-0.000 000 000 742 147 676 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 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
-0.000 000 000 742 147 676 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:

|-0.000 000 000 742 147 676 7| = 0.000 000 000 742 147 676 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: 0.000 000 000 742 147 676 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) 0.000 000 000 742 147 676 7 × 2 = 0 + 0.000 000 001 484 295 353 4;
  • 2) 0.000 000 001 484 295 353 4 × 2 = 0 + 0.000 000 002 968 590 706 8;
  • 3) 0.000 000 002 968 590 706 8 × 2 = 0 + 0.000 000 005 937 181 413 6;
  • 4) 0.000 000 005 937 181 413 6 × 2 = 0 + 0.000 000 011 874 362 827 2;
  • 5) 0.000 000 011 874 362 827 2 × 2 = 0 + 0.000 000 023 748 725 654 4;
  • 6) 0.000 000 023 748 725 654 4 × 2 = 0 + 0.000 000 047 497 451 308 8;
  • 7) 0.000 000 047 497 451 308 8 × 2 = 0 + 0.000 000 094 994 902 617 6;
  • 8) 0.000 000 094 994 902 617 6 × 2 = 0 + 0.000 000 189 989 805 235 2;
  • 9) 0.000 000 189 989 805 235 2 × 2 = 0 + 0.000 000 379 979 610 470 4;
  • 10) 0.000 000 379 979 610 470 4 × 2 = 0 + 0.000 000 759 959 220 940 8;
  • 11) 0.000 000 759 959 220 940 8 × 2 = 0 + 0.000 001 519 918 441 881 6;
  • 12) 0.000 001 519 918 441 881 6 × 2 = 0 + 0.000 003 039 836 883 763 2;
  • 13) 0.000 003 039 836 883 763 2 × 2 = 0 + 0.000 006 079 673 767 526 4;
  • 14) 0.000 006 079 673 767 526 4 × 2 = 0 + 0.000 012 159 347 535 052 8;
  • 15) 0.000 012 159 347 535 052 8 × 2 = 0 + 0.000 024 318 695 070 105 6;
  • 16) 0.000 024 318 695 070 105 6 × 2 = 0 + 0.000 048 637 390 140 211 2;
  • 17) 0.000 048 637 390 140 211 2 × 2 = 0 + 0.000 097 274 780 280 422 4;
  • 18) 0.000 097 274 780 280 422 4 × 2 = 0 + 0.000 194 549 560 560 844 8;
  • 19) 0.000 194 549 560 560 844 8 × 2 = 0 + 0.000 389 099 121 121 689 6;
  • 20) 0.000 389 099 121 121 689 6 × 2 = 0 + 0.000 778 198 242 243 379 2;
  • 21) 0.000 778 198 242 243 379 2 × 2 = 0 + 0.001 556 396 484 486 758 4;
  • 22) 0.001 556 396 484 486 758 4 × 2 = 0 + 0.003 112 792 968 973 516 8;
  • 23) 0.003 112 792 968 973 516 8 × 2 = 0 + 0.006 225 585 937 947 033 6;
  • 24) 0.006 225 585 937 947 033 6 × 2 = 0 + 0.012 451 171 875 894 067 2;
  • 25) 0.012 451 171 875 894 067 2 × 2 = 0 + 0.024 902 343 751 788 134 4;
  • 26) 0.024 902 343 751 788 134 4 × 2 = 0 + 0.049 804 687 503 576 268 8;
  • 27) 0.049 804 687 503 576 268 8 × 2 = 0 + 0.099 609 375 007 152 537 6;
  • 28) 0.099 609 375 007 152 537 6 × 2 = 0 + 0.199 218 750 014 305 075 2;
  • 29) 0.199 218 750 014 305 075 2 × 2 = 0 + 0.398 437 500 028 610 150 4;
  • 30) 0.398 437 500 028 610 150 4 × 2 = 0 + 0.796 875 000 057 220 300 8;
  • 31) 0.796 875 000 057 220 300 8 × 2 = 1 + 0.593 750 000 114 440 601 6;
  • 32) 0.593 750 000 114 440 601 6 × 2 = 1 + 0.187 500 000 228 881 203 2;
  • 33) 0.187 500 000 228 881 203 2 × 2 = 0 + 0.375 000 000 457 762 406 4;
  • 34) 0.375 000 000 457 762 406 4 × 2 = 0 + 0.750 000 000 915 524 812 8;
  • 35) 0.750 000 000 915 524 812 8 × 2 = 1 + 0.500 000 001 831 049 625 6;
  • 36) 0.500 000 001 831 049 625 6 × 2 = 1 + 0.000 000 003 662 099 251 2;
  • 37) 0.000 000 003 662 099 251 2 × 2 = 0 + 0.000 000 007 324 198 502 4;
  • 38) 0.000 000 007 324 198 502 4 × 2 = 0 + 0.000 000 014 648 397 004 8;
  • 39) 0.000 000 014 648 397 004 8 × 2 = 0 + 0.000 000 029 296 794 009 6;
  • 40) 0.000 000 029 296 794 009 6 × 2 = 0 + 0.000 000 058 593 588 019 2;
  • 41) 0.000 000 058 593 588 019 2 × 2 = 0 + 0.000 000 117 187 176 038 4;
  • 42) 0.000 000 117 187 176 038 4 × 2 = 0 + 0.000 000 234 374 352 076 8;
  • 43) 0.000 000 234 374 352 076 8 × 2 = 0 + 0.000 000 468 748 704 153 6;
  • 44) 0.000 000 468 748 704 153 6 × 2 = 0 + 0.000 000 937 497 408 307 2;
  • 45) 0.000 000 937 497 408 307 2 × 2 = 0 + 0.000 001 874 994 816 614 4;
  • 46) 0.000 001 874 994 816 614 4 × 2 = 0 + 0.000 003 749 989 633 228 8;
  • 47) 0.000 003 749 989 633 228 8 × 2 = 0 + 0.000 007 499 979 266 457 6;
  • 48) 0.000 007 499 979 266 457 6 × 2 = 0 + 0.000 014 999 958 532 915 2;
  • 49) 0.000 014 999 958 532 915 2 × 2 = 0 + 0.000 029 999 917 065 830 4;
  • 50) 0.000 029 999 917 065 830 4 × 2 = 0 + 0.000 059 999 834 131 660 8;
  • 51) 0.000 059 999 834 131 660 8 × 2 = 0 + 0.000 119 999 668 263 321 6;
  • 52) 0.000 119 999 668 263 321 6 × 2 = 0 + 0.000 239 999 336 526 643 2;
  • 53) 0.000 239 999 336 526 643 2 × 2 = 0 + 0.000 479 998 673 053 286 4;
  • 54) 0.000 479 998 673 053 286 4 × 2 = 0 + 0.000 959 997 346 106 572 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:


0.000 000 000 742 147 676 7(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 7(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 7(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 7 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