-0.000 000 000 742 139 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 139 7(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
-0.000 000 000 742 139 7(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:

|-0.000 000 000 742 139 7| = 0.000 000 000 742 139 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: 0.000 000 000 742 139 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) 0.000 000 000 742 139 7 × 2 = 0 + 0.000 000 001 484 279 4;
  • 2) 0.000 000 001 484 279 4 × 2 = 0 + 0.000 000 002 968 558 8;
  • 3) 0.000 000 002 968 558 8 × 2 = 0 + 0.000 000 005 937 117 6;
  • 4) 0.000 000 005 937 117 6 × 2 = 0 + 0.000 000 011 874 235 2;
  • 5) 0.000 000 011 874 235 2 × 2 = 0 + 0.000 000 023 748 470 4;
  • 6) 0.000 000 023 748 470 4 × 2 = 0 + 0.000 000 047 496 940 8;
  • 7) 0.000 000 047 496 940 8 × 2 = 0 + 0.000 000 094 993 881 6;
  • 8) 0.000 000 094 993 881 6 × 2 = 0 + 0.000 000 189 987 763 2;
  • 9) 0.000 000 189 987 763 2 × 2 = 0 + 0.000 000 379 975 526 4;
  • 10) 0.000 000 379 975 526 4 × 2 = 0 + 0.000 000 759 951 052 8;
  • 11) 0.000 000 759 951 052 8 × 2 = 0 + 0.000 001 519 902 105 6;
  • 12) 0.000 001 519 902 105 6 × 2 = 0 + 0.000 003 039 804 211 2;
  • 13) 0.000 003 039 804 211 2 × 2 = 0 + 0.000 006 079 608 422 4;
  • 14) 0.000 006 079 608 422 4 × 2 = 0 + 0.000 012 159 216 844 8;
  • 15) 0.000 012 159 216 844 8 × 2 = 0 + 0.000 024 318 433 689 6;
  • 16) 0.000 024 318 433 689 6 × 2 = 0 + 0.000 048 636 867 379 2;
  • 17) 0.000 048 636 867 379 2 × 2 = 0 + 0.000 097 273 734 758 4;
  • 18) 0.000 097 273 734 758 4 × 2 = 0 + 0.000 194 547 469 516 8;
  • 19) 0.000 194 547 469 516 8 × 2 = 0 + 0.000 389 094 939 033 6;
  • 20) 0.000 389 094 939 033 6 × 2 = 0 + 0.000 778 189 878 067 2;
  • 21) 0.000 778 189 878 067 2 × 2 = 0 + 0.001 556 379 756 134 4;
  • 22) 0.001 556 379 756 134 4 × 2 = 0 + 0.003 112 759 512 268 8;
  • 23) 0.003 112 759 512 268 8 × 2 = 0 + 0.006 225 519 024 537 6;
  • 24) 0.006 225 519 024 537 6 × 2 = 0 + 0.012 451 038 049 075 2;
  • 25) 0.012 451 038 049 075 2 × 2 = 0 + 0.024 902 076 098 150 4;
  • 26) 0.024 902 076 098 150 4 × 2 = 0 + 0.049 804 152 196 300 8;
  • 27) 0.049 804 152 196 300 8 × 2 = 0 + 0.099 608 304 392 601 6;
  • 28) 0.099 608 304 392 601 6 × 2 = 0 + 0.199 216 608 785 203 2;
  • 29) 0.199 216 608 785 203 2 × 2 = 0 + 0.398 433 217 570 406 4;
  • 30) 0.398 433 217 570 406 4 × 2 = 0 + 0.796 866 435 140 812 8;
  • 31) 0.796 866 435 140 812 8 × 2 = 1 + 0.593 732 870 281 625 6;
  • 32) 0.593 732 870 281 625 6 × 2 = 1 + 0.187 465 740 563 251 2;
  • 33) 0.187 465 740 563 251 2 × 2 = 0 + 0.374 931 481 126 502 4;
  • 34) 0.374 931 481 126 502 4 × 2 = 0 + 0.749 862 962 253 004 8;
  • 35) 0.749 862 962 253 004 8 × 2 = 1 + 0.499 725 924 506 009 6;
  • 36) 0.499 725 924 506 009 6 × 2 = 0 + 0.999 451 849 012 019 2;
  • 37) 0.999 451 849 012 019 2 × 2 = 1 + 0.998 903 698 024 038 4;
  • 38) 0.998 903 698 024 038 4 × 2 = 1 + 0.997 807 396 048 076 8;
  • 39) 0.997 807 396 048 076 8 × 2 = 1 + 0.995 614 792 096 153 6;
  • 40) 0.995 614 792 096 153 6 × 2 = 1 + 0.991 229 584 192 307 2;
  • 41) 0.991 229 584 192 307 2 × 2 = 1 + 0.982 459 168 384 614 4;
  • 42) 0.982 459 168 384 614 4 × 2 = 1 + 0.964 918 336 769 228 8;
  • 43) 0.964 918 336 769 228 8 × 2 = 1 + 0.929 836 673 538 457 6;
  • 44) 0.929 836 673 538 457 6 × 2 = 1 + 0.859 673 347 076 915 2;
  • 45) 0.859 673 347 076 915 2 × 2 = 1 + 0.719 346 694 153 830 4;
  • 46) 0.719 346 694 153 830 4 × 2 = 1 + 0.438 693 388 307 660 8;
  • 47) 0.438 693 388 307 660 8 × 2 = 0 + 0.877 386 776 615 321 6;
  • 48) 0.877 386 776 615 321 6 × 2 = 1 + 0.754 773 553 230 643 2;
  • 49) 0.754 773 553 230 643 2 × 2 = 1 + 0.509 547 106 461 286 4;
  • 50) 0.509 547 106 461 286 4 × 2 = 1 + 0.019 094 212 922 572 8;
  • 51) 0.019 094 212 922 572 8 × 2 = 0 + 0.038 188 425 845 145 6;
  • 52) 0.038 188 425 845 145 6 × 2 = 0 + 0.076 376 851 690 291 2;
  • 53) 0.076 376 851 690 291 2 × 2 = 0 + 0.152 753 703 380 582 4;
  • 54) 0.152 753 703 380 582 4 × 2 = 0 + 0.305 507 406 761 164 8;

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:


0.000 000 000 742 139 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1100 00(2)

6. Positive number before normalization:

0.000 000 000 742 139 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1100 00(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:


0.000 000 000 742 139 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1100 00(2) × 20 =

1.1001 0111 1111 1110 1110 000(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 1110 1110 000


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 0111 0000 =

100 1011 1111 1111 0111 0000


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 0111 0000


Decimal number -0.000 000 000 742 139 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0111 0000

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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