1.570 796 326 794 896 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.570 796 326 794 896 92(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
1.570 796 326 794 896 92(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: 0.570 796 326 794 896 92.

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) 0.570 796 326 794 896 92 × 2 = 1 + 0.141 592 653 589 793 84;
  • 2) 0.141 592 653 589 793 84 × 2 = 0 + 0.283 185 307 179 587 68;
  • 3) 0.283 185 307 179 587 68 × 2 = 0 + 0.566 370 614 359 175 36;
  • 4) 0.566 370 614 359 175 36 × 2 = 1 + 0.132 741 228 718 350 72;
  • 5) 0.132 741 228 718 350 72 × 2 = 0 + 0.265 482 457 436 701 44;
  • 6) 0.265 482 457 436 701 44 × 2 = 0 + 0.530 964 914 873 402 88;
  • 7) 0.530 964 914 873 402 88 × 2 = 1 + 0.061 929 829 746 805 76;
  • 8) 0.061 929 829 746 805 76 × 2 = 0 + 0.123 859 659 493 611 52;
  • 9) 0.123 859 659 493 611 52 × 2 = 0 + 0.247 719 318 987 223 04;
  • 10) 0.247 719 318 987 223 04 × 2 = 0 + 0.495 438 637 974 446 08;
  • 11) 0.495 438 637 974 446 08 × 2 = 0 + 0.990 877 275 948 892 16;
  • 12) 0.990 877 275 948 892 16 × 2 = 1 + 0.981 754 551 897 784 32;
  • 13) 0.981 754 551 897 784 32 × 2 = 1 + 0.963 509 103 795 568 64;
  • 14) 0.963 509 103 795 568 64 × 2 = 1 + 0.927 018 207 591 137 28;
  • 15) 0.927 018 207 591 137 28 × 2 = 1 + 0.854 036 415 182 274 56;
  • 16) 0.854 036 415 182 274 56 × 2 = 1 + 0.708 072 830 364 549 12;
  • 17) 0.708 072 830 364 549 12 × 2 = 1 + 0.416 145 660 729 098 24;
  • 18) 0.416 145 660 729 098 24 × 2 = 0 + 0.832 291 321 458 196 48;
  • 19) 0.832 291 321 458 196 48 × 2 = 1 + 0.664 582 642 916 392 96;
  • 20) 0.664 582 642 916 392 96 × 2 = 1 + 0.329 165 285 832 785 92;
  • 21) 0.329 165 285 832 785 92 × 2 = 0 + 0.658 330 571 665 571 84;
  • 22) 0.658 330 571 665 571 84 × 2 = 1 + 0.316 661 143 331 143 68;
  • 23) 0.316 661 143 331 143 68 × 2 = 0 + 0.633 322 286 662 287 36;
  • 24) 0.633 322 286 662 287 36 × 2 = 1 + 0.266 644 573 324 574 72;
  • 25) 0.266 644 573 324 574 72 × 2 = 0 + 0.533 289 146 649 149 44;
  • 26) 0.533 289 146 649 149 44 × 2 = 1 + 0.066 578 293 298 298 88;
  • 27) 0.066 578 293 298 298 88 × 2 = 0 + 0.133 156 586 596 597 76;
  • 28) 0.133 156 586 596 597 76 × 2 = 0 + 0.266 313 173 193 195 52;
  • 29) 0.266 313 173 193 195 52 × 2 = 0 + 0.532 626 346 386 391 04;
  • 30) 0.532 626 346 386 391 04 × 2 = 1 + 0.065 252 692 772 782 08;
  • 31) 0.065 252 692 772 782 08 × 2 = 0 + 0.130 505 385 545 564 16;
  • 32) 0.130 505 385 545 564 16 × 2 = 0 + 0.261 010 771 091 128 32;
  • 33) 0.261 010 771 091 128 32 × 2 = 0 + 0.522 021 542 182 256 64;
  • 34) 0.522 021 542 182 256 64 × 2 = 1 + 0.044 043 084 364 513 28;
  • 35) 0.044 043 084 364 513 28 × 2 = 0 + 0.088 086 168 729 026 56;
  • 36) 0.088 086 168 729 026 56 × 2 = 0 + 0.176 172 337 458 053 12;
  • 37) 0.176 172 337 458 053 12 × 2 = 0 + 0.352 344 674 916 106 24;
  • 38) 0.352 344 674 916 106 24 × 2 = 0 + 0.704 689 349 832 212 48;
  • 39) 0.704 689 349 832 212 48 × 2 = 1 + 0.409 378 699 664 424 96;
  • 40) 0.409 378 699 664 424 96 × 2 = 0 + 0.818 757 399 328 849 92;
  • 41) 0.818 757 399 328 849 92 × 2 = 1 + 0.637 514 798 657 699 84;
  • 42) 0.637 514 798 657 699 84 × 2 = 1 + 0.275 029 597 315 399 68;
  • 43) 0.275 029 597 315 399 68 × 2 = 0 + 0.550 059 194 630 799 36;
  • 44) 0.550 059 194 630 799 36 × 2 = 1 + 0.100 118 389 261 598 72;
  • 45) 0.100 118 389 261 598 72 × 2 = 0 + 0.200 236 778 523 197 44;
  • 46) 0.200 236 778 523 197 44 × 2 = 0 + 0.400 473 557 046 394 88;
  • 47) 0.400 473 557 046 394 88 × 2 = 0 + 0.800 947 114 092 789 76;
  • 48) 0.800 947 114 092 789 76 × 2 = 1 + 0.601 894 228 185 579 52;
  • 49) 0.601 894 228 185 579 52 × 2 = 1 + 0.203 788 456 371 159 04;
  • 50) 0.203 788 456 371 159 04 × 2 = 0 + 0.407 576 912 742 318 08;
  • 51) 0.407 576 912 742 318 08 × 2 = 0 + 0.815 153 825 484 636 16;
  • 52) 0.815 153 825 484 636 16 × 2 = 1 + 0.630 307 650 969 272 32;
  • 53) 0.630 307 650 969 272 32 × 2 = 1 + 0.260 615 301 938 544 64;

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:


0.570 796 326 794 896 92(10) =

0.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 1(2)

5. Positive number before normalization:

1.570 796 326 794 896 92(10) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 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:


1.570 796 326 794 896 92(10) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 1(2) =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 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):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 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. 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001 1 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001


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) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001


Decimal number 1.570 796 326 794 896 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1001

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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