;
  • 46) 0.507 812 500 076 810 129 469 603 84 × 2 = 1 + 0.015 625 000 153 620 258 939 207 68;
  • 47) 0.015 625 000 153 620 258 939 207 68 × 2 = 0 + 0.031 250 000 307 240 517 878 415 36;
  • 48) 0.031 250 000 307 240 517 878 415 36 × 2 = 0 + 0.062 500 000 614 481 035 756 830 72;
  • 49) 0.062 500 000 614 481 035 756 830 72 × 2 = 0 + 0.125 000 001 228 962 071 513 661 44;
  • 50) 0.125 000 001 228 962 071 513 661 44 × 2 = 0 + 0.250 000 002 457 924 143 027 322 88;
  • 51) 0.250 000 002 457 924 143 027 322 88 × 2 = 0 + 0.500 000 004 915 848 286 054 645 76;
  • 52) 0.500 000 004 915 848 286 054 645 76 × 2 = 1 + 0.000 000 009 831 696 572 109 291 52;
  • 53) 0.000 000 009 831 696 572 109 291 52 × 2 = 0 + 0.000 000 019 663 393 144 218 583 04;

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:


0.745 459 324 169 999 826 281 698 37(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 698 37(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:


1.745 459 324 169 999 826 281 698 37(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 1.745 459 324 169 999 826 281 698 37 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