-0.000 000 000 742 018 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 018(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
-0.000 000 000 742 018(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:

|-0.000 000 000 742 018| = 0.000 000 000 742 018


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: 0.000 000 000 742 018.

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) 0.000 000 000 742 018 × 2 = 0 + 0.000 000 001 484 036;
  • 2) 0.000 000 001 484 036 × 2 = 0 + 0.000 000 002 968 072;
  • 3) 0.000 000 002 968 072 × 2 = 0 + 0.000 000 005 936 144;
  • 4) 0.000 000 005 936 144 × 2 = 0 + 0.000 000 011 872 288;
  • 5) 0.000 000 011 872 288 × 2 = 0 + 0.000 000 023 744 576;
  • 6) 0.000 000 023 744 576 × 2 = 0 + 0.000 000 047 489 152;
  • 7) 0.000 000 047 489 152 × 2 = 0 + 0.000 000 094 978 304;
  • 8) 0.000 000 094 978 304 × 2 = 0 + 0.000 000 189 956 608;
  • 9) 0.000 000 189 956 608 × 2 = 0 + 0.000 000 379 913 216;
  • 10) 0.000 000 379 913 216 × 2 = 0 + 0.000 000 759 826 432;
  • 11) 0.000 000 759 826 432 × 2 = 0 + 0.000 001 519 652 864;
  • 12) 0.000 001 519 652 864 × 2 = 0 + 0.000 003 039 305 728;
  • 13) 0.000 003 039 305 728 × 2 = 0 + 0.000 006 078 611 456;
  • 14) 0.000 006 078 611 456 × 2 = 0 + 0.000 012 157 222 912;
  • 15) 0.000 012 157 222 912 × 2 = 0 + 0.000 024 314 445 824;
  • 16) 0.000 024 314 445 824 × 2 = 0 + 0.000 048 628 891 648;
  • 17) 0.000 048 628 891 648 × 2 = 0 + 0.000 097 257 783 296;
  • 18) 0.000 097 257 783 296 × 2 = 0 + 0.000 194 515 566 592;
  • 19) 0.000 194 515 566 592 × 2 = 0 + 0.000 389 031 133 184;
  • 20) 0.000 389 031 133 184 × 2 = 0 + 0.000 778 062 266 368;
  • 21) 0.000 778 062 266 368 × 2 = 0 + 0.001 556 124 532 736;
  • 22) 0.001 556 124 532 736 × 2 = 0 + 0.003 112 249 065 472;
  • 23) 0.003 112 249 065 472 × 2 = 0 + 0.006 224 498 130 944;
  • 24) 0.006 224 498 130 944 × 2 = 0 + 0.012 448 996 261 888;
  • 25) 0.012 448 996 261 888 × 2 = 0 + 0.024 897 992 523 776;
  • 26) 0.024 897 992 523 776 × 2 = 0 + 0.049 795 985 047 552;
  • 27) 0.049 795 985 047 552 × 2 = 0 + 0.099 591 970 095 104;
  • 28) 0.099 591 970 095 104 × 2 = 0 + 0.199 183 940 190 208;
  • 29) 0.199 183 940 190 208 × 2 = 0 + 0.398 367 880 380 416;
  • 30) 0.398 367 880 380 416 × 2 = 0 + 0.796 735 760 760 832;
  • 31) 0.796 735 760 760 832 × 2 = 1 + 0.593 471 521 521 664;
  • 32) 0.593 471 521 521 664 × 2 = 1 + 0.186 943 043 043 328;
  • 33) 0.186 943 043 043 328 × 2 = 0 + 0.373 886 086 086 656;
  • 34) 0.373 886 086 086 656 × 2 = 0 + 0.747 772 172 173 312;
  • 35) 0.747 772 172 173 312 × 2 = 1 + 0.495 544 344 346 624;
  • 36) 0.495 544 344 346 624 × 2 = 0 + 0.991 088 688 693 248;
  • 37) 0.991 088 688 693 248 × 2 = 1 + 0.982 177 377 386 496;
  • 38) 0.982 177 377 386 496 × 2 = 1 + 0.964 354 754 772 992;
  • 39) 0.964 354 754 772 992 × 2 = 1 + 0.928 709 509 545 984;
  • 40) 0.928 709 509 545 984 × 2 = 1 + 0.857 419 019 091 968;
  • 41) 0.857 419 019 091 968 × 2 = 1 + 0.714 838 038 183 936;
  • 42) 0.714 838 038 183 936 × 2 = 1 + 0.429 676 076 367 872;
  • 43) 0.429 676 076 367 872 × 2 = 0 + 0.859 352 152 735 744;
  • 44) 0.859 352 152 735 744 × 2 = 1 + 0.718 704 305 471 488;
  • 45) 0.718 704 305 471 488 × 2 = 1 + 0.437 408 610 942 976;
  • 46) 0.437 408 610 942 976 × 2 = 0 + 0.874 817 221 885 952;
  • 47) 0.874 817 221 885 952 × 2 = 1 + 0.749 634 443 771 904;
  • 48) 0.749 634 443 771 904 × 2 = 1 + 0.499 268 887 543 808;
  • 49) 0.499 268 887 543 808 × 2 = 0 + 0.998 537 775 087 616;
  • 50) 0.998 537 775 087 616 × 2 = 1 + 0.997 075 550 175 232;
  • 51) 0.997 075 550 175 232 × 2 = 1 + 0.994 151 100 350 464;
  • 52) 0.994 151 100 350 464 × 2 = 1 + 0.988 302 200 700 928;
  • 53) 0.988 302 200 700 928 × 2 = 1 + 0.976 604 401 401 856;
  • 54) 0.976 604 401 401 856 × 2 = 1 + 0.953 208 802 803 712;

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:


0.000 000 000 742 018(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1011 0111 11(2)

6. Positive number before normalization:

0.000 000 000 742 018(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1011 0111 11(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:


0.000 000 000 742 018(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1011 0111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1011 0111 11(2) × 20 =

1.1001 0111 1110 1101 1011 111(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 1110 1101 1011 111


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 0110 1101 1111 =

100 1011 1111 0110 1101 1111


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 0110 1101 1111


Decimal number -0.000 000 000 742 018 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0110 1101 1111

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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