-0.000 000 000 742 158 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 158(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
-0.000 000 000 742 158(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:

|-0.000 000 000 742 158| = 0.000 000 000 742 158


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: 0.000 000 000 742 158.

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) 0.000 000 000 742 158 × 2 = 0 + 0.000 000 001 484 316;
  • 2) 0.000 000 001 484 316 × 2 = 0 + 0.000 000 002 968 632;
  • 3) 0.000 000 002 968 632 × 2 = 0 + 0.000 000 005 937 264;
  • 4) 0.000 000 005 937 264 × 2 = 0 + 0.000 000 011 874 528;
  • 5) 0.000 000 011 874 528 × 2 = 0 + 0.000 000 023 749 056;
  • 6) 0.000 000 023 749 056 × 2 = 0 + 0.000 000 047 498 112;
  • 7) 0.000 000 047 498 112 × 2 = 0 + 0.000 000 094 996 224;
  • 8) 0.000 000 094 996 224 × 2 = 0 + 0.000 000 189 992 448;
  • 9) 0.000 000 189 992 448 × 2 = 0 + 0.000 000 379 984 896;
  • 10) 0.000 000 379 984 896 × 2 = 0 + 0.000 000 759 969 792;
  • 11) 0.000 000 759 969 792 × 2 = 0 + 0.000 001 519 939 584;
  • 12) 0.000 001 519 939 584 × 2 = 0 + 0.000 003 039 879 168;
  • 13) 0.000 003 039 879 168 × 2 = 0 + 0.000 006 079 758 336;
  • 14) 0.000 006 079 758 336 × 2 = 0 + 0.000 012 159 516 672;
  • 15) 0.000 012 159 516 672 × 2 = 0 + 0.000 024 319 033 344;
  • 16) 0.000 024 319 033 344 × 2 = 0 + 0.000 048 638 066 688;
  • 17) 0.000 048 638 066 688 × 2 = 0 + 0.000 097 276 133 376;
  • 18) 0.000 097 276 133 376 × 2 = 0 + 0.000 194 552 266 752;
  • 19) 0.000 194 552 266 752 × 2 = 0 + 0.000 389 104 533 504;
  • 20) 0.000 389 104 533 504 × 2 = 0 + 0.000 778 209 067 008;
  • 21) 0.000 778 209 067 008 × 2 = 0 + 0.001 556 418 134 016;
  • 22) 0.001 556 418 134 016 × 2 = 0 + 0.003 112 836 268 032;
  • 23) 0.003 112 836 268 032 × 2 = 0 + 0.006 225 672 536 064;
  • 24) 0.006 225 672 536 064 × 2 = 0 + 0.012 451 345 072 128;
  • 25) 0.012 451 345 072 128 × 2 = 0 + 0.024 902 690 144 256;
  • 26) 0.024 902 690 144 256 × 2 = 0 + 0.049 805 380 288 512;
  • 27) 0.049 805 380 288 512 × 2 = 0 + 0.099 610 760 577 024;
  • 28) 0.099 610 760 577 024 × 2 = 0 + 0.199 221 521 154 048;
  • 29) 0.199 221 521 154 048 × 2 = 0 + 0.398 443 042 308 096;
  • 30) 0.398 443 042 308 096 × 2 = 0 + 0.796 886 084 616 192;
  • 31) 0.796 886 084 616 192 × 2 = 1 + 0.593 772 169 232 384;
  • 32) 0.593 772 169 232 384 × 2 = 1 + 0.187 544 338 464 768;
  • 33) 0.187 544 338 464 768 × 2 = 0 + 0.375 088 676 929 536;
  • 34) 0.375 088 676 929 536 × 2 = 0 + 0.750 177 353 859 072;
  • 35) 0.750 177 353 859 072 × 2 = 1 + 0.500 354 707 718 144;
  • 36) 0.500 354 707 718 144 × 2 = 1 + 0.000 709 415 436 288;
  • 37) 0.000 709 415 436 288 × 2 = 0 + 0.001 418 830 872 576;
  • 38) 0.001 418 830 872 576 × 2 = 0 + 0.002 837 661 745 152;
  • 39) 0.002 837 661 745 152 × 2 = 0 + 0.005 675 323 490 304;
  • 40) 0.005 675 323 490 304 × 2 = 0 + 0.011 350 646 980 608;
  • 41) 0.011 350 646 980 608 × 2 = 0 + 0.022 701 293 961 216;
  • 42) 0.022 701 293 961 216 × 2 = 0 + 0.045 402 587 922 432;
  • 43) 0.045 402 587 922 432 × 2 = 0 + 0.090 805 175 844 864;
  • 44) 0.090 805 175 844 864 × 2 = 0 + 0.181 610 351 689 728;
  • 45) 0.181 610 351 689 728 × 2 = 0 + 0.363 220 703 379 456;
  • 46) 0.363 220 703 379 456 × 2 = 0 + 0.726 441 406 758 912;
  • 47) 0.726 441 406 758 912 × 2 = 1 + 0.452 882 813 517 824;
  • 48) 0.452 882 813 517 824 × 2 = 0 + 0.905 765 627 035 648;
  • 49) 0.905 765 627 035 648 × 2 = 1 + 0.811 531 254 071 296;
  • 50) 0.811 531 254 071 296 × 2 = 1 + 0.623 062 508 142 592;
  • 51) 0.623 062 508 142 592 × 2 = 1 + 0.246 125 016 285 184;
  • 52) 0.246 125 016 285 184 × 2 = 0 + 0.492 250 032 570 368;
  • 53) 0.492 250 032 570 368 × 2 = 0 + 0.984 500 065 140 736;
  • 54) 0.984 500 065 140 736 × 2 = 1 + 0.969 000 130 281 472;

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:


0.000 000 000 742 158(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 01(2)

6. Positive number before normalization:

0.000 000 000 742 158(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 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:


0.000 000 000 742 158(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0010 1110 01(2) × 20 =

1.1001 1000 0000 0001 0111 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 0001 0111 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 1011 1001 =

100 1100 0000 0000 1011 1001


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 1011 1001


Decimal number -0.000 000 000 742 158 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 1011 1001

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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