-0.000 000 000 742 189 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 189(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
-0.000 000 000 742 189(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:

|-0.000 000 000 742 189| = 0.000 000 000 742 189


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: 0.000 000 000 742 189.

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) 0.000 000 000 742 189 × 2 = 0 + 0.000 000 001 484 378;
  • 2) 0.000 000 001 484 378 × 2 = 0 + 0.000 000 002 968 756;
  • 3) 0.000 000 002 968 756 × 2 = 0 + 0.000 000 005 937 512;
  • 4) 0.000 000 005 937 512 × 2 = 0 + 0.000 000 011 875 024;
  • 5) 0.000 000 011 875 024 × 2 = 0 + 0.000 000 023 750 048;
  • 6) 0.000 000 023 750 048 × 2 = 0 + 0.000 000 047 500 096;
  • 7) 0.000 000 047 500 096 × 2 = 0 + 0.000 000 095 000 192;
  • 8) 0.000 000 095 000 192 × 2 = 0 + 0.000 000 190 000 384;
  • 9) 0.000 000 190 000 384 × 2 = 0 + 0.000 000 380 000 768;
  • 10) 0.000 000 380 000 768 × 2 = 0 + 0.000 000 760 001 536;
  • 11) 0.000 000 760 001 536 × 2 = 0 + 0.000 001 520 003 072;
  • 12) 0.000 001 520 003 072 × 2 = 0 + 0.000 003 040 006 144;
  • 13) 0.000 003 040 006 144 × 2 = 0 + 0.000 006 080 012 288;
  • 14) 0.000 006 080 012 288 × 2 = 0 + 0.000 012 160 024 576;
  • 15) 0.000 012 160 024 576 × 2 = 0 + 0.000 024 320 049 152;
  • 16) 0.000 024 320 049 152 × 2 = 0 + 0.000 048 640 098 304;
  • 17) 0.000 048 640 098 304 × 2 = 0 + 0.000 097 280 196 608;
  • 18) 0.000 097 280 196 608 × 2 = 0 + 0.000 194 560 393 216;
  • 19) 0.000 194 560 393 216 × 2 = 0 + 0.000 389 120 786 432;
  • 20) 0.000 389 120 786 432 × 2 = 0 + 0.000 778 241 572 864;
  • 21) 0.000 778 241 572 864 × 2 = 0 + 0.001 556 483 145 728;
  • 22) 0.001 556 483 145 728 × 2 = 0 + 0.003 112 966 291 456;
  • 23) 0.003 112 966 291 456 × 2 = 0 + 0.006 225 932 582 912;
  • 24) 0.006 225 932 582 912 × 2 = 0 + 0.012 451 865 165 824;
  • 25) 0.012 451 865 165 824 × 2 = 0 + 0.024 903 730 331 648;
  • 26) 0.024 903 730 331 648 × 2 = 0 + 0.049 807 460 663 296;
  • 27) 0.049 807 460 663 296 × 2 = 0 + 0.099 614 921 326 592;
  • 28) 0.099 614 921 326 592 × 2 = 0 + 0.199 229 842 653 184;
  • 29) 0.199 229 842 653 184 × 2 = 0 + 0.398 459 685 306 368;
  • 30) 0.398 459 685 306 368 × 2 = 0 + 0.796 919 370 612 736;
  • 31) 0.796 919 370 612 736 × 2 = 1 + 0.593 838 741 225 472;
  • 32) 0.593 838 741 225 472 × 2 = 1 + 0.187 677 482 450 944;
  • 33) 0.187 677 482 450 944 × 2 = 0 + 0.375 354 964 901 888;
  • 34) 0.375 354 964 901 888 × 2 = 0 + 0.750 709 929 803 776;
  • 35) 0.750 709 929 803 776 × 2 = 1 + 0.501 419 859 607 552;
  • 36) 0.501 419 859 607 552 × 2 = 1 + 0.002 839 719 215 104;
  • 37) 0.002 839 719 215 104 × 2 = 0 + 0.005 679 438 430 208;
  • 38) 0.005 679 438 430 208 × 2 = 0 + 0.011 358 876 860 416;
  • 39) 0.011 358 876 860 416 × 2 = 0 + 0.022 717 753 720 832;
  • 40) 0.022 717 753 720 832 × 2 = 0 + 0.045 435 507 441 664;
  • 41) 0.045 435 507 441 664 × 2 = 0 + 0.090 871 014 883 328;
  • 42) 0.090 871 014 883 328 × 2 = 0 + 0.181 742 029 766 656;
  • 43) 0.181 742 029 766 656 × 2 = 0 + 0.363 484 059 533 312;
  • 44) 0.363 484 059 533 312 × 2 = 0 + 0.726 968 119 066 624;
  • 45) 0.726 968 119 066 624 × 2 = 1 + 0.453 936 238 133 248;
  • 46) 0.453 936 238 133 248 × 2 = 0 + 0.907 872 476 266 496;
  • 47) 0.907 872 476 266 496 × 2 = 1 + 0.815 744 952 532 992;
  • 48) 0.815 744 952 532 992 × 2 = 1 + 0.631 489 905 065 984;
  • 49) 0.631 489 905 065 984 × 2 = 1 + 0.262 979 810 131 968;
  • 50) 0.262 979 810 131 968 × 2 = 0 + 0.525 959 620 263 936;
  • 51) 0.525 959 620 263 936 × 2 = 1 + 0.051 919 240 527 872;
  • 52) 0.051 919 240 527 872 × 2 = 0 + 0.103 838 481 055 744;
  • 53) 0.103 838 481 055 744 × 2 = 0 + 0.207 676 962 111 488;
  • 54) 0.207 676 962 111 488 × 2 = 0 + 0.415 353 924 222 976;

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:


0.000 000 000 742 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1011 1010 00(2)

6. Positive number before normalization:

0.000 000 000 742 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1011 1010 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:


0.000 000 000 742 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1011 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 1011 1010 00(2) × 20 =

1.1001 1000 0000 0101 1101 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 1000 0000 0101 1101 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 1100 0000 0010 1110 1000 =

100 1100 0000 0010 1110 1000


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 0010 1110 1000


Decimal number -0.000 000 000 742 189 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0010 1110 1000

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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