-0.000 000 000 740 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 1(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
-0.000 000 000 740 1(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:

|-0.000 000 000 740 1| = 0.000 000 000 740 1


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: 0.000 000 000 740 1.

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) 0.000 000 000 740 1 × 2 = 0 + 0.000 000 001 480 2;
  • 2) 0.000 000 001 480 2 × 2 = 0 + 0.000 000 002 960 4;
  • 3) 0.000 000 002 960 4 × 2 = 0 + 0.000 000 005 920 8;
  • 4) 0.000 000 005 920 8 × 2 = 0 + 0.000 000 011 841 6;
  • 5) 0.000 000 011 841 6 × 2 = 0 + 0.000 000 023 683 2;
  • 6) 0.000 000 023 683 2 × 2 = 0 + 0.000 000 047 366 4;
  • 7) 0.000 000 047 366 4 × 2 = 0 + 0.000 000 094 732 8;
  • 8) 0.000 000 094 732 8 × 2 = 0 + 0.000 000 189 465 6;
  • 9) 0.000 000 189 465 6 × 2 = 0 + 0.000 000 378 931 2;
  • 10) 0.000 000 378 931 2 × 2 = 0 + 0.000 000 757 862 4;
  • 11) 0.000 000 757 862 4 × 2 = 0 + 0.000 001 515 724 8;
  • 12) 0.000 001 515 724 8 × 2 = 0 + 0.000 003 031 449 6;
  • 13) 0.000 003 031 449 6 × 2 = 0 + 0.000 006 062 899 2;
  • 14) 0.000 006 062 899 2 × 2 = 0 + 0.000 012 125 798 4;
  • 15) 0.000 012 125 798 4 × 2 = 0 + 0.000 024 251 596 8;
  • 16) 0.000 024 251 596 8 × 2 = 0 + 0.000 048 503 193 6;
  • 17) 0.000 048 503 193 6 × 2 = 0 + 0.000 097 006 387 2;
  • 18) 0.000 097 006 387 2 × 2 = 0 + 0.000 194 012 774 4;
  • 19) 0.000 194 012 774 4 × 2 = 0 + 0.000 388 025 548 8;
  • 20) 0.000 388 025 548 8 × 2 = 0 + 0.000 776 051 097 6;
  • 21) 0.000 776 051 097 6 × 2 = 0 + 0.001 552 102 195 2;
  • 22) 0.001 552 102 195 2 × 2 = 0 + 0.003 104 204 390 4;
  • 23) 0.003 104 204 390 4 × 2 = 0 + 0.006 208 408 780 8;
  • 24) 0.006 208 408 780 8 × 2 = 0 + 0.012 416 817 561 6;
  • 25) 0.012 416 817 561 6 × 2 = 0 + 0.024 833 635 123 2;
  • 26) 0.024 833 635 123 2 × 2 = 0 + 0.049 667 270 246 4;
  • 27) 0.049 667 270 246 4 × 2 = 0 + 0.099 334 540 492 8;
  • 28) 0.099 334 540 492 8 × 2 = 0 + 0.198 669 080 985 6;
  • 29) 0.198 669 080 985 6 × 2 = 0 + 0.397 338 161 971 2;
  • 30) 0.397 338 161 971 2 × 2 = 0 + 0.794 676 323 942 4;
  • 31) 0.794 676 323 942 4 × 2 = 1 + 0.589 352 647 884 8;
  • 32) 0.589 352 647 884 8 × 2 = 1 + 0.178 705 295 769 6;
  • 33) 0.178 705 295 769 6 × 2 = 0 + 0.357 410 591 539 2;
  • 34) 0.357 410 591 539 2 × 2 = 0 + 0.714 821 183 078 4;
  • 35) 0.714 821 183 078 4 × 2 = 1 + 0.429 642 366 156 8;
  • 36) 0.429 642 366 156 8 × 2 = 0 + 0.859 284 732 313 6;
  • 37) 0.859 284 732 313 6 × 2 = 1 + 0.718 569 464 627 2;
  • 38) 0.718 569 464 627 2 × 2 = 1 + 0.437 138 929 254 4;
  • 39) 0.437 138 929 254 4 × 2 = 0 + 0.874 277 858 508 8;
  • 40) 0.874 277 858 508 8 × 2 = 1 + 0.748 555 717 017 6;
  • 41) 0.748 555 717 017 6 × 2 = 1 + 0.497 111 434 035 2;
  • 42) 0.497 111 434 035 2 × 2 = 0 + 0.994 222 868 070 4;
  • 43) 0.994 222 868 070 4 × 2 = 1 + 0.988 445 736 140 8;
  • 44) 0.988 445 736 140 8 × 2 = 1 + 0.976 891 472 281 6;
  • 45) 0.976 891 472 281 6 × 2 = 1 + 0.953 782 944 563 2;
  • 46) 0.953 782 944 563 2 × 2 = 1 + 0.907 565 889 126 4;
  • 47) 0.907 565 889 126 4 × 2 = 1 + 0.815 131 778 252 8;
  • 48) 0.815 131 778 252 8 × 2 = 1 + 0.630 263 556 505 6;
  • 49) 0.630 263 556 505 6 × 2 = 1 + 0.260 527 113 011 2;
  • 50) 0.260 527 113 011 2 × 2 = 0 + 0.521 054 226 022 4;
  • 51) 0.521 054 226 022 4 × 2 = 1 + 0.042 108 452 044 8;
  • 52) 0.042 108 452 044 8 × 2 = 0 + 0.084 216 904 089 6;
  • 53) 0.084 216 904 089 6 × 2 = 0 + 0.168 433 808 179 2;
  • 54) 0.168 433 808 179 2 × 2 = 0 + 0.336 867 616 358 4;

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:


0.000 000 000 740 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1011 1111 1010 00(2)

6. Positive number before normalization:

0.000 000 000 740 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1011 1111 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:


0.000 000 000 740 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1011 1111 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1011 1111 1010 00(2) × 20 =

1.1001 0110 1101 1111 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 0110 1101 1111 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 1011 0110 1111 1110 1000 =

100 1011 0110 1111 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 1011 0110 1111 1110 1000


Decimal number -0.000 000 000 740 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0110 1111 1110 1000

Share

» Convert decimal -0.000 000 000 734 9 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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