0.000 000 000 000 000 002 315 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 002 315(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 000 000 002 315(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 002 315.

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) 0.000 000 000 000 000 002 315 × 2 = 0 + 0.000 000 000 000 000 004 63;
  • 2) 0.000 000 000 000 000 004 63 × 2 = 0 + 0.000 000 000 000 000 009 26;
  • 3) 0.000 000 000 000 000 009 26 × 2 = 0 + 0.000 000 000 000 000 018 52;
  • 4) 0.000 000 000 000 000 018 52 × 2 = 0 + 0.000 000 000 000 000 037 04;
  • 5) 0.000 000 000 000 000 037 04 × 2 = 0 + 0.000 000 000 000 000 074 08;
  • 6) 0.000 000 000 000 000 074 08 × 2 = 0 + 0.000 000 000 000 000 148 16;
  • 7) 0.000 000 000 000 000 148 16 × 2 = 0 + 0.000 000 000 000 000 296 32;
  • 8) 0.000 000 000 000 000 296 32 × 2 = 0 + 0.000 000 000 000 000 592 64;
  • 9) 0.000 000 000 000 000 592 64 × 2 = 0 + 0.000 000 000 000 001 185 28;
  • 10) 0.000 000 000 000 001 185 28 × 2 = 0 + 0.000 000 000 000 002 370 56;
  • 11) 0.000 000 000 000 002 370 56 × 2 = 0 + 0.000 000 000 000 004 741 12;
  • 12) 0.000 000 000 000 004 741 12 × 2 = 0 + 0.000 000 000 000 009 482 24;
  • 13) 0.000 000 000 000 009 482 24 × 2 = 0 + 0.000 000 000 000 018 964 48;
  • 14) 0.000 000 000 000 018 964 48 × 2 = 0 + 0.000 000 000 000 037 928 96;
  • 15) 0.000 000 000 000 037 928 96 × 2 = 0 + 0.000 000 000 000 075 857 92;
  • 16) 0.000 000 000 000 075 857 92 × 2 = 0 + 0.000 000 000 000 151 715 84;
  • 17) 0.000 000 000 000 151 715 84 × 2 = 0 + 0.000 000 000 000 303 431 68;
  • 18) 0.000 000 000 000 303 431 68 × 2 = 0 + 0.000 000 000 000 606 863 36;
  • 19) 0.000 000 000 000 606 863 36 × 2 = 0 + 0.000 000 000 001 213 726 72;
  • 20) 0.000 000 000 001 213 726 72 × 2 = 0 + 0.000 000 000 002 427 453 44;
  • 21) 0.000 000 000 002 427 453 44 × 2 = 0 + 0.000 000 000 004 854 906 88;
  • 22) 0.000 000 000 004 854 906 88 × 2 = 0 + 0.000 000 000 009 709 813 76;
  • 23) 0.000 000 000 009 709 813 76 × 2 = 0 + 0.000 000 000 019 419 627 52;
  • 24) 0.000 000 000 019 419 627 52 × 2 = 0 + 0.000 000 000 038 839 255 04;
  • 25) 0.000 000 000 038 839 255 04 × 2 = 0 + 0.000 000 000 077 678 510 08;
  • 26) 0.000 000 000 077 678 510 08 × 2 = 0 + 0.000 000 000 155 357 020 16;
  • 27) 0.000 000 000 155 357 020 16 × 2 = 0 + 0.000 000 000 310 714 040 32;
  • 28) 0.000 000 000 310 714 040 32 × 2 = 0 + 0.000 000 000 621 428 080 64;
  • 29) 0.000 000 000 621 428 080 64 × 2 = 0 + 0.000 000 001 242 856 161 28;
  • 30) 0.000 000 001 242 856 161 28 × 2 = 0 + 0.000 000 002 485 712 322 56;
  • 31) 0.000 000 002 485 712 322 56 × 2 = 0 + 0.000 000 004 971 424 645 12;
  • 32) 0.000 000 004 971 424 645 12 × 2 = 0 + 0.000 000 009 942 849 290 24;
  • 33) 0.000 000 009 942 849 290 24 × 2 = 0 + 0.000 000 019 885 698 580 48;
  • 34) 0.000 000 019 885 698 580 48 × 2 = 0 + 0.000 000 039 771 397 160 96;
  • 35) 0.000 000 039 771 397 160 96 × 2 = 0 + 0.000 000 079 542 794 321 92;
  • 36) 0.000 000 079 542 794 321 92 × 2 = 0 + 0.000 000 159 085 588 643 84;
  • 37) 0.000 000 159 085 588 643 84 × 2 = 0 + 0.000 000 318 171 177 287 68;
  • 38) 0.000 000 318 171 177 287 68 × 2 = 0 + 0.000 000 636 342 354 575 36;
  • 39) 0.000 000 636 342 354 575 36 × 2 = 0 + 0.000 001 272 684 709 150 72;
  • 40) 0.000 001 272 684 709 150 72 × 2 = 0 + 0.000 002 545 369 418 301 44;
  • 41) 0.000 002 545 369 418 301 44 × 2 = 0 + 0.000 005 090 738 836 602 88;
  • 42) 0.000 005 090 738 836 602 88 × 2 = 0 + 0.000 010 181 477 673 205 76;
  • 43) 0.000 010 181 477 673 205 76 × 2 = 0 + 0.000 020 362 955 346 411 52;
  • 44) 0.000 020 362 955 346 411 52 × 2 = 0 + 0.000 040 725 910 692 823 04;
  • 45) 0.000 040 725 910 692 823 04 × 2 = 0 + 0.000 081 451 821 385 646 08;
  • 46) 0.000 081 451 821 385 646 08 × 2 = 0 + 0.000 162 903 642 771 292 16;
  • 47) 0.000 162 903 642 771 292 16 × 2 = 0 + 0.000 325 807 285 542 584 32;
  • 48) 0.000 325 807 285 542 584 32 × 2 = 0 + 0.000 651 614 571 085 168 64;
  • 49) 0.000 651 614 571 085 168 64 × 2 = 0 + 0.001 303 229 142 170 337 28;
  • 50) 0.001 303 229 142 170 337 28 × 2 = 0 + 0.002 606 458 284 340 674 56;
  • 51) 0.002 606 458 284 340 674 56 × 2 = 0 + 0.005 212 916 568 681 349 12;
  • 52) 0.005 212 916 568 681 349 12 × 2 = 0 + 0.010 425 833 137 362 698 24;
  • 53) 0.010 425 833 137 362 698 24 × 2 = 0 + 0.020 851 666 274 725 396 48;
  • 54) 0.020 851 666 274 725 396 48 × 2 = 0 + 0.041 703 332 549 450 792 96;
  • 55) 0.041 703 332 549 450 792 96 × 2 = 0 + 0.083 406 665 098 901 585 92;
  • 56) 0.083 406 665 098 901 585 92 × 2 = 0 + 0.166 813 330 197 803 171 84;
  • 57) 0.166 813 330 197 803 171 84 × 2 = 0 + 0.333 626 660 395 606 343 68;
  • 58) 0.333 626 660 395 606 343 68 × 2 = 0 + 0.667 253 320 791 212 687 36;
  • 59) 0.667 253 320 791 212 687 36 × 2 = 1 + 0.334 506 641 582 425 374 72;
  • 60) 0.334 506 641 582 425 374 72 × 2 = 0 + 0.669 013 283 164 850 749 44;
  • 61) 0.669 013 283 164 850 749 44 × 2 = 1 + 0.338 026 566 329 701 498 88;
  • 62) 0.338 026 566 329 701 498 88 × 2 = 0 + 0.676 053 132 659 402 997 76;
  • 63) 0.676 053 132 659 402 997 76 × 2 = 1 + 0.352 106 265 318 805 995 52;
  • 64) 0.352 106 265 318 805 995 52 × 2 = 0 + 0.704 212 530 637 611 991 04;
  • 65) 0.704 212 530 637 611 991 04 × 2 = 1 + 0.408 425 061 275 223 982 08;
  • 66) 0.408 425 061 275 223 982 08 × 2 = 0 + 0.816 850 122 550 447 964 16;
  • 67) 0.816 850 122 550 447 964 16 × 2 = 1 + 0.633 700 245 100 895 928 32;
  • 68) 0.633 700 245 100 895 928 32 × 2 = 1 + 0.267 400 490 201 791 856 64;
  • 69) 0.267 400 490 201 791 856 64 × 2 = 0 + 0.534 800 980 403 583 713 28;
  • 70) 0.534 800 980 403 583 713 28 × 2 = 1 + 0.069 601 960 807 167 426 56;
  • 71) 0.069 601 960 807 167 426 56 × 2 = 0 + 0.139 203 921 614 334 853 12;
  • 72) 0.139 203 921 614 334 853 12 × 2 = 0 + 0.278 407 843 228 669 706 24;
  • 73) 0.278 407 843 228 669 706 24 × 2 = 0 + 0.556 815 686 457 339 412 48;
  • 74) 0.556 815 686 457 339 412 48 × 2 = 1 + 0.113 631 372 914 678 824 96;
  • 75) 0.113 631 372 914 678 824 96 × 2 = 0 + 0.227 262 745 829 357 649 92;
  • 76) 0.227 262 745 829 357 649 92 × 2 = 0 + 0.454 525 491 658 715 299 84;
  • 77) 0.454 525 491 658 715 299 84 × 2 = 0 + 0.909 050 983 317 430 599 68;
  • 78) 0.909 050 983 317 430 599 68 × 2 = 1 + 0.818 101 966 634 861 199 36;
  • 79) 0.818 101 966 634 861 199 36 × 2 = 1 + 0.636 203 933 269 722 398 72;
  • 80) 0.636 203 933 269 722 398 72 × 2 = 1 + 0.272 407 866 539 444 797 44;
  • 81) 0.272 407 866 539 444 797 44 × 2 = 0 + 0.544 815 733 078 889 594 88;
  • 82) 0.544 815 733 078 889 594 88 × 2 = 1 + 0.089 631 466 157 779 189 76;

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:


0.000 000 000 000 000 002 315(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1011 0100 0100 0111 01(2)

5. Positive number before normalization:

0.000 000 000 000 000 002 315(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1011 0100 0100 0111 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 59 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 002 315(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1011 0100 0100 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1011 0100 0100 0111 01(2) × 20 =

1.0101 0101 1010 0010 0011 101(2) × 2-59


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -59

Mantissa (not normalized):
1.0101 0101 1010 0010 0011 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-59 + 2(8-1) - 1 =

(-59 + 127)(10) =

68(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 68 ÷ 2 = 34 + 0;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 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) =

68(10) =

0100 0100(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 1010 1101 0001 0001 1101 =

010 1010 1101 0001 0001 1101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0100 0100

Mantissa (23 bits) =
010 1010 1101 0001 0001 1101


Decimal number 0.000 000 000 000 000 002 315 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0100 0100 - 010 1010 1101 0001 0001 1101

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111