1.745 459 324 169 999 826 283 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 283 19(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
1.745 459 324 169 999 826 283 19(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: 0.745 459 324 169 999 826 283 19.

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) 0.745 459 324 169 999 826 283 19 × 2 = 1 + 0.490 918 648 339 999 652 566 38;
  • 2) 0.490 918 648 339 999 652 566 38 × 2 = 0 + 0.981 837 296 679 999 305 132 76;
  • 3) 0.981 837 296 679 999 305 132 76 × 2 = 1 + 0.963 674 593 359 998 610 265 52;
  • 4) 0.963 674 593 359 998 610 265 52 × 2 = 1 + 0.927 349 186 719 997 220 531 04;
  • 5) 0.927 349 186 719 997 220 531 04 × 2 = 1 + 0.854 698 373 439 994 441 062 08;
  • 6) 0.854 698 373 439 994 441 062 08 × 2 = 1 + 0.709 396 746 879 988 882 124 16;
  • 7) 0.709 396 746 879 988 882 124 16 × 2 = 1 + 0.418 793 493 759 977 764 248 32;
  • 8) 0.418 793 493 759 977 764 248 32 × 2 = 0 + 0.837 586 987 519 955 528 496 64;
  • 9) 0.837 586 987 519 955 528 496 64 × 2 = 1 + 0.675 173 975 039 911 056 993 28;
  • 10) 0.675 173 975 039 911 056 993 28 × 2 = 1 + 0.350 347 950 079 822 113 986 56;
  • 11) 0.350 347 950 079 822 113 986 56 × 2 = 0 + 0.700 695 900 159 644 227 973 12;
  • 12) 0.700 695 900 159 644 227 973 12 × 2 = 1 + 0.401 391 800 319 288 455 946 24;
  • 13) 0.401 391 800 319 288 455 946 24 × 2 = 0 + 0.802 783 600 638 576 911 892 48;
  • 14) 0.802 783 600 638 576 911 892 48 × 2 = 1 + 0.605 567 201 277 153 823 784 96;
  • 15) 0.605 567 201 277 153 823 784 96 × 2 = 1 + 0.211 134 402 554 307 647 569 92;
  • 16) 0.211 134 402 554 307 647 569 92 × 2 = 0 + 0.422 268 805 108 615 295 139 84;
  • 17) 0.422 268 805 108 615 295 139 84 × 2 = 0 + 0.844 537 610 217 230 590 279 68;
  • 18) 0.844 537 610 217 230 590 279 68 × 2 = 1 + 0.689 075 220 434 461 180 559 36;
  • 19) 0.689 075 220 434 461 180 559 36 × 2 = 1 + 0.378 150 440 868 922 361 118 72;
  • 20) 0.378 150 440 868 922 361 118 72 × 2 = 0 + 0.756 300 881 737 844 722 237 44;
  • 21) 0.756 300 881 737 844 722 237 44 × 2 = 1 + 0.512 601 763 475 689 444 474 88;
  • 22) 0.512 601 763 475 689 444 474 88 × 2 = 1 + 0.025 203 526 951 378 888 949 76;
  • 23) 0.025 203 526 951 378 888 949 76 × 2 = 0 + 0.050 407 053 902 757 777 899 52;
  • 24) 0.050 407 053 902 757 777 899 52 × 2 = 0 + 0.100 814 107 805 515 555 799 04;
  • 25) 0.100 814 107 805 515 555 799 04 × 2 = 0 + 0.201 628 215 611 031 111 598 08;
  • 26) 0.201 628 215 611 031 111 598 08 × 2 = 0 + 0.403 256 431 222 062 223 196 16;
  • 27) 0.403 256 431 222 062 223 196 16 × 2 = 0 + 0.806 512 862 444 124 446 392 32;
  • 28) 0.806 512 862 444 124 446 392 32 × 2 = 1 + 0.613 025 724 888 248 892 784 64;
  • 29) 0.613 025 724 888 248 892 784 64 × 2 = 1 + 0.226 051 449 776 497 785 569 28;
  • 30) 0.226 051 449 776 497 785 569 28 × 2 = 0 + 0.452 102 899 552 995 571 138 56;
  • 31) 0.452 102 899 552 995 571 138 56 × 2 = 0 + 0.904 205 799 105 991 142 277 12;
  • 32) 0.904 205 799 105 991 142 277 12 × 2 = 1 + 0.808 411 598 211 982 284 554 24;
  • 33) 0.808 411 598 211 982 284 554 24 × 2 = 1 + 0.616 823 196 423 964 569 108 48;
  • 34) 0.616 823 196 423 964 569 108 48 × 2 = 1 + 0.233 646 392 847 929 138 216 96;
  • 35) 0.233 646 392 847 929 138 216 96 × 2 = 0 + 0.467 292 785 695 858 276 433 92;
  • 36) 0.467 292 785 695 858 276 433 92 × 2 = 0 + 0.934 585 571 391 716 552 867 84;
  • 37) 0.934 585 571 391 716 552 867 84 × 2 = 1 + 0.869 171 142 783 433 105 735 68;
  • 38) 0.869 171 142 783 433 105 735 68 × 2 = 1 + 0.738 342 285 566 866 211 471 36;
  • 39) 0.738 342 285 566 866 211 471 36 × 2 = 1 + 0.476 684 571 133 732 422 942 72;
  • 40) 0.476 684 571 133 732 422 942 72 × 2 = 0 + 0.953 369 142 267 464 845 885 44;
  • 41) 0.953 369 142 267 464 845 885 44 × 2 = 1 + 0.906 738 284 534 929 691 770 88;
  • 42) 0.906 738 284 534 929 691 770 88 × 2 = 1 + 0.813 476 569 069 859 383 541 76;
  • 43) 0.813 476 569 069 859 383 541 76 × 2 = 1 + 0.626 953 138 139 718 767 083 52;
  • 44) 0.626 953 138 139 718 767 083 52 × 2 = 1 + 0.253 906 276 279 437 534 167 04;
  • 45) 0.253 906 276 279 437 534 167 04 × 2 = 0 + 0.507 812 552 558 875 068 334 08;
  • 46) 0.507 812 552 558 875 068 334 08 × 2 = 1 + 0.015 625 105 117 750 136 668 16;
  • 47) 0.015 625 105 117 750 136 668 16 × 2 = 0 + 0.031 250 210 235 500 273 336 32;
  • 48) 0.031 250 210 235 500 273 336 32 × 2 = 0 + 0.062 500 420 471 000 546 672 64;
  • 49) 0.062 500 420 471 000 546 672 64 × 2 = 0 + 0.125 000 840 942 001 093 345 28;
  • 50) 0.125 000 840 942 001 093 345 28 × 2 = 0 + 0.250 001 681 884 002 186 690 56;
  • 51) 0.250 001 681 884 002 186 690 56 × 2 = 0 + 0.500 003 363 768 004 373 381 12;
  • 52) 0.500 003 363 768 004 373 381 12 × 2 = 1 + 0.000 006 727 536 008 746 762 24;
  • 53) 0.000 006 727 536 008 746 762 24 × 2 = 0 + 0.000 013 455 072 017 493 524 48;

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:


0.745 459 324 169 999 826 283 19(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 283 19(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:


1.745 459 324 169 999 826 283 19(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 1.745 459 324 169 999 826 283 19 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