-0.000 000 000 742 148 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 03(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
-0.000 000 000 742 148 03(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:

|-0.000 000 000 742 148 03| = 0.000 000 000 742 148 03


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: 0.000 000 000 742 148 03.

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) 0.000 000 000 742 148 03 × 2 = 0 + 0.000 000 001 484 296 06;
  • 2) 0.000 000 001 484 296 06 × 2 = 0 + 0.000 000 002 968 592 12;
  • 3) 0.000 000 002 968 592 12 × 2 = 0 + 0.000 000 005 937 184 24;
  • 4) 0.000 000 005 937 184 24 × 2 = 0 + 0.000 000 011 874 368 48;
  • 5) 0.000 000 011 874 368 48 × 2 = 0 + 0.000 000 023 748 736 96;
  • 6) 0.000 000 023 748 736 96 × 2 = 0 + 0.000 000 047 497 473 92;
  • 7) 0.000 000 047 497 473 92 × 2 = 0 + 0.000 000 094 994 947 84;
  • 8) 0.000 000 094 994 947 84 × 2 = 0 + 0.000 000 189 989 895 68;
  • 9) 0.000 000 189 989 895 68 × 2 = 0 + 0.000 000 379 979 791 36;
  • 10) 0.000 000 379 979 791 36 × 2 = 0 + 0.000 000 759 959 582 72;
  • 11) 0.000 000 759 959 582 72 × 2 = 0 + 0.000 001 519 919 165 44;
  • 12) 0.000 001 519 919 165 44 × 2 = 0 + 0.000 003 039 838 330 88;
  • 13) 0.000 003 039 838 330 88 × 2 = 0 + 0.000 006 079 676 661 76;
  • 14) 0.000 006 079 676 661 76 × 2 = 0 + 0.000 012 159 353 323 52;
  • 15) 0.000 012 159 353 323 52 × 2 = 0 + 0.000 024 318 706 647 04;
  • 16) 0.000 024 318 706 647 04 × 2 = 0 + 0.000 048 637 413 294 08;
  • 17) 0.000 048 637 413 294 08 × 2 = 0 + 0.000 097 274 826 588 16;
  • 18) 0.000 097 274 826 588 16 × 2 = 0 + 0.000 194 549 653 176 32;
  • 19) 0.000 194 549 653 176 32 × 2 = 0 + 0.000 389 099 306 352 64;
  • 20) 0.000 389 099 306 352 64 × 2 = 0 + 0.000 778 198 612 705 28;
  • 21) 0.000 778 198 612 705 28 × 2 = 0 + 0.001 556 397 225 410 56;
  • 22) 0.001 556 397 225 410 56 × 2 = 0 + 0.003 112 794 450 821 12;
  • 23) 0.003 112 794 450 821 12 × 2 = 0 + 0.006 225 588 901 642 24;
  • 24) 0.006 225 588 901 642 24 × 2 = 0 + 0.012 451 177 803 284 48;
  • 25) 0.012 451 177 803 284 48 × 2 = 0 + 0.024 902 355 606 568 96;
  • 26) 0.024 902 355 606 568 96 × 2 = 0 + 0.049 804 711 213 137 92;
  • 27) 0.049 804 711 213 137 92 × 2 = 0 + 0.099 609 422 426 275 84;
  • 28) 0.099 609 422 426 275 84 × 2 = 0 + 0.199 218 844 852 551 68;
  • 29) 0.199 218 844 852 551 68 × 2 = 0 + 0.398 437 689 705 103 36;
  • 30) 0.398 437 689 705 103 36 × 2 = 0 + 0.796 875 379 410 206 72;
  • 31) 0.796 875 379 410 206 72 × 2 = 1 + 0.593 750 758 820 413 44;
  • 32) 0.593 750 758 820 413 44 × 2 = 1 + 0.187 501 517 640 826 88;
  • 33) 0.187 501 517 640 826 88 × 2 = 0 + 0.375 003 035 281 653 76;
  • 34) 0.375 003 035 281 653 76 × 2 = 0 + 0.750 006 070 563 307 52;
  • 35) 0.750 006 070 563 307 52 × 2 = 1 + 0.500 012 141 126 615 04;
  • 36) 0.500 012 141 126 615 04 × 2 = 1 + 0.000 024 282 253 230 08;
  • 37) 0.000 024 282 253 230 08 × 2 = 0 + 0.000 048 564 506 460 16;
  • 38) 0.000 048 564 506 460 16 × 2 = 0 + 0.000 097 129 012 920 32;
  • 39) 0.000 097 129 012 920 32 × 2 = 0 + 0.000 194 258 025 840 64;
  • 40) 0.000 194 258 025 840 64 × 2 = 0 + 0.000 388 516 051 681 28;
  • 41) 0.000 388 516 051 681 28 × 2 = 0 + 0.000 777 032 103 362 56;
  • 42) 0.000 777 032 103 362 56 × 2 = 0 + 0.001 554 064 206 725 12;
  • 43) 0.001 554 064 206 725 12 × 2 = 0 + 0.003 108 128 413 450 24;
  • 44) 0.003 108 128 413 450 24 × 2 = 0 + 0.006 216 256 826 900 48;
  • 45) 0.006 216 256 826 900 48 × 2 = 0 + 0.012 432 513 653 800 96;
  • 46) 0.012 432 513 653 800 96 × 2 = 0 + 0.024 865 027 307 601 92;
  • 47) 0.024 865 027 307 601 92 × 2 = 0 + 0.049 730 054 615 203 84;
  • 48) 0.049 730 054 615 203 84 × 2 = 0 + 0.099 460 109 230 407 68;
  • 49) 0.099 460 109 230 407 68 × 2 = 0 + 0.198 920 218 460 815 36;
  • 50) 0.198 920 218 460 815 36 × 2 = 0 + 0.397 840 436 921 630 72;
  • 51) 0.397 840 436 921 630 72 × 2 = 0 + 0.795 680 873 843 261 44;
  • 52) 0.795 680 873 843 261 44 × 2 = 1 + 0.591 361 747 686 522 88;
  • 53) 0.591 361 747 686 522 88 × 2 = 1 + 0.182 723 495 373 045 76;
  • 54) 0.182 723 495 373 045 76 × 2 = 0 + 0.365 446 990 746 091 52;

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:


0.000 000 000 742 148 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 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:


0.000 000 000 742 148 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) × 20 =

1.1001 1000 0000 0000 0000 110(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 0000 0000 110


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 0000 0110 =

100 1100 0000 0000 0000 0110


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 0000 0110


Decimal number -0.000 000 000 742 148 03 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0110

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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