0.000 000 000 050 402 949 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 050 402 949(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
0.000 000 000 050 402 949(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: 0.000 000 000 050 402 949.

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) 0.000 000 000 050 402 949 × 2 = 0 + 0.000 000 000 100 805 898;
  • 2) 0.000 000 000 100 805 898 × 2 = 0 + 0.000 000 000 201 611 796;
  • 3) 0.000 000 000 201 611 796 × 2 = 0 + 0.000 000 000 403 223 592;
  • 4) 0.000 000 000 403 223 592 × 2 = 0 + 0.000 000 000 806 447 184;
  • 5) 0.000 000 000 806 447 184 × 2 = 0 + 0.000 000 001 612 894 368;
  • 6) 0.000 000 001 612 894 368 × 2 = 0 + 0.000 000 003 225 788 736;
  • 7) 0.000 000 003 225 788 736 × 2 = 0 + 0.000 000 006 451 577 472;
  • 8) 0.000 000 006 451 577 472 × 2 = 0 + 0.000 000 012 903 154 944;
  • 9) 0.000 000 012 903 154 944 × 2 = 0 + 0.000 000 025 806 309 888;
  • 10) 0.000 000 025 806 309 888 × 2 = 0 + 0.000 000 051 612 619 776;
  • 11) 0.000 000 051 612 619 776 × 2 = 0 + 0.000 000 103 225 239 552;
  • 12) 0.000 000 103 225 239 552 × 2 = 0 + 0.000 000 206 450 479 104;
  • 13) 0.000 000 206 450 479 104 × 2 = 0 + 0.000 000 412 900 958 208;
  • 14) 0.000 000 412 900 958 208 × 2 = 0 + 0.000 000 825 801 916 416;
  • 15) 0.000 000 825 801 916 416 × 2 = 0 + 0.000 001 651 603 832 832;
  • 16) 0.000 001 651 603 832 832 × 2 = 0 + 0.000 003 303 207 665 664;
  • 17) 0.000 003 303 207 665 664 × 2 = 0 + 0.000 006 606 415 331 328;
  • 18) 0.000 006 606 415 331 328 × 2 = 0 + 0.000 013 212 830 662 656;
  • 19) 0.000 013 212 830 662 656 × 2 = 0 + 0.000 026 425 661 325 312;
  • 20) 0.000 026 425 661 325 312 × 2 = 0 + 0.000 052 851 322 650 624;
  • 21) 0.000 052 851 322 650 624 × 2 = 0 + 0.000 105 702 645 301 248;
  • 22) 0.000 105 702 645 301 248 × 2 = 0 + 0.000 211 405 290 602 496;
  • 23) 0.000 211 405 290 602 496 × 2 = 0 + 0.000 422 810 581 204 992;
  • 24) 0.000 422 810 581 204 992 × 2 = 0 + 0.000 845 621 162 409 984;
  • 25) 0.000 845 621 162 409 984 × 2 = 0 + 0.001 691 242 324 819 968;
  • 26) 0.001 691 242 324 819 968 × 2 = 0 + 0.003 382 484 649 639 936;
  • 27) 0.003 382 484 649 639 936 × 2 = 0 + 0.006 764 969 299 279 872;
  • 28) 0.006 764 969 299 279 872 × 2 = 0 + 0.013 529 938 598 559 744;
  • 29) 0.013 529 938 598 559 744 × 2 = 0 + 0.027 059 877 197 119 488;
  • 30) 0.027 059 877 197 119 488 × 2 = 0 + 0.054 119 754 394 238 976;
  • 31) 0.054 119 754 394 238 976 × 2 = 0 + 0.108 239 508 788 477 952;
  • 32) 0.108 239 508 788 477 952 × 2 = 0 + 0.216 479 017 576 955 904;
  • 33) 0.216 479 017 576 955 904 × 2 = 0 + 0.432 958 035 153 911 808;
  • 34) 0.432 958 035 153 911 808 × 2 = 0 + 0.865 916 070 307 823 616;
  • 35) 0.865 916 070 307 823 616 × 2 = 1 + 0.731 832 140 615 647 232;
  • 36) 0.731 832 140 615 647 232 × 2 = 1 + 0.463 664 281 231 294 464;
  • 37) 0.463 664 281 231 294 464 × 2 = 0 + 0.927 328 562 462 588 928;
  • 38) 0.927 328 562 462 588 928 × 2 = 1 + 0.854 657 124 925 177 856;
  • 39) 0.854 657 124 925 177 856 × 2 = 1 + 0.709 314 249 850 355 712;
  • 40) 0.709 314 249 850 355 712 × 2 = 1 + 0.418 628 499 700 711 424;
  • 41) 0.418 628 499 700 711 424 × 2 = 0 + 0.837 256 999 401 422 848;
  • 42) 0.837 256 999 401 422 848 × 2 = 1 + 0.674 513 998 802 845 696;
  • 43) 0.674 513 998 802 845 696 × 2 = 1 + 0.349 027 997 605 691 392;
  • 44) 0.349 027 997 605 691 392 × 2 = 0 + 0.698 055 995 211 382 784;
  • 45) 0.698 055 995 211 382 784 × 2 = 1 + 0.396 111 990 422 765 568;
  • 46) 0.396 111 990 422 765 568 × 2 = 0 + 0.792 223 980 845 531 136;
  • 47) 0.792 223 980 845 531 136 × 2 = 1 + 0.584 447 961 691 062 272;
  • 48) 0.584 447 961 691 062 272 × 2 = 1 + 0.168 895 923 382 124 544;
  • 49) 0.168 895 923 382 124 544 × 2 = 0 + 0.337 791 846 764 249 088;
  • 50) 0.337 791 846 764 249 088 × 2 = 0 + 0.675 583 693 528 498 176;
  • 51) 0.675 583 693 528 498 176 × 2 = 1 + 0.351 167 387 056 996 352;
  • 52) 0.351 167 387 056 996 352 × 2 = 0 + 0.702 334 774 113 992 704;
  • 53) 0.702 334 774 113 992 704 × 2 = 1 + 0.404 669 548 227 985 408;
  • 54) 0.404 669 548 227 985 408 × 2 = 0 + 0.809 339 096 455 970 816;
  • 55) 0.809 339 096 455 970 816 × 2 = 1 + 0.618 678 192 911 941 632;
  • 56) 0.618 678 192 911 941 632 × 2 = 1 + 0.237 356 385 823 883 264;
  • 57) 0.237 356 385 823 883 264 × 2 = 0 + 0.474 712 771 647 766 528;
  • 58) 0.474 712 771 647 766 528 × 2 = 0 + 0.949 425 543 295 533 056;

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:


0.000 000 000 050 402 949(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0010 1011 00(2)

5. Positive number before normalization:

0.000 000 000 050 402 949(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0010 1011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 35 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 050 402 949(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0010 1011 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0110 1011 0010 1011 00(2) × 20 =

1.1011 1011 0101 1001 0101 100(2) × 2-35


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): -35

Mantissa (not normalized):
1.1011 1011 0101 1001 0101 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-35 + 2(8-1) - 1 =

(-35 + 127)(10) =

92(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;
  • 92 ÷ 2 = 46 + 0;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

92(10) =

0101 1100(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. 101 1101 1010 1100 1010 1100 =

101 1101 1010 1100 1010 1100


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) =
0101 1100

Mantissa (23 bits) =
101 1101 1010 1100 1010 1100


Decimal number 0.000 000 000 050 402 949 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1100 - 101 1101 1010 1100 1010 1100

» Convert decimal 0.000 000 000 050 403 011 to 32 bit single precision IEEE 754 binary floating point representation standard

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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