-0.000 000 000 742 082 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 082(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
-0.000 000 000 742 082(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:

|-0.000 000 000 742 082| = 0.000 000 000 742 082


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: 0.000 000 000 742 082.

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) 0.000 000 000 742 082 × 2 = 0 + 0.000 000 001 484 164;
  • 2) 0.000 000 001 484 164 × 2 = 0 + 0.000 000 002 968 328;
  • 3) 0.000 000 002 968 328 × 2 = 0 + 0.000 000 005 936 656;
  • 4) 0.000 000 005 936 656 × 2 = 0 + 0.000 000 011 873 312;
  • 5) 0.000 000 011 873 312 × 2 = 0 + 0.000 000 023 746 624;
  • 6) 0.000 000 023 746 624 × 2 = 0 + 0.000 000 047 493 248;
  • 7) 0.000 000 047 493 248 × 2 = 0 + 0.000 000 094 986 496;
  • 8) 0.000 000 094 986 496 × 2 = 0 + 0.000 000 189 972 992;
  • 9) 0.000 000 189 972 992 × 2 = 0 + 0.000 000 379 945 984;
  • 10) 0.000 000 379 945 984 × 2 = 0 + 0.000 000 759 891 968;
  • 11) 0.000 000 759 891 968 × 2 = 0 + 0.000 001 519 783 936;
  • 12) 0.000 001 519 783 936 × 2 = 0 + 0.000 003 039 567 872;
  • 13) 0.000 003 039 567 872 × 2 = 0 + 0.000 006 079 135 744;
  • 14) 0.000 006 079 135 744 × 2 = 0 + 0.000 012 158 271 488;
  • 15) 0.000 012 158 271 488 × 2 = 0 + 0.000 024 316 542 976;
  • 16) 0.000 024 316 542 976 × 2 = 0 + 0.000 048 633 085 952;
  • 17) 0.000 048 633 085 952 × 2 = 0 + 0.000 097 266 171 904;
  • 18) 0.000 097 266 171 904 × 2 = 0 + 0.000 194 532 343 808;
  • 19) 0.000 194 532 343 808 × 2 = 0 + 0.000 389 064 687 616;
  • 20) 0.000 389 064 687 616 × 2 = 0 + 0.000 778 129 375 232;
  • 21) 0.000 778 129 375 232 × 2 = 0 + 0.001 556 258 750 464;
  • 22) 0.001 556 258 750 464 × 2 = 0 + 0.003 112 517 500 928;
  • 23) 0.003 112 517 500 928 × 2 = 0 + 0.006 225 035 001 856;
  • 24) 0.006 225 035 001 856 × 2 = 0 + 0.012 450 070 003 712;
  • 25) 0.012 450 070 003 712 × 2 = 0 + 0.024 900 140 007 424;
  • 26) 0.024 900 140 007 424 × 2 = 0 + 0.049 800 280 014 848;
  • 27) 0.049 800 280 014 848 × 2 = 0 + 0.099 600 560 029 696;
  • 28) 0.099 600 560 029 696 × 2 = 0 + 0.199 201 120 059 392;
  • 29) 0.199 201 120 059 392 × 2 = 0 + 0.398 402 240 118 784;
  • 30) 0.398 402 240 118 784 × 2 = 0 + 0.796 804 480 237 568;
  • 31) 0.796 804 480 237 568 × 2 = 1 + 0.593 608 960 475 136;
  • 32) 0.593 608 960 475 136 × 2 = 1 + 0.187 217 920 950 272;
  • 33) 0.187 217 920 950 272 × 2 = 0 + 0.374 435 841 900 544;
  • 34) 0.374 435 841 900 544 × 2 = 0 + 0.748 871 683 801 088;
  • 35) 0.748 871 683 801 088 × 2 = 1 + 0.497 743 367 602 176;
  • 36) 0.497 743 367 602 176 × 2 = 0 + 0.995 486 735 204 352;
  • 37) 0.995 486 735 204 352 × 2 = 1 + 0.990 973 470 408 704;
  • 38) 0.990 973 470 408 704 × 2 = 1 + 0.981 946 940 817 408;
  • 39) 0.981 946 940 817 408 × 2 = 1 + 0.963 893 881 634 816;
  • 40) 0.963 893 881 634 816 × 2 = 1 + 0.927 787 763 269 632;
  • 41) 0.927 787 763 269 632 × 2 = 1 + 0.855 575 526 539 264;
  • 42) 0.855 575 526 539 264 × 2 = 1 + 0.711 151 053 078 528;
  • 43) 0.711 151 053 078 528 × 2 = 1 + 0.422 302 106 157 056;
  • 44) 0.422 302 106 157 056 × 2 = 0 + 0.844 604 212 314 112;
  • 45) 0.844 604 212 314 112 × 2 = 1 + 0.689 208 424 628 224;
  • 46) 0.689 208 424 628 224 × 2 = 1 + 0.378 416 849 256 448;
  • 47) 0.378 416 849 256 448 × 2 = 0 + 0.756 833 698 512 896;
  • 48) 0.756 833 698 512 896 × 2 = 1 + 0.513 667 397 025 792;
  • 49) 0.513 667 397 025 792 × 2 = 1 + 0.027 334 794 051 584;
  • 50) 0.027 334 794 051 584 × 2 = 0 + 0.054 669 588 103 168;
  • 51) 0.054 669 588 103 168 × 2 = 0 + 0.109 339 176 206 336;
  • 52) 0.109 339 176 206 336 × 2 = 0 + 0.218 678 352 412 672;
  • 53) 0.218 678 352 412 672 × 2 = 0 + 0.437 356 704 825 344;
  • 54) 0.437 356 704 825 344 × 2 = 0 + 0.874 713 409 650 688;

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:


0.000 000 000 742 082(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1101 1000 00(2)

6. Positive number before normalization:

0.000 000 000 742 082(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1101 1000 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:


0.000 000 000 742 082(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1101 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1101 1000 00(2) × 20 =

1.1001 0111 1111 0110 1100 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 0110 1100 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 1011 0110 0000 =

100 1011 1111 1011 0110 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 1011 0110 0000


Decimal number -0.000 000 000 742 082 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1011 0110 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