-0.000 000 000 742 147 676 646 737 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 737 5(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
-0.000 000 000 742 147 676 646 737 5(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:

|-0.000 000 000 742 147 676 646 737 5| = 0.000 000 000 742 147 676 646 737 5


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: 0.000 000 000 742 147 676 646 737 5.

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) 0.000 000 000 742 147 676 646 737 5 × 2 = 0 + 0.000 000 001 484 295 353 293 475;
  • 2) 0.000 000 001 484 295 353 293 475 × 2 = 0 + 0.000 000 002 968 590 706 586 95;
  • 3) 0.000 000 002 968 590 706 586 95 × 2 = 0 + 0.000 000 005 937 181 413 173 9;
  • 4) 0.000 000 005 937 181 413 173 9 × 2 = 0 + 0.000 000 011 874 362 826 347 8;
  • 5) 0.000 000 011 874 362 826 347 8 × 2 = 0 + 0.000 000 023 748 725 652 695 6;
  • 6) 0.000 000 023 748 725 652 695 6 × 2 = 0 + 0.000 000 047 497 451 305 391 2;
  • 7) 0.000 000 047 497 451 305 391 2 × 2 = 0 + 0.000 000 094 994 902 610 782 4;
  • 8) 0.000 000 094 994 902 610 782 4 × 2 = 0 + 0.000 000 189 989 805 221 564 8;
  • 9) 0.000 000 189 989 805 221 564 8 × 2 = 0 + 0.000 000 379 979 610 443 129 6;
  • 10) 0.000 000 379 979 610 443 129 6 × 2 = 0 + 0.000 000 759 959 220 886 259 2;
  • 11) 0.000 000 759 959 220 886 259 2 × 2 = 0 + 0.000 001 519 918 441 772 518 4;
  • 12) 0.000 001 519 918 441 772 518 4 × 2 = 0 + 0.000 003 039 836 883 545 036 8;
  • 13) 0.000 003 039 836 883 545 036 8 × 2 = 0 + 0.000 006 079 673 767 090 073 6;
  • 14) 0.000 006 079 673 767 090 073 6 × 2 = 0 + 0.000 012 159 347 534 180 147 2;
  • 15) 0.000 012 159 347 534 180 147 2 × 2 = 0 + 0.000 024 318 695 068 360 294 4;
  • 16) 0.000 024 318 695 068 360 294 4 × 2 = 0 + 0.000 048 637 390 136 720 588 8;
  • 17) 0.000 048 637 390 136 720 588 8 × 2 = 0 + 0.000 097 274 780 273 441 177 6;
  • 18) 0.000 097 274 780 273 441 177 6 × 2 = 0 + 0.000 194 549 560 546 882 355 2;
  • 19) 0.000 194 549 560 546 882 355 2 × 2 = 0 + 0.000 389 099 121 093 764 710 4;
  • 20) 0.000 389 099 121 093 764 710 4 × 2 = 0 + 0.000 778 198 242 187 529 420 8;
  • 21) 0.000 778 198 242 187 529 420 8 × 2 = 0 + 0.001 556 396 484 375 058 841 6;
  • 22) 0.001 556 396 484 375 058 841 6 × 2 = 0 + 0.003 112 792 968 750 117 683 2;
  • 23) 0.003 112 792 968 750 117 683 2 × 2 = 0 + 0.006 225 585 937 500 235 366 4;
  • 24) 0.006 225 585 937 500 235 366 4 × 2 = 0 + 0.012 451 171 875 000 470 732 8;
  • 25) 0.012 451 171 875 000 470 732 8 × 2 = 0 + 0.024 902 343 750 000 941 465 6;
  • 26) 0.024 902 343 750 000 941 465 6 × 2 = 0 + 0.049 804 687 500 001 882 931 2;
  • 27) 0.049 804 687 500 001 882 931 2 × 2 = 0 + 0.099 609 375 000 003 765 862 4;
  • 28) 0.099 609 375 000 003 765 862 4 × 2 = 0 + 0.199 218 750 000 007 531 724 8;
  • 29) 0.199 218 750 000 007 531 724 8 × 2 = 0 + 0.398 437 500 000 015 063 449 6;
  • 30) 0.398 437 500 000 015 063 449 6 × 2 = 0 + 0.796 875 000 000 030 126 899 2;
  • 31) 0.796 875 000 000 030 126 899 2 × 2 = 1 + 0.593 750 000 000 060 253 798 4;
  • 32) 0.593 750 000 000 060 253 798 4 × 2 = 1 + 0.187 500 000 000 120 507 596 8;
  • 33) 0.187 500 000 000 120 507 596 8 × 2 = 0 + 0.375 000 000 000 241 015 193 6;
  • 34) 0.375 000 000 000 241 015 193 6 × 2 = 0 + 0.750 000 000 000 482 030 387 2;
  • 35) 0.750 000 000 000 482 030 387 2 × 2 = 1 + 0.500 000 000 000 964 060 774 4;
  • 36) 0.500 000 000 000 964 060 774 4 × 2 = 1 + 0.000 000 000 001 928 121 548 8;
  • 37) 0.000 000 000 001 928 121 548 8 × 2 = 0 + 0.000 000 000 003 856 243 097 6;
  • 38) 0.000 000 000 003 856 243 097 6 × 2 = 0 + 0.000 000 000 007 712 486 195 2;
  • 39) 0.000 000 000 007 712 486 195 2 × 2 = 0 + 0.000 000 000 015 424 972 390 4;
  • 40) 0.000 000 000 015 424 972 390 4 × 2 = 0 + 0.000 000 000 030 849 944 780 8;
  • 41) 0.000 000 000 030 849 944 780 8 × 2 = 0 + 0.000 000 000 061 699 889 561 6;
  • 42) 0.000 000 000 061 699 889 561 6 × 2 = 0 + 0.000 000 000 123 399 779 123 2;
  • 43) 0.000 000 000 123 399 779 123 2 × 2 = 0 + 0.000 000 000 246 799 558 246 4;
  • 44) 0.000 000 000 246 799 558 246 4 × 2 = 0 + 0.000 000 000 493 599 116 492 8;
  • 45) 0.000 000 000 493 599 116 492 8 × 2 = 0 + 0.000 000 000 987 198 232 985 6;
  • 46) 0.000 000 000 987 198 232 985 6 × 2 = 0 + 0.000 000 001 974 396 465 971 2;
  • 47) 0.000 000 001 974 396 465 971 2 × 2 = 0 + 0.000 000 003 948 792 931 942 4;
  • 48) 0.000 000 003 948 792 931 942 4 × 2 = 0 + 0.000 000 007 897 585 863 884 8;
  • 49) 0.000 000 007 897 585 863 884 8 × 2 = 0 + 0.000 000 015 795 171 727 769 6;
  • 50) 0.000 000 015 795 171 727 769 6 × 2 = 0 + 0.000 000 031 590 343 455 539 2;
  • 51) 0.000 000 031 590 343 455 539 2 × 2 = 0 + 0.000 000 063 180 686 911 078 4;
  • 52) 0.000 000 063 180 686 911 078 4 × 2 = 0 + 0.000 000 126 361 373 822 156 8;
  • 53) 0.000 000 126 361 373 822 156 8 × 2 = 0 + 0.000 000 252 722 747 644 313 6;
  • 54) 0.000 000 252 722 747 644 313 6 × 2 = 0 + 0.000 000 505 445 495 288 627 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:


0.000 000 000 742 147 676 646 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 676 646 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 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 1000 0000 0000 0000 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 737 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 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