1.745 459 324 169 999 825 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 825 69(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
1.745 459 324 169 999 825 69(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: 0.745 459 324 169 999 825 69.

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) 0.745 459 324 169 999 825 69 × 2 = 1 + 0.490 918 648 339 999 651 38;
  • 2) 0.490 918 648 339 999 651 38 × 2 = 0 + 0.981 837 296 679 999 302 76;
  • 3) 0.981 837 296 679 999 302 76 × 2 = 1 + 0.963 674 593 359 998 605 52;
  • 4) 0.963 674 593 359 998 605 52 × 2 = 1 + 0.927 349 186 719 997 211 04;
  • 5) 0.927 349 186 719 997 211 04 × 2 = 1 + 0.854 698 373 439 994 422 08;
  • 6) 0.854 698 373 439 994 422 08 × 2 = 1 + 0.709 396 746 879 988 844 16;
  • 7) 0.709 396 746 879 988 844 16 × 2 = 1 + 0.418 793 493 759 977 688 32;
  • 8) 0.418 793 493 759 977 688 32 × 2 = 0 + 0.837 586 987 519 955 376 64;
  • 9) 0.837 586 987 519 955 376 64 × 2 = 1 + 0.675 173 975 039 910 753 28;
  • 10) 0.675 173 975 039 910 753 28 × 2 = 1 + 0.350 347 950 079 821 506 56;
  • 11) 0.350 347 950 079 821 506 56 × 2 = 0 + 0.700 695 900 159 643 013 12;
  • 12) 0.700 695 900 159 643 013 12 × 2 = 1 + 0.401 391 800 319 286 026 24;
  • 13) 0.401 391 800 319 286 026 24 × 2 = 0 + 0.802 783 600 638 572 052 48;
  • 14) 0.802 783 600 638 572 052 48 × 2 = 1 + 0.605 567 201 277 144 104 96;
  • 15) 0.605 567 201 277 144 104 96 × 2 = 1 + 0.211 134 402 554 288 209 92;
  • 16) 0.211 134 402 554 288 209 92 × 2 = 0 + 0.422 268 805 108 576 419 84;
  • 17) 0.422 268 805 108 576 419 84 × 2 = 0 + 0.844 537 610 217 152 839 68;
  • 18) 0.844 537 610 217 152 839 68 × 2 = 1 + 0.689 075 220 434 305 679 36;
  • 19) 0.689 075 220 434 305 679 36 × 2 = 1 + 0.378 150 440 868 611 358 72;
  • 20) 0.378 150 440 868 611 358 72 × 2 = 0 + 0.756 300 881 737 222 717 44;
  • 21) 0.756 300 881 737 222 717 44 × 2 = 1 + 0.512 601 763 474 445 434 88;
  • 22) 0.512 601 763 474 445 434 88 × 2 = 1 + 0.025 203 526 948 890 869 76;
  • 23) 0.025 203 526 948 890 869 76 × 2 = 0 + 0.050 407 053 897 781 739 52;
  • 24) 0.050 407 053 897 781 739 52 × 2 = 0 + 0.100 814 107 795 563 479 04;
  • 25) 0.100 814 107 795 563 479 04 × 2 = 0 + 0.201 628 215 591 126 958 08;
  • 26) 0.201 628 215 591 126 958 08 × 2 = 0 + 0.403 256 431 182 253 916 16;
  • 27) 0.403 256 431 182 253 916 16 × 2 = 0 + 0.806 512 862 364 507 832 32;
  • 28) 0.806 512 862 364 507 832 32 × 2 = 1 + 0.613 025 724 729 015 664 64;
  • 29) 0.613 025 724 729 015 664 64 × 2 = 1 + 0.226 051 449 458 031 329 28;
  • 30) 0.226 051 449 458 031 329 28 × 2 = 0 + 0.452 102 898 916 062 658 56;
  • 31) 0.452 102 898 916 062 658 56 × 2 = 0 + 0.904 205 797 832 125 317 12;
  • 32) 0.904 205 797 832 125 317 12 × 2 = 1 + 0.808 411 595 664 250 634 24;
  • 33) 0.808 411 595 664 250 634 24 × 2 = 1 + 0.616 823 191 328 501 268 48;
  • 34) 0.616 823 191 328 501 268 48 × 2 = 1 + 0.233 646 382 657 002 536 96;
  • 35) 0.233 646 382 657 002 536 96 × 2 = 0 + 0.467 292 765 314 005 073 92;
  • 36) 0.467 292 765 314 005 073 92 × 2 = 0 + 0.934 585 530 628 010 147 84;
  • 37) 0.934 585 530 628 010 147 84 × 2 = 1 + 0.869 171 061 256 020 295 68;
  • 38) 0.869 171 061 256 020 295 68 × 2 = 1 + 0.738 342 122 512 040 591 36;
  • 39) 0.738 342 122 512 040 591 36 × 2 = 1 + 0.476 684 245 024 081 182 72;
  • 40) 0.476 684 245 024 081 182 72 × 2 = 0 + 0.953 368 490 048 162 365 44;
  • 41) 0.953 368 490 048 162 365 44 × 2 = 1 + 0.906 736 980 096 324 730 88;
  • 42) 0.906 736 980 096 324 730 88 × 2 = 1 + 0.813 473 960 192 649 461 76;
  • 43) 0.813 473 960 192 649 461 76 × 2 = 1 + 0.626 947 920 385 298 923 52;
  • 44) 0.626 947 920 385 298 923 52 × 2 = 1 + 0.253 895 840 770 597 847 04;
  • 45) 0.253 895 840 770 597 847 04 × 2 = 0 + 0.507 791 681 541 195 694 08;
  • 46) 0.507 791 681 541 195 694 08 × 2 = 1 + 0.015 583 363 082 391 388 16;
  • 47) 0.015 583 363 082 391 388 16 × 2 = 0 + 0.031 166 726 164 782 776 32;
  • 48) 0.031 166 726 164 782 776 32 × 2 = 0 + 0.062 333 452 329 565 552 64;
  • 49) 0.062 333 452 329 565 552 64 × 2 = 0 + 0.124 666 904 659 131 105 28;
  • 50) 0.124 666 904 659 131 105 28 × 2 = 0 + 0.249 333 809 318 262 210 56;
  • 51) 0.249 333 809 318 262 210 56 × 2 = 0 + 0.498 667 618 636 524 421 12;
  • 52) 0.498 667 618 636 524 421 12 × 2 = 0 + 0.997 335 237 273 048 842 24;
  • 53) 0.997 335 237 273 048 842 24 × 2 = 1 + 0.994 670 474 546 097 684 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:


0.745 459 324 169 999 825 69(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 825 69(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:


1.745 459 324 169 999 825 69(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 1.745 459 324 169 999 825 69 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