-0.000 000 000 742 149 51 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 51(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 51(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 51| = 0.000 000 000 742 149 51


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 51.

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 51 × 2 = 0 + 0.000 000 001 484 299 02;
  • 2) 0.000 000 001 484 299 02 × 2 = 0 + 0.000 000 002 968 598 04;
  • 3) 0.000 000 002 968 598 04 × 2 = 0 + 0.000 000 005 937 196 08;
  • 4) 0.000 000 005 937 196 08 × 2 = 0 + 0.000 000 011 874 392 16;
  • 5) 0.000 000 011 874 392 16 × 2 = 0 + 0.000 000 023 748 784 32;
  • 6) 0.000 000 023 748 784 32 × 2 = 0 + 0.000 000 047 497 568 64;
  • 7) 0.000 000 047 497 568 64 × 2 = 0 + 0.000 000 094 995 137 28;
  • 8) 0.000 000 094 995 137 28 × 2 = 0 + 0.000 000 189 990 274 56;
  • 9) 0.000 000 189 990 274 56 × 2 = 0 + 0.000 000 379 980 549 12;
  • 10) 0.000 000 379 980 549 12 × 2 = 0 + 0.000 000 759 961 098 24;
  • 11) 0.000 000 759 961 098 24 × 2 = 0 + 0.000 001 519 922 196 48;
  • 12) 0.000 001 519 922 196 48 × 2 = 0 + 0.000 003 039 844 392 96;
  • 13) 0.000 003 039 844 392 96 × 2 = 0 + 0.000 006 079 688 785 92;
  • 14) 0.000 006 079 688 785 92 × 2 = 0 + 0.000 012 159 377 571 84;
  • 15) 0.000 012 159 377 571 84 × 2 = 0 + 0.000 024 318 755 143 68;
  • 16) 0.000 024 318 755 143 68 × 2 = 0 + 0.000 048 637 510 287 36;
  • 17) 0.000 048 637 510 287 36 × 2 = 0 + 0.000 097 275 020 574 72;
  • 18) 0.000 097 275 020 574 72 × 2 = 0 + 0.000 194 550 041 149 44;
  • 19) 0.000 194 550 041 149 44 × 2 = 0 + 0.000 389 100 082 298 88;
  • 20) 0.000 389 100 082 298 88 × 2 = 0 + 0.000 778 200 164 597 76;
  • 21) 0.000 778 200 164 597 76 × 2 = 0 + 0.001 556 400 329 195 52;
  • 22) 0.001 556 400 329 195 52 × 2 = 0 + 0.003 112 800 658 391 04;
  • 23) 0.003 112 800 658 391 04 × 2 = 0 + 0.006 225 601 316 782 08;
  • 24) 0.006 225 601 316 782 08 × 2 = 0 + 0.012 451 202 633 564 16;
  • 25) 0.012 451 202 633 564 16 × 2 = 0 + 0.024 902 405 267 128 32;
  • 26) 0.024 902 405 267 128 32 × 2 = 0 + 0.049 804 810 534 256 64;
  • 27) 0.049 804 810 534 256 64 × 2 = 0 + 0.099 609 621 068 513 28;
  • 28) 0.099 609 621 068 513 28 × 2 = 0 + 0.199 219 242 137 026 56;
  • 29) 0.199 219 242 137 026 56 × 2 = 0 + 0.398 438 484 274 053 12;
  • 30) 0.398 438 484 274 053 12 × 2 = 0 + 0.796 876 968 548 106 24;
  • 31) 0.796 876 968 548 106 24 × 2 = 1 + 0.593 753 937 096 212 48;
  • 32) 0.593 753 937 096 212 48 × 2 = 1 + 0.187 507 874 192 424 96;
  • 33) 0.187 507 874 192 424 96 × 2 = 0 + 0.375 015 748 384 849 92;
  • 34) 0.375 015 748 384 849 92 × 2 = 0 + 0.750 031 496 769 699 84;
  • 35) 0.750 031 496 769 699 84 × 2 = 1 + 0.500 062 993 539 399 68;
  • 36) 0.500 062 993 539 399 68 × 2 = 1 + 0.000 125 987 078 799 36;
  • 37) 0.000 125 987 078 799 36 × 2 = 0 + 0.000 251 974 157 598 72;
  • 38) 0.000 251 974 157 598 72 × 2 = 0 + 0.000 503 948 315 197 44;
  • 39) 0.000 503 948 315 197 44 × 2 = 0 + 0.001 007 896 630 394 88;
  • 40) 0.001 007 896 630 394 88 × 2 = 0 + 0.002 015 793 260 789 76;
  • 41) 0.002 015 793 260 789 76 × 2 = 0 + 0.004 031 586 521 579 52;
  • 42) 0.004 031 586 521 579 52 × 2 = 0 + 0.008 063 173 043 159 04;
  • 43) 0.008 063 173 043 159 04 × 2 = 0 + 0.016 126 346 086 318 08;
  • 44) 0.016 126 346 086 318 08 × 2 = 0 + 0.032 252 692 172 636 16;
  • 45) 0.032 252 692 172 636 16 × 2 = 0 + 0.064 505 384 345 272 32;
  • 46) 0.064 505 384 345 272 32 × 2 = 0 + 0.129 010 768 690 544 64;
  • 47) 0.129 010 768 690 544 64 × 2 = 0 + 0.258 021 537 381 089 28;
  • 48) 0.258 021 537 381 089 28 × 2 = 0 + 0.516 043 074 762 178 56;
  • 49) 0.516 043 074 762 178 56 × 2 = 1 + 0.032 086 149 524 357 12;
  • 50) 0.032 086 149 524 357 12 × 2 = 0 + 0.064 172 299 048 714 24;
  • 51) 0.064 172 299 048 714 24 × 2 = 0 + 0.128 344 598 097 428 48;
  • 52) 0.128 344 598 097 428 48 × 2 = 0 + 0.256 689 196 194 856 96;
  • 53) 0.256 689 196 194 856 96 × 2 = 0 + 0.513 378 392 389 713 92;
  • 54) 0.513 378 392 389 713 92 × 2 = 1 + 0.026 756 784 779 427 84;

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 51(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 149 51(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(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 51(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 01(2) × 20 =

1.1001 1000 0000 0000 0100 001(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 001


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 0001 =

100 1100 0000 0000 0010 0001


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 0001


Decimal number -0.000 000 000 742 149 51 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 0001

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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