-0.000 000 000 742 147 675 55 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 55(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
-0.000 000 000 742 147 675 55(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. Start with the positive version of the number:

|-0.000 000 000 742 147 675 55| = 0.000 000 000 742 147 675 55


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: 0.000 000 000 742 147 675 55.

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) 0.000 000 000 742 147 675 55 × 2 = 0 + 0.000 000 001 484 295 351 1;
  • 2) 0.000 000 001 484 295 351 1 × 2 = 0 + 0.000 000 002 968 590 702 2;
  • 3) 0.000 000 002 968 590 702 2 × 2 = 0 + 0.000 000 005 937 181 404 4;
  • 4) 0.000 000 005 937 181 404 4 × 2 = 0 + 0.000 000 011 874 362 808 8;
  • 5) 0.000 000 011 874 362 808 8 × 2 = 0 + 0.000 000 023 748 725 617 6;
  • 6) 0.000 000 023 748 725 617 6 × 2 = 0 + 0.000 000 047 497 451 235 2;
  • 7) 0.000 000 047 497 451 235 2 × 2 = 0 + 0.000 000 094 994 902 470 4;
  • 8) 0.000 000 094 994 902 470 4 × 2 = 0 + 0.000 000 189 989 804 940 8;
  • 9) 0.000 000 189 989 804 940 8 × 2 = 0 + 0.000 000 379 979 609 881 6;
  • 10) 0.000 000 379 979 609 881 6 × 2 = 0 + 0.000 000 759 959 219 763 2;
  • 11) 0.000 000 759 959 219 763 2 × 2 = 0 + 0.000 001 519 918 439 526 4;
  • 12) 0.000 001 519 918 439 526 4 × 2 = 0 + 0.000 003 039 836 879 052 8;
  • 13) 0.000 003 039 836 879 052 8 × 2 = 0 + 0.000 006 079 673 758 105 6;
  • 14) 0.000 006 079 673 758 105 6 × 2 = 0 + 0.000 012 159 347 516 211 2;
  • 15) 0.000 012 159 347 516 211 2 × 2 = 0 + 0.000 024 318 695 032 422 4;
  • 16) 0.000 024 318 695 032 422 4 × 2 = 0 + 0.000 048 637 390 064 844 8;
  • 17) 0.000 048 637 390 064 844 8 × 2 = 0 + 0.000 097 274 780 129 689 6;
  • 18) 0.000 097 274 780 129 689 6 × 2 = 0 + 0.000 194 549 560 259 379 2;
  • 19) 0.000 194 549 560 259 379 2 × 2 = 0 + 0.000 389 099 120 518 758 4;
  • 20) 0.000 389 099 120 518 758 4 × 2 = 0 + 0.000 778 198 241 037 516 8;
  • 21) 0.000 778 198 241 037 516 8 × 2 = 0 + 0.001 556 396 482 075 033 6;
  • 22) 0.001 556 396 482 075 033 6 × 2 = 0 + 0.003 112 792 964 150 067 2;
  • 23) 0.003 112 792 964 150 067 2 × 2 = 0 + 0.006 225 585 928 300 134 4;
  • 24) 0.006 225 585 928 300 134 4 × 2 = 0 + 0.012 451 171 856 600 268 8;
  • 25) 0.012 451 171 856 600 268 8 × 2 = 0 + 0.024 902 343 713 200 537 6;
  • 26) 0.024 902 343 713 200 537 6 × 2 = 0 + 0.049 804 687 426 401 075 2;
  • 27) 0.049 804 687 426 401 075 2 × 2 = 0 + 0.099 609 374 852 802 150 4;
  • 28) 0.099 609 374 852 802 150 4 × 2 = 0 + 0.199 218 749 705 604 300 8;
  • 29) 0.199 218 749 705 604 300 8 × 2 = 0 + 0.398 437 499 411 208 601 6;
  • 30) 0.398 437 499 411 208 601 6 × 2 = 0 + 0.796 874 998 822 417 203 2;
  • 31) 0.796 874 998 822 417 203 2 × 2 = 1 + 0.593 749 997 644 834 406 4;
  • 32) 0.593 749 997 644 834 406 4 × 2 = 1 + 0.187 499 995 289 668 812 8;
  • 33) 0.187 499 995 289 668 812 8 × 2 = 0 + 0.374 999 990 579 337 625 6;
  • 34) 0.374 999 990 579 337 625 6 × 2 = 0 + 0.749 999 981 158 675 251 2;
  • 35) 0.749 999 981 158 675 251 2 × 2 = 1 + 0.499 999 962 317 350 502 4;
  • 36) 0.499 999 962 317 350 502 4 × 2 = 0 + 0.999 999 924 634 701 004 8;
  • 37) 0.999 999 924 634 701 004 8 × 2 = 1 + 0.999 999 849 269 402 009 6;
  • 38) 0.999 999 849 269 402 009 6 × 2 = 1 + 0.999 999 698 538 804 019 2;
  • 39) 0.999 999 698 538 804 019 2 × 2 = 1 + 0.999 999 397 077 608 038 4;
  • 40) 0.999 999 397 077 608 038 4 × 2 = 1 + 0.999 998 794 155 216 076 8;
  • 41) 0.999 998 794 155 216 076 8 × 2 = 1 + 0.999 997 588 310 432 153 6;
  • 42) 0.999 997 588 310 432 153 6 × 2 = 1 + 0.999 995 176 620 864 307 2;
  • 43) 0.999 995 176 620 864 307 2 × 2 = 1 + 0.999 990 353 241 728 614 4;
  • 44) 0.999 990 353 241 728 614 4 × 2 = 1 + 0.999 980 706 483 457 228 8;
  • 45) 0.999 980 706 483 457 228 8 × 2 = 1 + 0.999 961 412 966 914 457 6;
  • 46) 0.999 961 412 966 914 457 6 × 2 = 1 + 0.999 922 825 933 828 915 2;
  • 47) 0.999 922 825 933 828 915 2 × 2 = 1 + 0.999 845 651 867 657 830 4;
  • 48) 0.999 845 651 867 657 830 4 × 2 = 1 + 0.999 691 303 735 315 660 8;
  • 49) 0.999 691 303 735 315 660 8 × 2 = 1 + 0.999 382 607 470 631 321 6;
  • 50) 0.999 382 607 470 631 321 6 × 2 = 1 + 0.998 765 214 941 262 643 2;
  • 51) 0.998 765 214 941 262 643 2 × 2 = 1 + 0.997 530 429 882 525 286 4;
  • 52) 0.997 530 429 882 525 286 4 × 2 = 1 + 0.995 060 859 765 050 572 8;
  • 53) 0.995 060 859 765 050 572 8 × 2 = 1 + 0.990 121 719 530 101 145 6;
  • 54) 0.990 121 719 530 101 145 6 × 2 = 1 + 0.980 243 439 060 202 291 2;

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).


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:


0.000 000 000 742 147 675 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 675 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 675 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 675 55 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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