-0.000 000 000 742 149 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 79(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
-0.000 000 000 742 149 79(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:

|-0.000 000 000 742 149 79| = 0.000 000 000 742 149 79


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: 0.000 000 000 742 149 79.

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) 0.000 000 000 742 149 79 × 2 = 0 + 0.000 000 001 484 299 58;
  • 2) 0.000 000 001 484 299 58 × 2 = 0 + 0.000 000 002 968 599 16;
  • 3) 0.000 000 002 968 599 16 × 2 = 0 + 0.000 000 005 937 198 32;
  • 4) 0.000 000 005 937 198 32 × 2 = 0 + 0.000 000 011 874 396 64;
  • 5) 0.000 000 011 874 396 64 × 2 = 0 + 0.000 000 023 748 793 28;
  • 6) 0.000 000 023 748 793 28 × 2 = 0 + 0.000 000 047 497 586 56;
  • 7) 0.000 000 047 497 586 56 × 2 = 0 + 0.000 000 094 995 173 12;
  • 8) 0.000 000 094 995 173 12 × 2 = 0 + 0.000 000 189 990 346 24;
  • 9) 0.000 000 189 990 346 24 × 2 = 0 + 0.000 000 379 980 692 48;
  • 10) 0.000 000 379 980 692 48 × 2 = 0 + 0.000 000 759 961 384 96;
  • 11) 0.000 000 759 961 384 96 × 2 = 0 + 0.000 001 519 922 769 92;
  • 12) 0.000 001 519 922 769 92 × 2 = 0 + 0.000 003 039 845 539 84;
  • 13) 0.000 003 039 845 539 84 × 2 = 0 + 0.000 006 079 691 079 68;
  • 14) 0.000 006 079 691 079 68 × 2 = 0 + 0.000 012 159 382 159 36;
  • 15) 0.000 012 159 382 159 36 × 2 = 0 + 0.000 024 318 764 318 72;
  • 16) 0.000 024 318 764 318 72 × 2 = 0 + 0.000 048 637 528 637 44;
  • 17) 0.000 048 637 528 637 44 × 2 = 0 + 0.000 097 275 057 274 88;
  • 18) 0.000 097 275 057 274 88 × 2 = 0 + 0.000 194 550 114 549 76;
  • 19) 0.000 194 550 114 549 76 × 2 = 0 + 0.000 389 100 229 099 52;
  • 20) 0.000 389 100 229 099 52 × 2 = 0 + 0.000 778 200 458 199 04;
  • 21) 0.000 778 200 458 199 04 × 2 = 0 + 0.001 556 400 916 398 08;
  • 22) 0.001 556 400 916 398 08 × 2 = 0 + 0.003 112 801 832 796 16;
  • 23) 0.003 112 801 832 796 16 × 2 = 0 + 0.006 225 603 665 592 32;
  • 24) 0.006 225 603 665 592 32 × 2 = 0 + 0.012 451 207 331 184 64;
  • 25) 0.012 451 207 331 184 64 × 2 = 0 + 0.024 902 414 662 369 28;
  • 26) 0.024 902 414 662 369 28 × 2 = 0 + 0.049 804 829 324 738 56;
  • 27) 0.049 804 829 324 738 56 × 2 = 0 + 0.099 609 658 649 477 12;
  • 28) 0.099 609 658 649 477 12 × 2 = 0 + 0.199 219 317 298 954 24;
  • 29) 0.199 219 317 298 954 24 × 2 = 0 + 0.398 438 634 597 908 48;
  • 30) 0.398 438 634 597 908 48 × 2 = 0 + 0.796 877 269 195 816 96;
  • 31) 0.796 877 269 195 816 96 × 2 = 1 + 0.593 754 538 391 633 92;
  • 32) 0.593 754 538 391 633 92 × 2 = 1 + 0.187 509 076 783 267 84;
  • 33) 0.187 509 076 783 267 84 × 2 = 0 + 0.375 018 153 566 535 68;
  • 34) 0.375 018 153 566 535 68 × 2 = 0 + 0.750 036 307 133 071 36;
  • 35) 0.750 036 307 133 071 36 × 2 = 1 + 0.500 072 614 266 142 72;
  • 36) 0.500 072 614 266 142 72 × 2 = 1 + 0.000 145 228 532 285 44;
  • 37) 0.000 145 228 532 285 44 × 2 = 0 + 0.000 290 457 064 570 88;
  • 38) 0.000 290 457 064 570 88 × 2 = 0 + 0.000 580 914 129 141 76;
  • 39) 0.000 580 914 129 141 76 × 2 = 0 + 0.001 161 828 258 283 52;
  • 40) 0.001 161 828 258 283 52 × 2 = 0 + 0.002 323 656 516 567 04;
  • 41) 0.002 323 656 516 567 04 × 2 = 0 + 0.004 647 313 033 134 08;
  • 42) 0.004 647 313 033 134 08 × 2 = 0 + 0.009 294 626 066 268 16;
  • 43) 0.009 294 626 066 268 16 × 2 = 0 + 0.018 589 252 132 536 32;
  • 44) 0.018 589 252 132 536 32 × 2 = 0 + 0.037 178 504 265 072 64;
  • 45) 0.037 178 504 265 072 64 × 2 = 0 + 0.074 357 008 530 145 28;
  • 46) 0.074 357 008 530 145 28 × 2 = 0 + 0.148 714 017 060 290 56;
  • 47) 0.148 714 017 060 290 56 × 2 = 0 + 0.297 428 034 120 581 12;
  • 48) 0.297 428 034 120 581 12 × 2 = 0 + 0.594 856 068 241 162 24;
  • 49) 0.594 856 068 241 162 24 × 2 = 1 + 0.189 712 136 482 324 48;
  • 50) 0.189 712 136 482 324 48 × 2 = 0 + 0.379 424 272 964 648 96;
  • 51) 0.379 424 272 964 648 96 × 2 = 0 + 0.758 848 545 929 297 92;
  • 52) 0.758 848 545 929 297 92 × 2 = 1 + 0.517 697 091 858 595 84;
  • 53) 0.517 697 091 858 595 84 × 2 = 1 + 0.035 394 183 717 191 68;
  • 54) 0.035 394 183 717 191 68 × 2 = 0 + 0.070 788 367 434 383 36;

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:


0.000 000 000 742 149 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 10(2)

6. Positive number before normalization:

0.000 000 000 742 149 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 10(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:


0.000 000 000 742 149 79(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1001 10(2) × 20 =

1.1001 1000 0000 0000 0100 110(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 0100 110


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 0010 0110 =

100 1100 0000 0000 0010 0110


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 0010 0110


Decimal number -0.000 000 000 742 149 79 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 0110

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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