-0.000 000 000 742 138 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 138(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
-0.000 000 000 742 138(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:

|-0.000 000 000 742 138| = 0.000 000 000 742 138


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: 0.000 000 000 742 138.

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) 0.000 000 000 742 138 × 2 = 0 + 0.000 000 001 484 276;
  • 2) 0.000 000 001 484 276 × 2 = 0 + 0.000 000 002 968 552;
  • 3) 0.000 000 002 968 552 × 2 = 0 + 0.000 000 005 937 104;
  • 4) 0.000 000 005 937 104 × 2 = 0 + 0.000 000 011 874 208;
  • 5) 0.000 000 011 874 208 × 2 = 0 + 0.000 000 023 748 416;
  • 6) 0.000 000 023 748 416 × 2 = 0 + 0.000 000 047 496 832;
  • 7) 0.000 000 047 496 832 × 2 = 0 + 0.000 000 094 993 664;
  • 8) 0.000 000 094 993 664 × 2 = 0 + 0.000 000 189 987 328;
  • 9) 0.000 000 189 987 328 × 2 = 0 + 0.000 000 379 974 656;
  • 10) 0.000 000 379 974 656 × 2 = 0 + 0.000 000 759 949 312;
  • 11) 0.000 000 759 949 312 × 2 = 0 + 0.000 001 519 898 624;
  • 12) 0.000 001 519 898 624 × 2 = 0 + 0.000 003 039 797 248;
  • 13) 0.000 003 039 797 248 × 2 = 0 + 0.000 006 079 594 496;
  • 14) 0.000 006 079 594 496 × 2 = 0 + 0.000 012 159 188 992;
  • 15) 0.000 012 159 188 992 × 2 = 0 + 0.000 024 318 377 984;
  • 16) 0.000 024 318 377 984 × 2 = 0 + 0.000 048 636 755 968;
  • 17) 0.000 048 636 755 968 × 2 = 0 + 0.000 097 273 511 936;
  • 18) 0.000 097 273 511 936 × 2 = 0 + 0.000 194 547 023 872;
  • 19) 0.000 194 547 023 872 × 2 = 0 + 0.000 389 094 047 744;
  • 20) 0.000 389 094 047 744 × 2 = 0 + 0.000 778 188 095 488;
  • 21) 0.000 778 188 095 488 × 2 = 0 + 0.001 556 376 190 976;
  • 22) 0.001 556 376 190 976 × 2 = 0 + 0.003 112 752 381 952;
  • 23) 0.003 112 752 381 952 × 2 = 0 + 0.006 225 504 763 904;
  • 24) 0.006 225 504 763 904 × 2 = 0 + 0.012 451 009 527 808;
  • 25) 0.012 451 009 527 808 × 2 = 0 + 0.024 902 019 055 616;
  • 26) 0.024 902 019 055 616 × 2 = 0 + 0.049 804 038 111 232;
  • 27) 0.049 804 038 111 232 × 2 = 0 + 0.099 608 076 222 464;
  • 28) 0.099 608 076 222 464 × 2 = 0 + 0.199 216 152 444 928;
  • 29) 0.199 216 152 444 928 × 2 = 0 + 0.398 432 304 889 856;
  • 30) 0.398 432 304 889 856 × 2 = 0 + 0.796 864 609 779 712;
  • 31) 0.796 864 609 779 712 × 2 = 1 + 0.593 729 219 559 424;
  • 32) 0.593 729 219 559 424 × 2 = 1 + 0.187 458 439 118 848;
  • 33) 0.187 458 439 118 848 × 2 = 0 + 0.374 916 878 237 696;
  • 34) 0.374 916 878 237 696 × 2 = 0 + 0.749 833 756 475 392;
  • 35) 0.749 833 756 475 392 × 2 = 1 + 0.499 667 512 950 784;
  • 36) 0.499 667 512 950 784 × 2 = 0 + 0.999 335 025 901 568;
  • 37) 0.999 335 025 901 568 × 2 = 1 + 0.998 670 051 803 136;
  • 38) 0.998 670 051 803 136 × 2 = 1 + 0.997 340 103 606 272;
  • 39) 0.997 340 103 606 272 × 2 = 1 + 0.994 680 207 212 544;
  • 40) 0.994 680 207 212 544 × 2 = 1 + 0.989 360 414 425 088;
  • 41) 0.989 360 414 425 088 × 2 = 1 + 0.978 720 828 850 176;
  • 42) 0.978 720 828 850 176 × 2 = 1 + 0.957 441 657 700 352;
  • 43) 0.957 441 657 700 352 × 2 = 1 + 0.914 883 315 400 704;
  • 44) 0.914 883 315 400 704 × 2 = 1 + 0.829 766 630 801 408;
  • 45) 0.829 766 630 801 408 × 2 = 1 + 0.659 533 261 602 816;
  • 46) 0.659 533 261 602 816 × 2 = 1 + 0.319 066 523 205 632;
  • 47) 0.319 066 523 205 632 × 2 = 0 + 0.638 133 046 411 264;
  • 48) 0.638 133 046 411 264 × 2 = 1 + 0.276 266 092 822 528;
  • 49) 0.276 266 092 822 528 × 2 = 0 + 0.552 532 185 645 056;
  • 50) 0.552 532 185 645 056 × 2 = 1 + 0.105 064 371 290 112;
  • 51) 0.105 064 371 290 112 × 2 = 0 + 0.210 128 742 580 224;
  • 52) 0.210 128 742 580 224 × 2 = 0 + 0.420 257 485 160 448;
  • 53) 0.420 257 485 160 448 × 2 = 0 + 0.840 514 970 320 896;
  • 54) 0.840 514 970 320 896 × 2 = 1 + 0.681 029 940 641 792;

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:


0.000 000 000 742 138(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0100 01(2)

6. Positive number before normalization:

0.000 000 000 742 138(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0100 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:


0.000 000 000 742 138(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0100 01(2) × 20 =

1.1001 0111 1111 1110 1010 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 0111 1111 1110 1010 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 1011 1111 1111 0101 0001 =

100 1011 1111 1111 0101 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 1011 1111 1111 0101 0001


Decimal number -0.000 000 000 742 138 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0101 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