3.141 592 653 589 793 238 465 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 465 17(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
3.141 592 653 589 793 238 465 17(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 238 465 17.

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) 0.141 592 653 589 793 238 465 17 × 2 = 0 + 0.283 185 307 179 586 476 930 34;
  • 2) 0.283 185 307 179 586 476 930 34 × 2 = 0 + 0.566 370 614 359 172 953 860 68;
  • 3) 0.566 370 614 359 172 953 860 68 × 2 = 1 + 0.132 741 228 718 345 907 721 36;
  • 4) 0.132 741 228 718 345 907 721 36 × 2 = 0 + 0.265 482 457 436 691 815 442 72;
  • 5) 0.265 482 457 436 691 815 442 72 × 2 = 0 + 0.530 964 914 873 383 630 885 44;
  • 6) 0.530 964 914 873 383 630 885 44 × 2 = 1 + 0.061 929 829 746 767 261 770 88;
  • 7) 0.061 929 829 746 767 261 770 88 × 2 = 0 + 0.123 859 659 493 534 523 541 76;
  • 8) 0.123 859 659 493 534 523 541 76 × 2 = 0 + 0.247 719 318 987 069 047 083 52;
  • 9) 0.247 719 318 987 069 047 083 52 × 2 = 0 + 0.495 438 637 974 138 094 167 04;
  • 10) 0.495 438 637 974 138 094 167 04 × 2 = 0 + 0.990 877 275 948 276 188 334 08;
  • 11) 0.990 877 275 948 276 188 334 08 × 2 = 1 + 0.981 754 551 896 552 376 668 16;
  • 12) 0.981 754 551 896 552 376 668 16 × 2 = 1 + 0.963 509 103 793 104 753 336 32;
  • 13) 0.963 509 103 793 104 753 336 32 × 2 = 1 + 0.927 018 207 586 209 506 672 64;
  • 14) 0.927 018 207 586 209 506 672 64 × 2 = 1 + 0.854 036 415 172 419 013 345 28;
  • 15) 0.854 036 415 172 419 013 345 28 × 2 = 1 + 0.708 072 830 344 838 026 690 56;
  • 16) 0.708 072 830 344 838 026 690 56 × 2 = 1 + 0.416 145 660 689 676 053 381 12;
  • 17) 0.416 145 660 689 676 053 381 12 × 2 = 0 + 0.832 291 321 379 352 106 762 24;
  • 18) 0.832 291 321 379 352 106 762 24 × 2 = 1 + 0.664 582 642 758 704 213 524 48;
  • 19) 0.664 582 642 758 704 213 524 48 × 2 = 1 + 0.329 165 285 517 408 427 048 96;
  • 20) 0.329 165 285 517 408 427 048 96 × 2 = 0 + 0.658 330 571 034 816 854 097 92;
  • 21) 0.658 330 571 034 816 854 097 92 × 2 = 1 + 0.316 661 142 069 633 708 195 84;
  • 22) 0.316 661 142 069 633 708 195 84 × 2 = 0 + 0.633 322 284 139 267 416 391 68;
  • 23) 0.633 322 284 139 267 416 391 68 × 2 = 1 + 0.266 644 568 278 534 832 783 36;
  • 24) 0.266 644 568 278 534 832 783 36 × 2 = 0 + 0.533 289 136 557 069 665 566 72;
  • 25) 0.533 289 136 557 069 665 566 72 × 2 = 1 + 0.066 578 273 114 139 331 133 44;
  • 26) 0.066 578 273 114 139 331 133 44 × 2 = 0 + 0.133 156 546 228 278 662 266 88;
  • 27) 0.133 156 546 228 278 662 266 88 × 2 = 0 + 0.266 313 092 456 557 324 533 76;
  • 28) 0.266 313 092 456 557 324 533 76 × 2 = 0 + 0.532 626 184 913 114 649 067 52;
  • 29) 0.532 626 184 913 114 649 067 52 × 2 = 1 + 0.065 252 369 826 229 298 135 04;
  • 30) 0.065 252 369 826 229 298 135 04 × 2 = 0 + 0.130 504 739 652 458 596 270 08;
  • 31) 0.130 504 739 652 458 596 270 08 × 2 = 0 + 0.261 009 479 304 917 192 540 16;
  • 32) 0.261 009 479 304 917 192 540 16 × 2 = 0 + 0.522 018 958 609 834 385 080 32;
  • 33) 0.522 018 958 609 834 385 080 32 × 2 = 1 + 0.044 037 917 219 668 770 160 64;
  • 34) 0.044 037 917 219 668 770 160 64 × 2 = 0 + 0.088 075 834 439 337 540 321 28;
  • 35) 0.088 075 834 439 337 540 321 28 × 2 = 0 + 0.176 151 668 878 675 080 642 56;
  • 36) 0.176 151 668 878 675 080 642 56 × 2 = 0 + 0.352 303 337 757 350 161 285 12;
  • 37) 0.352 303 337 757 350 161 285 12 × 2 = 0 + 0.704 606 675 514 700 322 570 24;
  • 38) 0.704 606 675 514 700 322 570 24 × 2 = 1 + 0.409 213 351 029 400 645 140 48;
  • 39) 0.409 213 351 029 400 645 140 48 × 2 = 0 + 0.818 426 702 058 801 290 280 96;
  • 40) 0.818 426 702 058 801 290 280 96 × 2 = 1 + 0.636 853 404 117 602 580 561 92;
  • 41) 0.636 853 404 117 602 580 561 92 × 2 = 1 + 0.273 706 808 235 205 161 123 84;
  • 42) 0.273 706 808 235 205 161 123 84 × 2 = 0 + 0.547 413 616 470 410 322 247 68;
  • 43) 0.547 413 616 470 410 322 247 68 × 2 = 1 + 0.094 827 232 940 820 644 495 36;
  • 44) 0.094 827 232 940 820 644 495 36 × 2 = 0 + 0.189 654 465 881 641 288 990 72;
  • 45) 0.189 654 465 881 641 288 990 72 × 2 = 0 + 0.379 308 931 763 282 577 981 44;
  • 46) 0.379 308 931 763 282 577 981 44 × 2 = 0 + 0.758 617 863 526 565 155 962 88;
  • 47) 0.758 617 863 526 565 155 962 88 × 2 = 1 + 0.517 235 727 053 130 311 925 76;
  • 48) 0.517 235 727 053 130 311 925 76 × 2 = 1 + 0.034 471 454 106 260 623 851 52;
  • 49) 0.034 471 454 106 260 623 851 52 × 2 = 0 + 0.068 942 908 212 521 247 703 04;
  • 50) 0.068 942 908 212 521 247 703 04 × 2 = 0 + 0.137 885 816 425 042 495 406 08;
  • 51) 0.137 885 816 425 042 495 406 08 × 2 = 0 + 0.275 771 632 850 084 990 812 16;
  • 52) 0.275 771 632 850 084 990 812 16 × 2 = 0 + 0.551 543 265 700 169 981 624 32;
  • 53) 0.551 543 265 700 169 981 624 32 × 2 = 1 + 0.103 086 531 400 339 963 248 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:


0.141 592 653 589 793 238 465 17(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 465 17(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 589 793 238 465 17(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 01(2) × 21


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): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 01


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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 1000 01 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


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) =
100 0000 0000

Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


Decimal number 3.141 592 653 589 793 238 465 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

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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