-0.000 000 000 742 174 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 174(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
-0.000 000 000 742 174(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:

|-0.000 000 000 742 174| = 0.000 000 000 742 174


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: 0.000 000 000 742 174.

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) 0.000 000 000 742 174 × 2 = 0 + 0.000 000 001 484 348;
  • 2) 0.000 000 001 484 348 × 2 = 0 + 0.000 000 002 968 696;
  • 3) 0.000 000 002 968 696 × 2 = 0 + 0.000 000 005 937 392;
  • 4) 0.000 000 005 937 392 × 2 = 0 + 0.000 000 011 874 784;
  • 5) 0.000 000 011 874 784 × 2 = 0 + 0.000 000 023 749 568;
  • 6) 0.000 000 023 749 568 × 2 = 0 + 0.000 000 047 499 136;
  • 7) 0.000 000 047 499 136 × 2 = 0 + 0.000 000 094 998 272;
  • 8) 0.000 000 094 998 272 × 2 = 0 + 0.000 000 189 996 544;
  • 9) 0.000 000 189 996 544 × 2 = 0 + 0.000 000 379 993 088;
  • 10) 0.000 000 379 993 088 × 2 = 0 + 0.000 000 759 986 176;
  • 11) 0.000 000 759 986 176 × 2 = 0 + 0.000 001 519 972 352;
  • 12) 0.000 001 519 972 352 × 2 = 0 + 0.000 003 039 944 704;
  • 13) 0.000 003 039 944 704 × 2 = 0 + 0.000 006 079 889 408;
  • 14) 0.000 006 079 889 408 × 2 = 0 + 0.000 012 159 778 816;
  • 15) 0.000 012 159 778 816 × 2 = 0 + 0.000 024 319 557 632;
  • 16) 0.000 024 319 557 632 × 2 = 0 + 0.000 048 639 115 264;
  • 17) 0.000 048 639 115 264 × 2 = 0 + 0.000 097 278 230 528;
  • 18) 0.000 097 278 230 528 × 2 = 0 + 0.000 194 556 461 056;
  • 19) 0.000 194 556 461 056 × 2 = 0 + 0.000 389 112 922 112;
  • 20) 0.000 389 112 922 112 × 2 = 0 + 0.000 778 225 844 224;
  • 21) 0.000 778 225 844 224 × 2 = 0 + 0.001 556 451 688 448;
  • 22) 0.001 556 451 688 448 × 2 = 0 + 0.003 112 903 376 896;
  • 23) 0.003 112 903 376 896 × 2 = 0 + 0.006 225 806 753 792;
  • 24) 0.006 225 806 753 792 × 2 = 0 + 0.012 451 613 507 584;
  • 25) 0.012 451 613 507 584 × 2 = 0 + 0.024 903 227 015 168;
  • 26) 0.024 903 227 015 168 × 2 = 0 + 0.049 806 454 030 336;
  • 27) 0.049 806 454 030 336 × 2 = 0 + 0.099 612 908 060 672;
  • 28) 0.099 612 908 060 672 × 2 = 0 + 0.199 225 816 121 344;
  • 29) 0.199 225 816 121 344 × 2 = 0 + 0.398 451 632 242 688;
  • 30) 0.398 451 632 242 688 × 2 = 0 + 0.796 903 264 485 376;
  • 31) 0.796 903 264 485 376 × 2 = 1 + 0.593 806 528 970 752;
  • 32) 0.593 806 528 970 752 × 2 = 1 + 0.187 613 057 941 504;
  • 33) 0.187 613 057 941 504 × 2 = 0 + 0.375 226 115 883 008;
  • 34) 0.375 226 115 883 008 × 2 = 0 + 0.750 452 231 766 016;
  • 35) 0.750 452 231 766 016 × 2 = 1 + 0.500 904 463 532 032;
  • 36) 0.500 904 463 532 032 × 2 = 1 + 0.001 808 927 064 064;
  • 37) 0.001 808 927 064 064 × 2 = 0 + 0.003 617 854 128 128;
  • 38) 0.003 617 854 128 128 × 2 = 0 + 0.007 235 708 256 256;
  • 39) 0.007 235 708 256 256 × 2 = 0 + 0.014 471 416 512 512;
  • 40) 0.014 471 416 512 512 × 2 = 0 + 0.028 942 833 025 024;
  • 41) 0.028 942 833 025 024 × 2 = 0 + 0.057 885 666 050 048;
  • 42) 0.057 885 666 050 048 × 2 = 0 + 0.115 771 332 100 096;
  • 43) 0.115 771 332 100 096 × 2 = 0 + 0.231 542 664 200 192;
  • 44) 0.231 542 664 200 192 × 2 = 0 + 0.463 085 328 400 384;
  • 45) 0.463 085 328 400 384 × 2 = 0 + 0.926 170 656 800 768;
  • 46) 0.926 170 656 800 768 × 2 = 1 + 0.852 341 313 601 536;
  • 47) 0.852 341 313 601 536 × 2 = 1 + 0.704 682 627 203 072;
  • 48) 0.704 682 627 203 072 × 2 = 1 + 0.409 365 254 406 144;
  • 49) 0.409 365 254 406 144 × 2 = 0 + 0.818 730 508 812 288;
  • 50) 0.818 730 508 812 288 × 2 = 1 + 0.637 461 017 624 576;
  • 51) 0.637 461 017 624 576 × 2 = 1 + 0.274 922 035 249 152;
  • 52) 0.274 922 035 249 152 × 2 = 0 + 0.549 844 070 498 304;
  • 53) 0.549 844 070 498 304 × 2 = 1 + 0.099 688 140 996 608;
  • 54) 0.099 688 140 996 608 × 2 = 0 + 0.199 376 281 993 216;

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:


0.000 000 000 742 174(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0111 0110 10(2)

6. Positive number before normalization:

0.000 000 000 742 174(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0111 0110 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:


0.000 000 000 742 174(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0111 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0111 0110 10(2) × 20 =

1.1001 1000 0000 0011 1011 010(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 0011 1011 010


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 0001 1101 1010 =

100 1100 0000 0001 1101 1010


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 0001 1101 1010


Decimal number -0.000 000 000 742 174 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0001 1101 1010

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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