-0.000 000 000 694 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 694(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
-0.000 000 000 694(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:

|-0.000 000 000 694| = 0.000 000 000 694


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: 0.000 000 000 694.

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) 0.000 000 000 694 × 2 = 0 + 0.000 000 001 388;
  • 2) 0.000 000 001 388 × 2 = 0 + 0.000 000 002 776;
  • 3) 0.000 000 002 776 × 2 = 0 + 0.000 000 005 552;
  • 4) 0.000 000 005 552 × 2 = 0 + 0.000 000 011 104;
  • 5) 0.000 000 011 104 × 2 = 0 + 0.000 000 022 208;
  • 6) 0.000 000 022 208 × 2 = 0 + 0.000 000 044 416;
  • 7) 0.000 000 044 416 × 2 = 0 + 0.000 000 088 832;
  • 8) 0.000 000 088 832 × 2 = 0 + 0.000 000 177 664;
  • 9) 0.000 000 177 664 × 2 = 0 + 0.000 000 355 328;
  • 10) 0.000 000 355 328 × 2 = 0 + 0.000 000 710 656;
  • 11) 0.000 000 710 656 × 2 = 0 + 0.000 001 421 312;
  • 12) 0.000 001 421 312 × 2 = 0 + 0.000 002 842 624;
  • 13) 0.000 002 842 624 × 2 = 0 + 0.000 005 685 248;
  • 14) 0.000 005 685 248 × 2 = 0 + 0.000 011 370 496;
  • 15) 0.000 011 370 496 × 2 = 0 + 0.000 022 740 992;
  • 16) 0.000 022 740 992 × 2 = 0 + 0.000 045 481 984;
  • 17) 0.000 045 481 984 × 2 = 0 + 0.000 090 963 968;
  • 18) 0.000 090 963 968 × 2 = 0 + 0.000 181 927 936;
  • 19) 0.000 181 927 936 × 2 = 0 + 0.000 363 855 872;
  • 20) 0.000 363 855 872 × 2 = 0 + 0.000 727 711 744;
  • 21) 0.000 727 711 744 × 2 = 0 + 0.001 455 423 488;
  • 22) 0.001 455 423 488 × 2 = 0 + 0.002 910 846 976;
  • 23) 0.002 910 846 976 × 2 = 0 + 0.005 821 693 952;
  • 24) 0.005 821 693 952 × 2 = 0 + 0.011 643 387 904;
  • 25) 0.011 643 387 904 × 2 = 0 + 0.023 286 775 808;
  • 26) 0.023 286 775 808 × 2 = 0 + 0.046 573 551 616;
  • 27) 0.046 573 551 616 × 2 = 0 + 0.093 147 103 232;
  • 28) 0.093 147 103 232 × 2 = 0 + 0.186 294 206 464;
  • 29) 0.186 294 206 464 × 2 = 0 + 0.372 588 412 928;
  • 30) 0.372 588 412 928 × 2 = 0 + 0.745 176 825 856;
  • 31) 0.745 176 825 856 × 2 = 1 + 0.490 353 651 712;
  • 32) 0.490 353 651 712 × 2 = 0 + 0.980 707 303 424;
  • 33) 0.980 707 303 424 × 2 = 1 + 0.961 414 606 848;
  • 34) 0.961 414 606 848 × 2 = 1 + 0.922 829 213 696;
  • 35) 0.922 829 213 696 × 2 = 1 + 0.845 658 427 392;
  • 36) 0.845 658 427 392 × 2 = 1 + 0.691 316 854 784;
  • 37) 0.691 316 854 784 × 2 = 1 + 0.382 633 709 568;
  • 38) 0.382 633 709 568 × 2 = 0 + 0.765 267 419 136;
  • 39) 0.765 267 419 136 × 2 = 1 + 0.530 534 838 272;
  • 40) 0.530 534 838 272 × 2 = 1 + 0.061 069 676 544;
  • 41) 0.061 069 676 544 × 2 = 0 + 0.122 139 353 088;
  • 42) 0.122 139 353 088 × 2 = 0 + 0.244 278 706 176;
  • 43) 0.244 278 706 176 × 2 = 0 + 0.488 557 412 352;
  • 44) 0.488 557 412 352 × 2 = 0 + 0.977 114 824 704;
  • 45) 0.977 114 824 704 × 2 = 1 + 0.954 229 649 408;
  • 46) 0.954 229 649 408 × 2 = 1 + 0.908 459 298 816;
  • 47) 0.908 459 298 816 × 2 = 1 + 0.816 918 597 632;
  • 48) 0.816 918 597 632 × 2 = 1 + 0.633 837 195 264;
  • 49) 0.633 837 195 264 × 2 = 1 + 0.267 674 390 528;
  • 50) 0.267 674 390 528 × 2 = 0 + 0.535 348 781 056;
  • 51) 0.535 348 781 056 × 2 = 1 + 0.070 697 562 112;
  • 52) 0.070 697 562 112 × 2 = 0 + 0.141 395 124 224;
  • 53) 0.141 395 124 224 × 2 = 0 + 0.282 790 248 448;
  • 54) 0.282 790 248 448 × 2 = 0 + 0.565 580 496 896;

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:


0.000 000 000 694(10) =

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

6. Positive number before normalization:

0.000 000 000 694(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1111 1011 0000 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:


0.000 000 000 694(10) =

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

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

1.0111 1101 1000 0111 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):
1.0111 1101 1000 0111 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. 011 1110 1100 0011 1110 1000 =

011 1110 1100 0011 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) =
011 1110 1100 0011 1110 1000


Decimal number -0.000 000 000 694 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 1110 1100 0011 1110 1000

Share

» Convert decimal -0.000 000 000 748 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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