-0.000 000 000 742 147 679 42 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 679 42(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
-0.000 000 000 742 147 679 42(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:

|-0.000 000 000 742 147 679 42| = 0.000 000 000 742 147 679 42


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: 0.000 000 000 742 147 679 42.

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) 0.000 000 000 742 147 679 42 × 2 = 0 + 0.000 000 001 484 295 358 84;
  • 2) 0.000 000 001 484 295 358 84 × 2 = 0 + 0.000 000 002 968 590 717 68;
  • 3) 0.000 000 002 968 590 717 68 × 2 = 0 + 0.000 000 005 937 181 435 36;
  • 4) 0.000 000 005 937 181 435 36 × 2 = 0 + 0.000 000 011 874 362 870 72;
  • 5) 0.000 000 011 874 362 870 72 × 2 = 0 + 0.000 000 023 748 725 741 44;
  • 6) 0.000 000 023 748 725 741 44 × 2 = 0 + 0.000 000 047 497 451 482 88;
  • 7) 0.000 000 047 497 451 482 88 × 2 = 0 + 0.000 000 094 994 902 965 76;
  • 8) 0.000 000 094 994 902 965 76 × 2 = 0 + 0.000 000 189 989 805 931 52;
  • 9) 0.000 000 189 989 805 931 52 × 2 = 0 + 0.000 000 379 979 611 863 04;
  • 10) 0.000 000 379 979 611 863 04 × 2 = 0 + 0.000 000 759 959 223 726 08;
  • 11) 0.000 000 759 959 223 726 08 × 2 = 0 + 0.000 001 519 918 447 452 16;
  • 12) 0.000 001 519 918 447 452 16 × 2 = 0 + 0.000 003 039 836 894 904 32;
  • 13) 0.000 003 039 836 894 904 32 × 2 = 0 + 0.000 006 079 673 789 808 64;
  • 14) 0.000 006 079 673 789 808 64 × 2 = 0 + 0.000 012 159 347 579 617 28;
  • 15) 0.000 012 159 347 579 617 28 × 2 = 0 + 0.000 024 318 695 159 234 56;
  • 16) 0.000 024 318 695 159 234 56 × 2 = 0 + 0.000 048 637 390 318 469 12;
  • 17) 0.000 048 637 390 318 469 12 × 2 = 0 + 0.000 097 274 780 636 938 24;
  • 18) 0.000 097 274 780 636 938 24 × 2 = 0 + 0.000 194 549 561 273 876 48;
  • 19) 0.000 194 549 561 273 876 48 × 2 = 0 + 0.000 389 099 122 547 752 96;
  • 20) 0.000 389 099 122 547 752 96 × 2 = 0 + 0.000 778 198 245 095 505 92;
  • 21) 0.000 778 198 245 095 505 92 × 2 = 0 + 0.001 556 396 490 191 011 84;
  • 22) 0.001 556 396 490 191 011 84 × 2 = 0 + 0.003 112 792 980 382 023 68;
  • 23) 0.003 112 792 980 382 023 68 × 2 = 0 + 0.006 225 585 960 764 047 36;
  • 24) 0.006 225 585 960 764 047 36 × 2 = 0 + 0.012 451 171 921 528 094 72;
  • 25) 0.012 451 171 921 528 094 72 × 2 = 0 + 0.024 902 343 843 056 189 44;
  • 26) 0.024 902 343 843 056 189 44 × 2 = 0 + 0.049 804 687 686 112 378 88;
  • 27) 0.049 804 687 686 112 378 88 × 2 = 0 + 0.099 609 375 372 224 757 76;
  • 28) 0.099 609 375 372 224 757 76 × 2 = 0 + 0.199 218 750 744 449 515 52;
  • 29) 0.199 218 750 744 449 515 52 × 2 = 0 + 0.398 437 501 488 899 031 04;
  • 30) 0.398 437 501 488 899 031 04 × 2 = 0 + 0.796 875 002 977 798 062 08;
  • 31) 0.796 875 002 977 798 062 08 × 2 = 1 + 0.593 750 005 955 596 124 16;
  • 32) 0.593 750 005 955 596 124 16 × 2 = 1 + 0.187 500 011 911 192 248 32;
  • 33) 0.187 500 011 911 192 248 32 × 2 = 0 + 0.375 000 023 822 384 496 64;
  • 34) 0.375 000 023 822 384 496 64 × 2 = 0 + 0.750 000 047 644 768 993 28;
  • 35) 0.750 000 047 644 768 993 28 × 2 = 1 + 0.500 000 095 289 537 986 56;
  • 36) 0.500 000 095 289 537 986 56 × 2 = 1 + 0.000 000 190 579 075 973 12;
  • 37) 0.000 000 190 579 075 973 12 × 2 = 0 + 0.000 000 381 158 151 946 24;
  • 38) 0.000 000 381 158 151 946 24 × 2 = 0 + 0.000 000 762 316 303 892 48;
  • 39) 0.000 000 762 316 303 892 48 × 2 = 0 + 0.000 001 524 632 607 784 96;
  • 40) 0.000 001 524 632 607 784 96 × 2 = 0 + 0.000 003 049 265 215 569 92;
  • 41) 0.000 003 049 265 215 569 92 × 2 = 0 + 0.000 006 098 530 431 139 84;
  • 42) 0.000 006 098 530 431 139 84 × 2 = 0 + 0.000 012 197 060 862 279 68;
  • 43) 0.000 012 197 060 862 279 68 × 2 = 0 + 0.000 024 394 121 724 559 36;
  • 44) 0.000 024 394 121 724 559 36 × 2 = 0 + 0.000 048 788 243 449 118 72;
  • 45) 0.000 048 788 243 449 118 72 × 2 = 0 + 0.000 097 576 486 898 237 44;
  • 46) 0.000 097 576 486 898 237 44 × 2 = 0 + 0.000 195 152 973 796 474 88;
  • 47) 0.000 195 152 973 796 474 88 × 2 = 0 + 0.000 390 305 947 592 949 76;
  • 48) 0.000 390 305 947 592 949 76 × 2 = 0 + 0.000 780 611 895 185 899 52;
  • 49) 0.000 780 611 895 185 899 52 × 2 = 0 + 0.001 561 223 790 371 799 04;
  • 50) 0.001 561 223 790 371 799 04 × 2 = 0 + 0.003 122 447 580 743 598 08;
  • 51) 0.003 122 447 580 743 598 08 × 2 = 0 + 0.006 244 895 161 487 196 16;
  • 52) 0.006 244 895 161 487 196 16 × 2 = 0 + 0.012 489 790 322 974 392 32;
  • 53) 0.012 489 790 322 974 392 32 × 2 = 0 + 0.024 979 580 645 948 784 64;
  • 54) 0.024 979 580 645 948 784 64 × 2 = 0 + 0.049 959 161 291 897 569 28;

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:


0.000 000 000 742 147 679 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 679 42(10) =

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


0.000 000 000 742 147 679 42(10) =

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

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

1.1001 1000 0000 0000 0000 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 0000 0000 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 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 679 42 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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