32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 000 106 524 7 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 106 524 7(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 106 524 7| = 0.000 000 000 106 524 7

2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 106 524 7.

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) 0.000 000 000 106 524 7 × 2 = 0 + 0.000 000 000 213 049 4;
  • 2) 0.000 000 000 213 049 4 × 2 = 0 + 0.000 000 000 426 098 8;
  • 3) 0.000 000 000 426 098 8 × 2 = 0 + 0.000 000 000 852 197 6;
  • 4) 0.000 000 000 852 197 6 × 2 = 0 + 0.000 000 001 704 395 2;
  • 5) 0.000 000 001 704 395 2 × 2 = 0 + 0.000 000 003 408 790 4;
  • 6) 0.000 000 003 408 790 4 × 2 = 0 + 0.000 000 006 817 580 8;
  • 7) 0.000 000 006 817 580 8 × 2 = 0 + 0.000 000 013 635 161 6;
  • 8) 0.000 000 013 635 161 6 × 2 = 0 + 0.000 000 027 270 323 2;
  • 9) 0.000 000 027 270 323 2 × 2 = 0 + 0.000 000 054 540 646 4;
  • 10) 0.000 000 054 540 646 4 × 2 = 0 + 0.000 000 109 081 292 8;
  • 11) 0.000 000 109 081 292 8 × 2 = 0 + 0.000 000 218 162 585 6;
  • 12) 0.000 000 218 162 585 6 × 2 = 0 + 0.000 000 436 325 171 2;
  • 13) 0.000 000 436 325 171 2 × 2 = 0 + 0.000 000 872 650 342 4;
  • 14) 0.000 000 872 650 342 4 × 2 = 0 + 0.000 001 745 300 684 8;
  • 15) 0.000 001 745 300 684 8 × 2 = 0 + 0.000 003 490 601 369 6;
  • 16) 0.000 003 490 601 369 6 × 2 = 0 + 0.000 006 981 202 739 2;
  • 17) 0.000 006 981 202 739 2 × 2 = 0 + 0.000 013 962 405 478 4;
  • 18) 0.000 013 962 405 478 4 × 2 = 0 + 0.000 027 924 810 956 8;
  • 19) 0.000 027 924 810 956 8 × 2 = 0 + 0.000 055 849 621 913 6;
  • 20) 0.000 055 849 621 913 6 × 2 = 0 + 0.000 111 699 243 827 2;
  • 21) 0.000 111 699 243 827 2 × 2 = 0 + 0.000 223 398 487 654 4;
  • 22) 0.000 223 398 487 654 4 × 2 = 0 + 0.000 446 796 975 308 8;
  • 23) 0.000 446 796 975 308 8 × 2 = 0 + 0.000 893 593 950 617 6;
  • 24) 0.000 893 593 950 617 6 × 2 = 0 + 0.001 787 187 901 235 2;
  • 25) 0.001 787 187 901 235 2 × 2 = 0 + 0.003 574 375 802 470 4;
  • 26) 0.003 574 375 802 470 4 × 2 = 0 + 0.007 148 751 604 940 8;
  • 27) 0.007 148 751 604 940 8 × 2 = 0 + 0.014 297 503 209 881 6;
  • 28) 0.014 297 503 209 881 6 × 2 = 0 + 0.028 595 006 419 763 2;
  • 29) 0.028 595 006 419 763 2 × 2 = 0 + 0.057 190 012 839 526 4;
  • 30) 0.057 190 012 839 526 4 × 2 = 0 + 0.114 380 025 679 052 8;
  • 31) 0.114 380 025 679 052 8 × 2 = 0 + 0.228 760 051 358 105 6;
  • 32) 0.228 760 051 358 105 6 × 2 = 0 + 0.457 520 102 716 211 2;
  • 33) 0.457 520 102 716 211 2 × 2 = 0 + 0.915 040 205 432 422 4;
  • 34) 0.915 040 205 432 422 4 × 2 = 1 + 0.830 080 410 864 844 8;
  • 35) 0.830 080 410 864 844 8 × 2 = 1 + 0.660 160 821 729 689 6;
  • 36) 0.660 160 821 729 689 6 × 2 = 1 + 0.320 321 643 459 379 2;
  • 37) 0.320 321 643 459 379 2 × 2 = 0 + 0.640 643 286 918 758 4;
  • 38) 0.640 643 286 918 758 4 × 2 = 1 + 0.281 286 573 837 516 8;
  • 39) 0.281 286 573 837 516 8 × 2 = 0 + 0.562 573 147 675 033 6;
  • 40) 0.562 573 147 675 033 6 × 2 = 1 + 0.125 146 295 350 067 2;
  • 41) 0.125 146 295 350 067 2 × 2 = 0 + 0.250 292 590 700 134 4;
  • 42) 0.250 292 590 700 134 4 × 2 = 0 + 0.500 585 181 400 268 8;
  • 43) 0.500 585 181 400 268 8 × 2 = 1 + 0.001 170 362 800 537 6;
  • 44) 0.001 170 362 800 537 6 × 2 = 0 + 0.002 340 725 601 075 2;
  • 45) 0.002 340 725 601 075 2 × 2 = 0 + 0.004 681 451 202 150 4;
  • 46) 0.004 681 451 202 150 4 × 2 = 0 + 0.009 362 902 404 300 8;
  • 47) 0.009 362 902 404 300 8 × 2 = 0 + 0.018 725 804 808 601 6;
  • 48) 0.018 725 804 808 601 6 × 2 = 0 + 0.037 451 609 617 203 2;
  • 49) 0.037 451 609 617 203 2 × 2 = 0 + 0.074 903 219 234 406 4;
  • 50) 0.074 903 219 234 406 4 × 2 = 0 + 0.149 806 438 468 812 8;
  • 51) 0.149 806 438 468 812 8 × 2 = 0 + 0.299 612 876 937 625 6;
  • 52) 0.299 612 876 937 625 6 × 2 = 0 + 0.599 225 753 875 251 2;
  • 53) 0.599 225 753 875 251 2 × 2 = 1 + 0.198 451 507 750 502 4;
  • 54) 0.198 451 507 750 502 4 × 2 = 0 + 0.396 903 015 501 004 8;
  • 55) 0.396 903 015 501 004 8 × 2 = 0 + 0.793 806 031 002 009 6;
  • 56) 0.793 806 031 002 009 6 × 2 = 1 + 0.587 612 062 004 019 2;
  • 57) 0.587 612 062 004 019 2 × 2 = 1 + 0.175 224 124 008 038 4;

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...)


5. 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:


0.000 000 000 106 524 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0111 0101 0010 0000 0000 1001 1(2)


6. Positive number before normalization:

0.000 000 000 106 524 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0111 0101 0010 0000 0000 1001 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 34 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 106 524 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0111 0101 0010 0000 0000 1001 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0111 0101 0010 0000 0000 1001 1(2) × 20 =

1.1101 0100 1000 0000 0010 011(2) × 2-34


8. 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 1 (a negative number)

Exponent (unadjusted): -34

Mantissa (not normalized):
1.1101 0100 1000 0000 0010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-34 + 2(8-1) - 1 =

(-34 + 127)(10) =

93(10)


10. 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;
  • 93 ÷ 2 = 46 + 1;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

93(10) =

0101 1101(2)


12. 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. 110 1010 0100 0000 0001 0011 =

110 1010 0100 0000 0001 0011


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

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

Exponent (8 bits) =
0101 1101

Mantissa (23 bits) =
110 1010 0100 0000 0001 0011


The base ten decimal number -0.000 000 000 106 524 7 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0101 1101 - 110 1010 0100 0000 0001 0011

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

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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